Sort by Largest Digit(s)
up vote
23
down vote
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Challenge:
Given a list of integer, sort descending by their single largest digit(s). The order for numbers with the same largest digit are then sorted by second largest digit, etc.
We ignore duplicated digits in numbers. And if all digits in a number are the same, the order of those numbers in the list can be in any way you'd like.
Example:
Input: [123, 478, -904, 62778, 0, -73, 8491, 3120, 6458, -7738, 373]
Possible outputs: [8491, -904, 62778, 478, -7738, 6458, 373, -73, 3120, 123, 0]
[8491, -904, 62778, 478, -7738, 6458, -73, 373, 3120, 123, 0]
Why? Here are the relevant digits the numbers were sorted on:
Output:
[8491, -904, 62778, 478, -7738, 6458, 373, -73, 3120, 123, 0 ]
Relevant digits they were sorted on:
[[9,8], [9,4], [8,7,6], [8,7,4], [8,7,3], [8,6], [7,3], [7,3], [3,2,1,0], [3,2,1], [0]]
Challenge rules:
- We ignore duplicated digits, so
478
and-7738
will be ordered as478, -7738
, because the largest digits are[8,7,4]
and[8,7,3]
, and not[8,7,4]
and[8,7,7,3]
. - If multiple numbers have the same digits, the order of those can be either way. So
373
and-73
can be sorted as both373, -73
or-73, 373
(digits are[7,3]
for both of these numbers). - If a number contains no more digits to check, it will be placed at the back of the relevant numbers. So
123
and3120
will be sorted as3120, 123
, because the largest digits[3,2,1]
are the same, but0
comes beforenone
. - You can assume all numbers in the input are in the range
[-999999,999999]
. - Just one of the possible outputs is enough as result, but you are allowed to output all possible outputs where sublists can be in any permutation if you want (although I doubt it would save bytes in any language).
General rules:
- This is code-golf, so shortest answer in bytes wins.
Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.
Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.
Default Loopholes are forbidden.- If possible, please add a link with a test for your code (i.e. TIO).
- Also, adding an explanation for your answer is highly recommended.
Test cases:
Input: [123, 478, -904, 62778, 0, -73, 8491, 3120, 6458, -7738, 373]
Possible outputs: [8491, -904, 62778, 478, -7738, 6458, 373, -73, 3120, 123, 0]
[8491, -904, 62778, 478, -7738, 6458, -73, 373, 3120, 123, 0]
Input: [11, -312, 902, 23, 321, 2132, 34202, -34, -382]
Possible outputs: [902, -382, 34202, -34, -312, 321, 2132, 23, 11]
[902, -382, 34202, -34, 2132, -312, 321, 23, 11]
etc. The sublist [-312, 321, 2132] can be in any permutation
Input: [9, 44, 2212, 4, 6, 6, 1, 2, 192, 21, 29384, 0]
Possible outputs: [29384, 192, 9, 6, 6, 4, 44, 2212, 21, 2, 1, 0]
[29384, 192, 9, 6, 6, 44, 4, 2212, 21, 2, 1, 0]
etc. The sublists [4, 44] and [2212, 21] can be in any permutation
Input: [44, -88, 9, 233, -3, 14, 101, 77, 555, 67]
Output: [9, -88, 67, 77, 555, 14, 44, 233, -3, 101]
code-golf number integer sorting
add a comment |
up vote
23
down vote
favorite
Challenge:
Given a list of integer, sort descending by their single largest digit(s). The order for numbers with the same largest digit are then sorted by second largest digit, etc.
We ignore duplicated digits in numbers. And if all digits in a number are the same, the order of those numbers in the list can be in any way you'd like.
Example:
Input: [123, 478, -904, 62778, 0, -73, 8491, 3120, 6458, -7738, 373]
Possible outputs: [8491, -904, 62778, 478, -7738, 6458, 373, -73, 3120, 123, 0]
[8491, -904, 62778, 478, -7738, 6458, -73, 373, 3120, 123, 0]
Why? Here are the relevant digits the numbers were sorted on:
Output:
[8491, -904, 62778, 478, -7738, 6458, 373, -73, 3120, 123, 0 ]
Relevant digits they were sorted on:
[[9,8], [9,4], [8,7,6], [8,7,4], [8,7,3], [8,6], [7,3], [7,3], [3,2,1,0], [3,2,1], [0]]
Challenge rules:
- We ignore duplicated digits, so
478
and-7738
will be ordered as478, -7738
, because the largest digits are[8,7,4]
and[8,7,3]
, and not[8,7,4]
and[8,7,7,3]
. - If multiple numbers have the same digits, the order of those can be either way. So
373
and-73
can be sorted as both373, -73
or-73, 373
(digits are[7,3]
for both of these numbers). - If a number contains no more digits to check, it will be placed at the back of the relevant numbers. So
123
and3120
will be sorted as3120, 123
, because the largest digits[3,2,1]
are the same, but0
comes beforenone
. - You can assume all numbers in the input are in the range
[-999999,999999]
. - Just one of the possible outputs is enough as result, but you are allowed to output all possible outputs where sublists can be in any permutation if you want (although I doubt it would save bytes in any language).
General rules:
- This is code-golf, so shortest answer in bytes wins.
Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.
Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.
Default Loopholes are forbidden.- If possible, please add a link with a test for your code (i.e. TIO).
- Also, adding an explanation for your answer is highly recommended.
Test cases:
Input: [123, 478, -904, 62778, 0, -73, 8491, 3120, 6458, -7738, 373]
Possible outputs: [8491, -904, 62778, 478, -7738, 6458, 373, -73, 3120, 123, 0]
[8491, -904, 62778, 478, -7738, 6458, -73, 373, 3120, 123, 0]
Input: [11, -312, 902, 23, 321, 2132, 34202, -34, -382]
Possible outputs: [902, -382, 34202, -34, -312, 321, 2132, 23, 11]
[902, -382, 34202, -34, 2132, -312, 321, 23, 11]
etc. The sublist [-312, 321, 2132] can be in any permutation
Input: [9, 44, 2212, 4, 6, 6, 1, 2, 192, 21, 29384, 0]
Possible outputs: [29384, 192, 9, 6, 6, 4, 44, 2212, 21, 2, 1, 0]
[29384, 192, 9, 6, 6, 44, 4, 2212, 21, 2, 1, 0]
etc. The sublists [4, 44] and [2212, 21] can be in any permutation
Input: [44, -88, 9, 233, -3, 14, 101, 77, 555, 67]
Output: [9, -88, 67, 77, 555, 14, 44, 233, -3, 101]
code-golf number integer sorting
add a comment |
up vote
23
down vote
favorite
up vote
23
down vote
favorite
Challenge:
Given a list of integer, sort descending by their single largest digit(s). The order for numbers with the same largest digit are then sorted by second largest digit, etc.
We ignore duplicated digits in numbers. And if all digits in a number are the same, the order of those numbers in the list can be in any way you'd like.
Example:
Input: [123, 478, -904, 62778, 0, -73, 8491, 3120, 6458, -7738, 373]
Possible outputs: [8491, -904, 62778, 478, -7738, 6458, 373, -73, 3120, 123, 0]
[8491, -904, 62778, 478, -7738, 6458, -73, 373, 3120, 123, 0]
Why? Here are the relevant digits the numbers were sorted on:
Output:
[8491, -904, 62778, 478, -7738, 6458, 373, -73, 3120, 123, 0 ]
Relevant digits they were sorted on:
[[9,8], [9,4], [8,7,6], [8,7,4], [8,7,3], [8,6], [7,3], [7,3], [3,2,1,0], [3,2,1], [0]]
Challenge rules:
- We ignore duplicated digits, so
478
and-7738
will be ordered as478, -7738
, because the largest digits are[8,7,4]
and[8,7,3]
, and not[8,7,4]
and[8,7,7,3]
. - If multiple numbers have the same digits, the order of those can be either way. So
373
and-73
can be sorted as both373, -73
or-73, 373
(digits are[7,3]
for both of these numbers). - If a number contains no more digits to check, it will be placed at the back of the relevant numbers. So
123
and3120
will be sorted as3120, 123
, because the largest digits[3,2,1]
are the same, but0
comes beforenone
. - You can assume all numbers in the input are in the range
[-999999,999999]
. - Just one of the possible outputs is enough as result, but you are allowed to output all possible outputs where sublists can be in any permutation if you want (although I doubt it would save bytes in any language).
General rules:
- This is code-golf, so shortest answer in bytes wins.
Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.
Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.
Default Loopholes are forbidden.- If possible, please add a link with a test for your code (i.e. TIO).
- Also, adding an explanation for your answer is highly recommended.
Test cases:
Input: [123, 478, -904, 62778, 0, -73, 8491, 3120, 6458, -7738, 373]
Possible outputs: [8491, -904, 62778, 478, -7738, 6458, 373, -73, 3120, 123, 0]
[8491, -904, 62778, 478, -7738, 6458, -73, 373, 3120, 123, 0]
Input: [11, -312, 902, 23, 321, 2132, 34202, -34, -382]
Possible outputs: [902, -382, 34202, -34, -312, 321, 2132, 23, 11]
[902, -382, 34202, -34, 2132, -312, 321, 23, 11]
etc. The sublist [-312, 321, 2132] can be in any permutation
Input: [9, 44, 2212, 4, 6, 6, 1, 2, 192, 21, 29384, 0]
Possible outputs: [29384, 192, 9, 6, 6, 4, 44, 2212, 21, 2, 1, 0]
[29384, 192, 9, 6, 6, 44, 4, 2212, 21, 2, 1, 0]
etc. The sublists [4, 44] and [2212, 21] can be in any permutation
Input: [44, -88, 9, 233, -3, 14, 101, 77, 555, 67]
Output: [9, -88, 67, 77, 555, 14, 44, 233, -3, 101]
code-golf number integer sorting
Challenge:
Given a list of integer, sort descending by their single largest digit(s). The order for numbers with the same largest digit are then sorted by second largest digit, etc.
We ignore duplicated digits in numbers. And if all digits in a number are the same, the order of those numbers in the list can be in any way you'd like.
Example:
Input: [123, 478, -904, 62778, 0, -73, 8491, 3120, 6458, -7738, 373]
Possible outputs: [8491, -904, 62778, 478, -7738, 6458, 373, -73, 3120, 123, 0]
[8491, -904, 62778, 478, -7738, 6458, -73, 373, 3120, 123, 0]
Why? Here are the relevant digits the numbers were sorted on:
Output:
[8491, -904, 62778, 478, -7738, 6458, 373, -73, 3120, 123, 0 ]
Relevant digits they were sorted on:
[[9,8], [9,4], [8,7,6], [8,7,4], [8,7,3], [8,6], [7,3], [7,3], [3,2,1,0], [3,2,1], [0]]
Challenge rules:
- We ignore duplicated digits, so
478
and-7738
will be ordered as478, -7738
, because the largest digits are[8,7,4]
and[8,7,3]
, and not[8,7,4]
and[8,7,7,3]
. - If multiple numbers have the same digits, the order of those can be either way. So
373
and-73
can be sorted as both373, -73
or-73, 373
(digits are[7,3]
for both of these numbers). - If a number contains no more digits to check, it will be placed at the back of the relevant numbers. So
123
and3120
will be sorted as3120, 123
, because the largest digits[3,2,1]
are the same, but0
comes beforenone
. - You can assume all numbers in the input are in the range
[-999999,999999]
. - Just one of the possible outputs is enough as result, but you are allowed to output all possible outputs where sublists can be in any permutation if you want (although I doubt it would save bytes in any language).
General rules:
- This is code-golf, so shortest answer in bytes wins.
Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.
Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.
Default Loopholes are forbidden.- If possible, please add a link with a test for your code (i.e. TIO).
- Also, adding an explanation for your answer is highly recommended.
Test cases:
Input: [123, 478, -904, 62778, 0, -73, 8491, 3120, 6458, -7738, 373]
Possible outputs: [8491, -904, 62778, 478, -7738, 6458, 373, -73, 3120, 123, 0]
[8491, -904, 62778, 478, -7738, 6458, -73, 373, 3120, 123, 0]
Input: [11, -312, 902, 23, 321, 2132, 34202, -34, -382]
Possible outputs: [902, -382, 34202, -34, -312, 321, 2132, 23, 11]
[902, -382, 34202, -34, 2132, -312, 321, 23, 11]
etc. The sublist [-312, 321, 2132] can be in any permutation
Input: [9, 44, 2212, 4, 6, 6, 1, 2, 192, 21, 29384, 0]
Possible outputs: [29384, 192, 9, 6, 6, 4, 44, 2212, 21, 2, 1, 0]
[29384, 192, 9, 6, 6, 44, 4, 2212, 21, 2, 1, 0]
etc. The sublists [4, 44] and [2212, 21] can be in any permutation
Input: [44, -88, 9, 233, -3, 14, 101, 77, 555, 67]
Output: [9, -88, 67, 77, 555, 14, 44, 233, -3, 101]
code-golf number integer sorting
code-golf number integer sorting
asked Nov 15 at 9:41
Kevin Cruijssen
34k554181
34k554181
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25 Answers
25
active
oldest
votes
up vote
6
down vote
05AB1E, 5 bytes
ΣêR}R
Try it online!
or as a Test suite
Explanation
Σ } # sort input by
ê # its sorted unique characters
R # reversed (to sort descending)
R # reverse the result (to sort descending)
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up vote
6
down vote
R, 97 95 bytes
function(x)x[rev(order(sapply(Map(sort,Map(unique,strsplit(paste(x),"")),T),Reduce,f=paste0)))]
Try it online!
This challenge seems to have been pessimized for R. Explanation of the original version (start at 1. and work up):
f <- function(x) {
x[ # 8. Input vector in
rev( # 7. Reversed
order( # 6. Lexicographical order
sapply( # 5. Paste....
Map(sort, # 4. Sort each using...
Map(unique, # 3. Deduplicate each
strsplit( # 2. Split each string into characters
paste(x), # 1. Coerce each number to string
"")),
T), # 4. ...descending sort.
paste,collapse="") # 5. ...back into strings
)
)
]
}
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up vote
6
down vote
Perl 6, 36 34 33 31 bytes
-1 byte thanks to Jo King
-2 bytes thanks to Phil H
*.sort:{sort 1,|set -<<m:g/d/}
Try it online!
Explanation
{ } # Map each number, e.g. -373
m:g/d/ # Extract digits: (3, 7, 3)
-<< # Negate each digit: (-3, -7, -3)
set # Convert to set to remove duplicates
| # Pass as list of pairs: (-3 => True, -7 => True)
1, # Prepend 1 for "none": (1, -3 => True, -7 => True)
sort # Sort (compares 1 and pair by string value): (-7 => True, -3 => True, 1)
*.sort: # Sort lexicographically
1
Nice! -2 bytes for swappingm:g/d./
for.abs.comb
: tio.run/…
– Phil H
2 days ago
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up vote
6
down vote
Python 2, 60 55 54 bytes
-1 byte thanks to Jonas Ausevicius.
def f(l):l.sort(cmp,lambda n:sorted(set(`n`))[::-1],1)
Try it online!
Ungolfed
def f(l):
l.sort( # Sort the list in place
cmp = cmp, # ... compare with the builtin function cmp
key = k, # ... on the function k
reverse = 1 # ... in reverse
) # As the arguments are used in the right order, no names are necessary.
k = lambda n:sorted( # sort
set(`n`) # ... the set of digits
)[::-1] # reverse the result
# As '-' is smaller than the digits,
# it will be sorted to the back and ignored for sorting
Try it online!
5
None
can be replaced withcmp
insort
function
– Jonas Ausevicius
Nov 15 at 10:14
The [::-1] can be exchanged for a double negation, I think.
– DonQuiKong
2 days ago
@DonQuiKong that would be quite a bit longer though, as the digits are all strings, and would need to be converted to ints for this.
– ovs
yesterday
@JonasAusevicius Thanks a lot.
– ovs
yesterday
add a comment |
up vote
5
down vote
Python 2, 58 60 bytes
lambda a:sorted(a,key=lambda x:sorted(set(`x`))[::-1])[::-1]
Try it online!
add a comment |
up vote
5
down vote
Brachylog, 9 bytes
{ȧdṫo₁}ᵒ¹
Note: due to how ordering works in brachylog, it does not work on number correctly. This is fixed by casting the number to a string (ṫ
) at the cost of 1 byte.
Try it online!
2
What do you mean by "Due to how ordering works in brachylog, it does not work as intended."? I've tried all four test cases, and its giving the correct results (unless I accidentally looked past something).
– Kevin Cruijssen
Nov 15 at 10:12
@KevinCruijssen Theṫ
(to string) fixes the issue. Ordering digits in a number descending works as follows. Order from smallest to largest then reverse. The problem is that the number3120
ordered from smallest to largest is0123
which is equal to123
which reversed is321
and not3210
– Kroppeb
Nov 15 at 11:00
2
Ah ok, so your current code is working due to the added toString (ṫ
). As mentioned by @Arnauld, I thought your comment meant your current code doesn't work. It might be better to mention it like: "This could have been 8 bytes by removing theṫ
(toString), but unfortunately it does not work as intended due to how ordering works in Brachylog."
– Kevin Cruijssen
Nov 15 at 11:12
Looking at what I wrote it seems that my brain got distracted midsentence. Fixed it.
– Kroppeb
Nov 15 at 11:31
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up vote
5
down vote
Pyth, 7 6 bytes
-1 byte by @Sok
_o_{S`
Pyth, which uses only printable ASCII, is at a bit of a disadvantage here. Optimally encoded this would be 6*log(95)/log(256) = 4.927
bytes, beating 05AB1E.
Explained:
o Sort the implicit input by lambda N:
_ reversed
{ uniquified
S sorted
' string representation [of N]
_ then reverse the result.
Try it here.
2
The trailingN
can be left out to save 1 byte - all lambda-type functions infer the presence of the principle lambda variable if any arguments are missing from the end. Examples includem
inferringd
,f
inferringT
,u
inferringG
...
– Sok
Nov 15 at 12:08
add a comment |
up vote
4
down vote
Jelly, 8 bytes
ADṢUQµÞU
Try it online!
How it works
ADṢUQµÞU Main link (monad). Input: integer list
µÞU Sort by (reversed):
AD Absolute value converted to decimal digits
ṢUQ Sort, reverse, take unique values
2
I just implemented this then found your post. I went with normal reverses,Ṛ
, rather than the upends,U
. Note, however, that you do not need theD
since sort,Ṣ
, is implemented with aniterable(z, make_digits=True)
call inside. So that wasAṢQṚµÞṚ
for 7.
– Jonathan Allan
Nov 15 at 18:16
add a comment |
up vote
3
down vote
MathGolf, 7 6 bytes
áÉ░▀zx
Try it online! or as a test suite.
Explanation
After looking at Emigna's 05AB1E solution, I found that I didn't need the absolute operator (and my previous answer was actually incorrect because of that operator). Now the main difference is that I convert to string and get unique characters instead of using the 1-byte operator in 05AB1E.
áÉ Sort by the value generated from mapping each element using the next 3 instructions
░ Convert to string
▀ Get unique characters
z Sort reversed (last instruction of block)
x Reverse list (needed because I don't have a sort-reversed by mapping)
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up vote
3
down vote
C (gcc), 114 111 bytes
a;v(n){n=n<0?-n:n;for(a=0;n;n/=10)a|=1<<n%10;n=a;}c(int*a,int*b){a=v(*a)<v(*b);}f(a,n)int*a,n;{qsort(a,n,4,c);}
Try it online!
Explanation:
f() uses qsort() to sort provided array in place. Using comparison function c() to compare numbers which evaluates numbers using v().
v() calculates a higher number if bigger digits are present in parameter.
[Edit 1]
Improved by 3 bytes. 2 byte credits to Kevin. Thanks
You can golfn>0
ton
I think in your loop of your methodv
.
– Kevin Cruijssen
Nov 15 at 11:55
Suggestfor(a=0;n=abs(n);
instead ofn=n<?-n:n;for(a=0;n;
– ceilingcat
2 days ago
f()'s argument listint*a,n
can be shortened toint*a
.
– gastropner
26 mins ago
add a comment |
up vote
3
down vote
Stax, 6 7 bytes
èó≥ü≤♥¥
Run and debug it
This seems to give incorrect results. For example, in the test case of your TIO it outputs-904 8491 478 62778 6458 -7738 -73 373 123 3120 0
instead of the intended8491 -904 62778 478 -7738 6458 373 -73 3120 123 0
or8491 -904 62778 478 -7738 6458 -73 373 3120 123 0
. This test case is also used in the example, and to explain the rules, so I would take a look at that to understand it better. It seems you are only sorting by the single largest digit once, without any of the other rules?
– Kevin Cruijssen
2 days ago
@KevinCruijssen: Yes, my apologies. I mis-read the problem statement. I've adjusted the program to handle the stated requirements. This program is accepting the input integers as quoted strings. That's usually acceptable, but if not I might need to add another byte.
– recursive
2 days ago
Looks good now, +1 from me. And yes, inputting as strings is completely fine.
– Kevin Cruijssen
2 days ago
add a comment |
up vote
2
down vote
J, 17 bytes
{~[::~.@:~@":@|
Try it online!
Explanation:
@| - find the absolute value and
@": - convert to string and
@:~ - sort down and
~. - keep only the unique symbols
: - grade down the entire list of strings
[: - function composition
{~ - use the graded-down list to index into the input
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up vote
2
down vote
JavaScript (SpiderMonkey), 68 bytes
Thanks for @Arnauld for reminding me again that SpiderMonkey uses stable sort, so -4 bytes for removing ||-1
.
A=>A.sort((x,y,F=n=>[...new Set(""+n)].sort().reverse())=>F(x)<F(y))
Try it online!
JavaScript (Node.js), 72 bytes
A=>A.sort((x,y,F=n=>[...new Set(""+n)].sort().reverse())=>F(x)<F(y)||-1)
Try it online!
Or 68 bytes with SpiderMonkey.
– Arnauld
Nov 15 at 11:04
1
@Arnauld oh stable sort again ;P
– Shieru Asakoto
Nov 15 at 12:17
Actually, I think V8 is using at least 2 different sort algorithms. It seems like it's stable if the size of the array is less than or equal to $10$.
– Arnauld
Nov 15 at 12:37
1
@Arnauld V8 use quick sort before Chrome 70. The quick sort algorithm perform an insertion sort when array size is small enough. And latest Chrome had changed to stable sort for matching other browsers' (IE/Firefox/Safari) behavior.
– tsh
2 days ago
add a comment |
up vote
2
down vote
Java (JDK), 98 bytes
l->l.sort((a,b)->{int r=0,i=58;for(;r==0&i-->48;)r=(b.indexOf(i)>>9)-(a.indexOf(i)>>9);return r;})
Try it online!
Explanation
l-> // Consumer<List<String>>
l.sort( // Use the incorporated sort method which uses a...
(a,b)->{ // Comparator of Strings
int r=0, // define r as the result, initiated to 0
i=58; // i as the codepoint to test for.
for(;r==0&i-->48;) // for each digit codepoint from '9' to '0',
// and while no difference was found.
r= // set r as the difference between
(b.indexOf(i)>>9)- // was the digit found in b? then 0 else -1 using the bit-shift operator
(a.indexOf(i)>>9); // and was the digit found in a? then 0 else -1.
return r; // return the comparison result.
}
)
Note:
I needed a way to map numbers to either 0/1
or 0/-1
.
indexOf
has the nice property that it's consistently returning -1
for characters not found. -1
right-shifted by any number is always -1
. Any positive number right-shifted by a big enough number will always produce 0
.
So here we are:
input input.indexOf('9') input.indexOf('9')>>9
"999" 0 0
"111119" 5 0
"123456" -1 -1
1
Ah, yeah, that's what I mean. ;p Nice golf of using>>9
instead of>>32
due to the limited range of the numbers.
– Kevin Cruijssen
Nov 15 at 12:53
add a comment |
up vote
2
down vote
Japt, 12 bytes
ñ_a ì â ñnÃw
All test cases
Explanation:
ñ_ Ãw :Sort Descending by:
a : Get the absolute value
ì : Get the digits
â : Remove duplicates
ñn : Sort the digits in descending order
add a comment |
up vote
2
down vote
Haskell, 54 52 bytes
import Data.List
f=r.sortOn(r.sort.nub.show);r=reverse
Try it online!
Definingr=reverse
saves two bytes. We also allow anonymous functions, so thef=
does not need to ve counted.
– Laikoni
Nov 15 at 23:33
I moved the import and f= to the TIO header. Is that OK?
– Martin Lütke
Nov 15 at 23:34
Same byte count, but maybe of some interest:f=r$r id.nub.show;r=(reverse.).sortOn
.
– Laikoni
Nov 15 at 23:43
1
The import actually needs to be counted.
– Laikoni
Nov 15 at 23:45
2
You may want to have a look at our Guide to golfing rules in Haskell.
– Laikoni
Nov 15 at 23:52
|
show 1 more comment
up vote
2
down vote
APL (Dyalog Extended), 19 bytes
{⍵[⍒∪¨(∨'¯'~⍨⍕)¨⍵]}
Try it online!
Fixed at a cost of +2 bytes thanks to OP.
I think you are missing an 'uniquify' somewhere? If I try the example test case in your TIO for example,¯7738
is placed before478
, but it should be after it: digits[8,7,4]
come before digits[8,7,3]
.
– Kevin Cruijssen
2 days ago
Thanks, @KevinCruijssen
– Zacharý
2 days ago
add a comment |
up vote
1
down vote
Ruby, 55 bytes
->a{a.sort_by{|x|x.abs.digits.sort.reverse|}.reverse}
Try it online!
add a comment |
up vote
1
down vote
Perl 5 -nl
, 68 bytes
$a{eval"9876543210=~y/$_//dcr"}=$_}{say$a{$_}for reverse sort keys%a
Try it online!
add a comment |
up vote
1
down vote
Julia 1.0, 50 bytes
x->sort(x,by=y->sort(unique([digits(-abs(y));1])))
Try it online!
add a comment |
up vote
1
down vote
APL(NARS), 366 chars, 732 bytes
_gb←⍬
∇a _s w;t
t←_gb[a]⋄_gb[a]←_gb[w]⋄_gb[w]←t
∇
∇(_f _q)w;l;r;ls;i
(l r)←w⋄→0×⍳l≥r⋄l _s⌊2÷⍨l+r⋄ls←i←l⋄→3
→3×⍳∼0<_gb[i]_f _gb[l]⋄ls+←1⋄ls _s i
→2×⍳r≥i+←1
l _s ls⋄_f _q l(ls-1)⋄_f _q(ls+1)r
∇
∇r←(a qsort)w
r←¯1⋄→0×⍳1≠⍴⍴w⋄_gb←w⋄a _q 1(↑⍴w)⋄r←_gb
∇
f←{∪t[⍒t←⍎¨⍕∣⍵]}
∇r←a c b;x;y;i;m
x←f a⋄y←f b⋄r←i←0⋄m←(↑⍴x)⌊(↑⍴y)⋄→3
→0×⍳x[i]<y[i]⋄→3×⍳∼x[i]>y[i]⋄r←1⋄→0
→2×⍳m≥i+←1⋄r←(↑⍴x)>(↑⍴y)
∇
For the qsort operator, it is one traslation in APL of algo page 139 K&R Linguaggio C.
I think in it there is using data as C with pointers...
Test
c qsort 123, 478, ¯904, 62778, 0, ¯73, 8491, 3120, 6458, ¯7738, 373
8491 ¯904 62778 478 ¯7738 6458 ¯73 373 3120 123 0
c qsort 11, ¯312, 902, 23, 321, 2132, 34202, ¯34, ¯382
902 ¯382 34202 ¯34 321 ¯312 2132 23 11
c qsort 9, 44, 2212, 4, 6, 6, 1, 2, 192, 21, 29384, 0
29384 192 9 6 6 4 44 2212 21 2 1 0
c qsort 44, ¯88, 9, 233, ¯3, 14, 101, 77, 555, 67
9 ¯88 67 77 555 14 44 233 ¯3 101
add a comment |
up vote
1
down vote
Powershell, 44 bytes
$args|sort{$_-split'(.)'-ne'-'|sort -u -d}-d
Test script:
$f = {
$args|sort{$_-split'(.)'-ne'-'|sort -u -d}-d
}
@(
,( (123, 478, -904, 62778, 0, -73, 8491, 3120, 6458, -7738, 373),
(8491, -904, 62778, 478, -7738, 6458, 373, -73, 3120, 123, 0),
(8491, -904, 62778, 478, -7738, 6458, -73, 373, 3120, 123, 0) )
,( (11, -312, 902, 23, 321, 2132, 34202, -34, -382),
(902, -382, 34202, -34, -312, 321, 2132, 23, 11),
(902, -382, 34202, -34, 2132, -312, 321, 23, 11) )
,( (9, 44, 2212, 4, 6, 6, 1, 2, 192, 21, 29384, 0),
(29384, 192, 9, 6, 6, 4, 44, 2212, 21, 2, 1, 0),
(29384, 192, 9, 6, 6, 44, 4, 2212, 21, 2, 1, 0),
(29384, 192, 9, 6, 6, 44, 4, 21, 2212, 2, 1, 0) )
,( (44, -88, 9, 233, -3, 14, 101, 77, 555, 67),
,(9, -88, 67, 77, 555, 14, 44, 233, -3, 101) )
) | % {
$a, $expected = $_
$result = &$f @a
$true-in($expected|%{"$result"-eq"$_"})
"$result"
}
Output:
True
8491 -904 62778 478 -7738 6458 -73 373 3120 123 0
True
902 -382 34202 -34 2132 -312 321 23 11
True
29384 192 9 6 6 44 4 21 2212 2 1 0
True
9 -88 67 77 555 14 44 233 -3 101
add a comment |
up vote
1
down vote
PHP, 87 86 84 bytes
while(--$argc)$a[_.strrev(count_chars($n=$argv[++$i],3))]=$n;krsort($a);print_r($a);
Run with -nr
or try it online.
Replace ++$i
with $argc
(+1 byte) to suppress the Notice (and render -n
obosolete).
breakdown
while(--$argc) # loop through command line arguments
$a[ # key=
_. # 3. prepend non-numeric char for non-numeric sort
strrev( # 2. reverse =^= sort descending
count_chars($n=$argv[++$i],3) # 1. get characters used in argument
)
]=$n; # value=argument
krsort($a); # sort by key descending
print_r($a); # print
-
is "smaller" than the digits, so it has no affect on the sorting.
add a comment |
up vote
1
down vote
Common Lisp, 88 bytes
(sort(read)'string> :key(lambda(x)(sort(remove-duplicates(format()"~d"(abs x)))'char>)))
Try it online!
Good old verbose Common Lisp!
Explanation:
(sort ; sort
(read) ; what to sort: a list of numbers, read on input stream
'string> ; comparison predicate (remember: this is a typed language!)
:key (lambda (x) ; how to get an element to sort; get a number
(sort (remove-duplicates ; then sort the unique digits (characters)
(format() "~d" (abs x))) ; from its string representation
'char>))) ; with the appropriate comparison operator for characters
add a comment |
up vote
1
down vote
C# (Visual C# Interactive Compiler), 75 74 bytes
-1 thanks @ASCII-only
x=>x.OrderByDescending(y=>String.Concat((y+"").Distinct().OrderBy(z=>-z)))
Try it online!
In C#, strings are considered "enumerables" of characters. I use this to my advantage by first converting each number to a string. LINQ is then leveraged to get the unique characters (digits) sorted in reverse order. I convert each sorted character array back into a string and use that as the sort key to order the whole list.
Looks like you'll be able to get away with not adding-
, looks like order of those doesn't really matter?
– ASCII-only
21 hours ago
Without the-
, test case #2 returns... 321 2132 ...
which seems incorrect?
– dana
15 hours ago
nah, read the example more carefully
– ASCII-only
9 hours ago
OK - I think your right. Thanks for the tip!
– dana
9 hours ago
add a comment |
25 Answers
25
active
oldest
votes
25 Answers
25
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
05AB1E, 5 bytes
ΣêR}R
Try it online!
or as a Test suite
Explanation
Σ } # sort input by
ê # its sorted unique characters
R # reversed (to sort descending)
R # reverse the result (to sort descending)
add a comment |
up vote
6
down vote
05AB1E, 5 bytes
ΣêR}R
Try it online!
or as a Test suite
Explanation
Σ } # sort input by
ê # its sorted unique characters
R # reversed (to sort descending)
R # reverse the result (to sort descending)
add a comment |
up vote
6
down vote
up vote
6
down vote
05AB1E, 5 bytes
ΣêR}R
Try it online!
or as a Test suite
Explanation
Σ } # sort input by
ê # its sorted unique characters
R # reversed (to sort descending)
R # reverse the result (to sort descending)
05AB1E, 5 bytes
ΣêR}R
Try it online!
or as a Test suite
Explanation
Σ } # sort input by
ê # its sorted unique characters
R # reversed (to sort descending)
R # reverse the result (to sort descending)
edited Nov 15 at 10:04
answered Nov 15 at 9:47
Emigna
44.8k432136
44.8k432136
add a comment |
add a comment |
up vote
6
down vote
R, 97 95 bytes
function(x)x[rev(order(sapply(Map(sort,Map(unique,strsplit(paste(x),"")),T),Reduce,f=paste0)))]
Try it online!
This challenge seems to have been pessimized for R. Explanation of the original version (start at 1. and work up):
f <- function(x) {
x[ # 8. Input vector in
rev( # 7. Reversed
order( # 6. Lexicographical order
sapply( # 5. Paste....
Map(sort, # 4. Sort each using...
Map(unique, # 3. Deduplicate each
strsplit( # 2. Split each string into characters
paste(x), # 1. Coerce each number to string
"")),
T), # 4. ...descending sort.
paste,collapse="") # 5. ...back into strings
)
)
]
}
add a comment |
up vote
6
down vote
R, 97 95 bytes
function(x)x[rev(order(sapply(Map(sort,Map(unique,strsplit(paste(x),"")),T),Reduce,f=paste0)))]
Try it online!
This challenge seems to have been pessimized for R. Explanation of the original version (start at 1. and work up):
f <- function(x) {
x[ # 8. Input vector in
rev( # 7. Reversed
order( # 6. Lexicographical order
sapply( # 5. Paste....
Map(sort, # 4. Sort each using...
Map(unique, # 3. Deduplicate each
strsplit( # 2. Split each string into characters
paste(x), # 1. Coerce each number to string
"")),
T), # 4. ...descending sort.
paste,collapse="") # 5. ...back into strings
)
)
]
}
add a comment |
up vote
6
down vote
up vote
6
down vote
R, 97 95 bytes
function(x)x[rev(order(sapply(Map(sort,Map(unique,strsplit(paste(x),"")),T),Reduce,f=paste0)))]
Try it online!
This challenge seems to have been pessimized for R. Explanation of the original version (start at 1. and work up):
f <- function(x) {
x[ # 8. Input vector in
rev( # 7. Reversed
order( # 6. Lexicographical order
sapply( # 5. Paste....
Map(sort, # 4. Sort each using...
Map(unique, # 3. Deduplicate each
strsplit( # 2. Split each string into characters
paste(x), # 1. Coerce each number to string
"")),
T), # 4. ...descending sort.
paste,collapse="") # 5. ...back into strings
)
)
]
}
R, 97 95 bytes
function(x)x[rev(order(sapply(Map(sort,Map(unique,strsplit(paste(x),"")),T),Reduce,f=paste0)))]
Try it online!
This challenge seems to have been pessimized for R. Explanation of the original version (start at 1. and work up):
f <- function(x) {
x[ # 8. Input vector in
rev( # 7. Reversed
order( # 6. Lexicographical order
sapply( # 5. Paste....
Map(sort, # 4. Sort each using...
Map(unique, # 3. Deduplicate each
strsplit( # 2. Split each string into characters
paste(x), # 1. Coerce each number to string
"")),
T), # 4. ...descending sort.
paste,collapse="") # 5. ...back into strings
)
)
]
}
edited Nov 15 at 17:38
answered Nov 15 at 14:56
ngm
2,99923
2,99923
add a comment |
add a comment |
up vote
6
down vote
Perl 6, 36 34 33 31 bytes
-1 byte thanks to Jo King
-2 bytes thanks to Phil H
*.sort:{sort 1,|set -<<m:g/d/}
Try it online!
Explanation
{ } # Map each number, e.g. -373
m:g/d/ # Extract digits: (3, 7, 3)
-<< # Negate each digit: (-3, -7, -3)
set # Convert to set to remove duplicates
| # Pass as list of pairs: (-3 => True, -7 => True)
1, # Prepend 1 for "none": (1, -3 => True, -7 => True)
sort # Sort (compares 1 and pair by string value): (-7 => True, -3 => True, 1)
*.sort: # Sort lexicographically
1
Nice! -2 bytes for swappingm:g/d./
for.abs.comb
: tio.run/…
– Phil H
2 days ago
add a comment |
up vote
6
down vote
Perl 6, 36 34 33 31 bytes
-1 byte thanks to Jo King
-2 bytes thanks to Phil H
*.sort:{sort 1,|set -<<m:g/d/}
Try it online!
Explanation
{ } # Map each number, e.g. -373
m:g/d/ # Extract digits: (3, 7, 3)
-<< # Negate each digit: (-3, -7, -3)
set # Convert to set to remove duplicates
| # Pass as list of pairs: (-3 => True, -7 => True)
1, # Prepend 1 for "none": (1, -3 => True, -7 => True)
sort # Sort (compares 1 and pair by string value): (-7 => True, -3 => True, 1)
*.sort: # Sort lexicographically
1
Nice! -2 bytes for swappingm:g/d./
for.abs.comb
: tio.run/…
– Phil H
2 days ago
add a comment |
up vote
6
down vote
up vote
6
down vote
Perl 6, 36 34 33 31 bytes
-1 byte thanks to Jo King
-2 bytes thanks to Phil H
*.sort:{sort 1,|set -<<m:g/d/}
Try it online!
Explanation
{ } # Map each number, e.g. -373
m:g/d/ # Extract digits: (3, 7, 3)
-<< # Negate each digit: (-3, -7, -3)
set # Convert to set to remove duplicates
| # Pass as list of pairs: (-3 => True, -7 => True)
1, # Prepend 1 for "none": (1, -3 => True, -7 => True)
sort # Sort (compares 1 and pair by string value): (-7 => True, -3 => True, 1)
*.sort: # Sort lexicographically
Perl 6, 36 34 33 31 bytes
-1 byte thanks to Jo King
-2 bytes thanks to Phil H
*.sort:{sort 1,|set -<<m:g/d/}
Try it online!
Explanation
{ } # Map each number, e.g. -373
m:g/d/ # Extract digits: (3, 7, 3)
-<< # Negate each digit: (-3, -7, -3)
set # Convert to set to remove duplicates
| # Pass as list of pairs: (-3 => True, -7 => True)
1, # Prepend 1 for "none": (1, -3 => True, -7 => True)
sort # Sort (compares 1 and pair by string value): (-7 => True, -3 => True, 1)
*.sort: # Sort lexicographically
edited 2 days ago
answered Nov 15 at 10:52
nwellnhof
5,9281121
5,9281121
1
Nice! -2 bytes for swappingm:g/d./
for.abs.comb
: tio.run/…
– Phil H
2 days ago
add a comment |
1
Nice! -2 bytes for swappingm:g/d./
for.abs.comb
: tio.run/…
– Phil H
2 days ago
1
1
Nice! -2 bytes for swapping
m:g/d./
for .abs.comb
: tio.run/…– Phil H
2 days ago
Nice! -2 bytes for swapping
m:g/d./
for .abs.comb
: tio.run/…– Phil H
2 days ago
add a comment |
up vote
6
down vote
Python 2, 60 55 54 bytes
-1 byte thanks to Jonas Ausevicius.
def f(l):l.sort(cmp,lambda n:sorted(set(`n`))[::-1],1)
Try it online!
Ungolfed
def f(l):
l.sort( # Sort the list in place
cmp = cmp, # ... compare with the builtin function cmp
key = k, # ... on the function k
reverse = 1 # ... in reverse
) # As the arguments are used in the right order, no names are necessary.
k = lambda n:sorted( # sort
set(`n`) # ... the set of digits
)[::-1] # reverse the result
# As '-' is smaller than the digits,
# it will be sorted to the back and ignored for sorting
Try it online!
5
None
can be replaced withcmp
insort
function
– Jonas Ausevicius
Nov 15 at 10:14
The [::-1] can be exchanged for a double negation, I think.
– DonQuiKong
2 days ago
@DonQuiKong that would be quite a bit longer though, as the digits are all strings, and would need to be converted to ints for this.
– ovs
yesterday
@JonasAusevicius Thanks a lot.
– ovs
yesterday
add a comment |
up vote
6
down vote
Python 2, 60 55 54 bytes
-1 byte thanks to Jonas Ausevicius.
def f(l):l.sort(cmp,lambda n:sorted(set(`n`))[::-1],1)
Try it online!
Ungolfed
def f(l):
l.sort( # Sort the list in place
cmp = cmp, # ... compare with the builtin function cmp
key = k, # ... on the function k
reverse = 1 # ... in reverse
) # As the arguments are used in the right order, no names are necessary.
k = lambda n:sorted( # sort
set(`n`) # ... the set of digits
)[::-1] # reverse the result
# As '-' is smaller than the digits,
# it will be sorted to the back and ignored for sorting
Try it online!
5
None
can be replaced withcmp
insort
function
– Jonas Ausevicius
Nov 15 at 10:14
The [::-1] can be exchanged for a double negation, I think.
– DonQuiKong
2 days ago
@DonQuiKong that would be quite a bit longer though, as the digits are all strings, and would need to be converted to ints for this.
– ovs
yesterday
@JonasAusevicius Thanks a lot.
– ovs
yesterday
add a comment |
up vote
6
down vote
up vote
6
down vote
Python 2, 60 55 54 bytes
-1 byte thanks to Jonas Ausevicius.
def f(l):l.sort(cmp,lambda n:sorted(set(`n`))[::-1],1)
Try it online!
Ungolfed
def f(l):
l.sort( # Sort the list in place
cmp = cmp, # ... compare with the builtin function cmp
key = k, # ... on the function k
reverse = 1 # ... in reverse
) # As the arguments are used in the right order, no names are necessary.
k = lambda n:sorted( # sort
set(`n`) # ... the set of digits
)[::-1] # reverse the result
# As '-' is smaller than the digits,
# it will be sorted to the back and ignored for sorting
Try it online!
Python 2, 60 55 54 bytes
-1 byte thanks to Jonas Ausevicius.
def f(l):l.sort(cmp,lambda n:sorted(set(`n`))[::-1],1)
Try it online!
Ungolfed
def f(l):
l.sort( # Sort the list in place
cmp = cmp, # ... compare with the builtin function cmp
key = k, # ... on the function k
reverse = 1 # ... in reverse
) # As the arguments are used in the right order, no names are necessary.
k = lambda n:sorted( # sort
set(`n`) # ... the set of digits
)[::-1] # reverse the result
# As '-' is smaller than the digits,
# it will be sorted to the back and ignored for sorting
Try it online!
edited yesterday
answered Nov 15 at 9:54
ovs
18.3k21059
18.3k21059
5
None
can be replaced withcmp
insort
function
– Jonas Ausevicius
Nov 15 at 10:14
The [::-1] can be exchanged for a double negation, I think.
– DonQuiKong
2 days ago
@DonQuiKong that would be quite a bit longer though, as the digits are all strings, and would need to be converted to ints for this.
– ovs
yesterday
@JonasAusevicius Thanks a lot.
– ovs
yesterday
add a comment |
5
None
can be replaced withcmp
insort
function
– Jonas Ausevicius
Nov 15 at 10:14
The [::-1] can be exchanged for a double negation, I think.
– DonQuiKong
2 days ago
@DonQuiKong that would be quite a bit longer though, as the digits are all strings, and would need to be converted to ints for this.
– ovs
yesterday
@JonasAusevicius Thanks a lot.
– ovs
yesterday
5
5
None
can be replaced with cmp
in sort
function– Jonas Ausevicius
Nov 15 at 10:14
None
can be replaced with cmp
in sort
function– Jonas Ausevicius
Nov 15 at 10:14
The [::-1] can be exchanged for a double negation, I think.
– DonQuiKong
2 days ago
The [::-1] can be exchanged for a double negation, I think.
– DonQuiKong
2 days ago
@DonQuiKong that would be quite a bit longer though, as the digits are all strings, and would need to be converted to ints for this.
– ovs
yesterday
@DonQuiKong that would be quite a bit longer though, as the digits are all strings, and would need to be converted to ints for this.
– ovs
yesterday
@JonasAusevicius Thanks a lot.
– ovs
yesterday
@JonasAusevicius Thanks a lot.
– ovs
yesterday
add a comment |
up vote
5
down vote
Python 2, 58 60 bytes
lambda a:sorted(a,key=lambda x:sorted(set(`x`))[::-1])[::-1]
Try it online!
add a comment |
up vote
5
down vote
Python 2, 58 60 bytes
lambda a:sorted(a,key=lambda x:sorted(set(`x`))[::-1])[::-1]
Try it online!
add a comment |
up vote
5
down vote
up vote
5
down vote
Python 2, 58 60 bytes
lambda a:sorted(a,key=lambda x:sorted(set(`x`))[::-1])[::-1]
Try it online!
Python 2, 58 60 bytes
lambda a:sorted(a,key=lambda x:sorted(set(`x`))[::-1])[::-1]
Try it online!
answered Nov 15 at 10:02
Jonas Ausevicius
1113
1113
add a comment |
add a comment |
up vote
5
down vote
Brachylog, 9 bytes
{ȧdṫo₁}ᵒ¹
Note: due to how ordering works in brachylog, it does not work on number correctly. This is fixed by casting the number to a string (ṫ
) at the cost of 1 byte.
Try it online!
2
What do you mean by "Due to how ordering works in brachylog, it does not work as intended."? I've tried all four test cases, and its giving the correct results (unless I accidentally looked past something).
– Kevin Cruijssen
Nov 15 at 10:12
@KevinCruijssen Theṫ
(to string) fixes the issue. Ordering digits in a number descending works as follows. Order from smallest to largest then reverse. The problem is that the number3120
ordered from smallest to largest is0123
which is equal to123
which reversed is321
and not3210
– Kroppeb
Nov 15 at 11:00
2
Ah ok, so your current code is working due to the added toString (ṫ
). As mentioned by @Arnauld, I thought your comment meant your current code doesn't work. It might be better to mention it like: "This could have been 8 bytes by removing theṫ
(toString), but unfortunately it does not work as intended due to how ordering works in Brachylog."
– Kevin Cruijssen
Nov 15 at 11:12
Looking at what I wrote it seems that my brain got distracted midsentence. Fixed it.
– Kroppeb
Nov 15 at 11:31
add a comment |
up vote
5
down vote
Brachylog, 9 bytes
{ȧdṫo₁}ᵒ¹
Note: due to how ordering works in brachylog, it does not work on number correctly. This is fixed by casting the number to a string (ṫ
) at the cost of 1 byte.
Try it online!
2
What do you mean by "Due to how ordering works in brachylog, it does not work as intended."? I've tried all four test cases, and its giving the correct results (unless I accidentally looked past something).
– Kevin Cruijssen
Nov 15 at 10:12
@KevinCruijssen Theṫ
(to string) fixes the issue. Ordering digits in a number descending works as follows. Order from smallest to largest then reverse. The problem is that the number3120
ordered from smallest to largest is0123
which is equal to123
which reversed is321
and not3210
– Kroppeb
Nov 15 at 11:00
2
Ah ok, so your current code is working due to the added toString (ṫ
). As mentioned by @Arnauld, I thought your comment meant your current code doesn't work. It might be better to mention it like: "This could have been 8 bytes by removing theṫ
(toString), but unfortunately it does not work as intended due to how ordering works in Brachylog."
– Kevin Cruijssen
Nov 15 at 11:12
Looking at what I wrote it seems that my brain got distracted midsentence. Fixed it.
– Kroppeb
Nov 15 at 11:31
add a comment |
up vote
5
down vote
up vote
5
down vote
Brachylog, 9 bytes
{ȧdṫo₁}ᵒ¹
Note: due to how ordering works in brachylog, it does not work on number correctly. This is fixed by casting the number to a string (ṫ
) at the cost of 1 byte.
Try it online!
Brachylog, 9 bytes
{ȧdṫo₁}ᵒ¹
Note: due to how ordering works in brachylog, it does not work on number correctly. This is fixed by casting the number to a string (ṫ
) at the cost of 1 byte.
Try it online!
edited Nov 15 at 11:30
answered Nov 15 at 9:59
Kroppeb
1,00628
1,00628
2
What do you mean by "Due to how ordering works in brachylog, it does not work as intended."? I've tried all four test cases, and its giving the correct results (unless I accidentally looked past something).
– Kevin Cruijssen
Nov 15 at 10:12
@KevinCruijssen Theṫ
(to string) fixes the issue. Ordering digits in a number descending works as follows. Order from smallest to largest then reverse. The problem is that the number3120
ordered from smallest to largest is0123
which is equal to123
which reversed is321
and not3210
– Kroppeb
Nov 15 at 11:00
2
Ah ok, so your current code is working due to the added toString (ṫ
). As mentioned by @Arnauld, I thought your comment meant your current code doesn't work. It might be better to mention it like: "This could have been 8 bytes by removing theṫ
(toString), but unfortunately it does not work as intended due to how ordering works in Brachylog."
– Kevin Cruijssen
Nov 15 at 11:12
Looking at what I wrote it seems that my brain got distracted midsentence. Fixed it.
– Kroppeb
Nov 15 at 11:31
add a comment |
2
What do you mean by "Due to how ordering works in brachylog, it does not work as intended."? I've tried all four test cases, and its giving the correct results (unless I accidentally looked past something).
– Kevin Cruijssen
Nov 15 at 10:12
@KevinCruijssen Theṫ
(to string) fixes the issue. Ordering digits in a number descending works as follows. Order from smallest to largest then reverse. The problem is that the number3120
ordered from smallest to largest is0123
which is equal to123
which reversed is321
and not3210
– Kroppeb
Nov 15 at 11:00
2
Ah ok, so your current code is working due to the added toString (ṫ
). As mentioned by @Arnauld, I thought your comment meant your current code doesn't work. It might be better to mention it like: "This could have been 8 bytes by removing theṫ
(toString), but unfortunately it does not work as intended due to how ordering works in Brachylog."
– Kevin Cruijssen
Nov 15 at 11:12
Looking at what I wrote it seems that my brain got distracted midsentence. Fixed it.
– Kroppeb
Nov 15 at 11:31
2
2
What do you mean by "Due to how ordering works in brachylog, it does not work as intended."? I've tried all four test cases, and its giving the correct results (unless I accidentally looked past something).
– Kevin Cruijssen
Nov 15 at 10:12
What do you mean by "Due to how ordering works in brachylog, it does not work as intended."? I've tried all four test cases, and its giving the correct results (unless I accidentally looked past something).
– Kevin Cruijssen
Nov 15 at 10:12
@KevinCruijssen The
ṫ
(to string) fixes the issue. Ordering digits in a number descending works as follows. Order from smallest to largest then reverse. The problem is that the number 3120
ordered from smallest to largest is 0123
which is equal to 123
which reversed is 321
and not 3210
– Kroppeb
Nov 15 at 11:00
@KevinCruijssen The
ṫ
(to string) fixes the issue. Ordering digits in a number descending works as follows. Order from smallest to largest then reverse. The problem is that the number 3120
ordered from smallest to largest is 0123
which is equal to 123
which reversed is 321
and not 3210
– Kroppeb
Nov 15 at 11:00
2
2
Ah ok, so your current code is working due to the added toString (
ṫ
). As mentioned by @Arnauld, I thought your comment meant your current code doesn't work. It might be better to mention it like: "This could have been 8 bytes by removing the ṫ
(toString), but unfortunately it does not work as intended due to how ordering works in Brachylog."– Kevin Cruijssen
Nov 15 at 11:12
Ah ok, so your current code is working due to the added toString (
ṫ
). As mentioned by @Arnauld, I thought your comment meant your current code doesn't work. It might be better to mention it like: "This could have been 8 bytes by removing the ṫ
(toString), but unfortunately it does not work as intended due to how ordering works in Brachylog."– Kevin Cruijssen
Nov 15 at 11:12
Looking at what I wrote it seems that my brain got distracted midsentence. Fixed it.
– Kroppeb
Nov 15 at 11:31
Looking at what I wrote it seems that my brain got distracted midsentence. Fixed it.
– Kroppeb
Nov 15 at 11:31
add a comment |
up vote
5
down vote
Pyth, 7 6 bytes
-1 byte by @Sok
_o_{S`
Pyth, which uses only printable ASCII, is at a bit of a disadvantage here. Optimally encoded this would be 6*log(95)/log(256) = 4.927
bytes, beating 05AB1E.
Explained:
o Sort the implicit input by lambda N:
_ reversed
{ uniquified
S sorted
' string representation [of N]
_ then reverse the result.
Try it here.
2
The trailingN
can be left out to save 1 byte - all lambda-type functions infer the presence of the principle lambda variable if any arguments are missing from the end. Examples includem
inferringd
,f
inferringT
,u
inferringG
...
– Sok
Nov 15 at 12:08
add a comment |
up vote
5
down vote
Pyth, 7 6 bytes
-1 byte by @Sok
_o_{S`
Pyth, which uses only printable ASCII, is at a bit of a disadvantage here. Optimally encoded this would be 6*log(95)/log(256) = 4.927
bytes, beating 05AB1E.
Explained:
o Sort the implicit input by lambda N:
_ reversed
{ uniquified
S sorted
' string representation [of N]
_ then reverse the result.
Try it here.
2
The trailingN
can be left out to save 1 byte - all lambda-type functions infer the presence of the principle lambda variable if any arguments are missing from the end. Examples includem
inferringd
,f
inferringT
,u
inferringG
...
– Sok
Nov 15 at 12:08
add a comment |
up vote
5
down vote
up vote
5
down vote
Pyth, 7 6 bytes
-1 byte by @Sok
_o_{S`
Pyth, which uses only printable ASCII, is at a bit of a disadvantage here. Optimally encoded this would be 6*log(95)/log(256) = 4.927
bytes, beating 05AB1E.
Explained:
o Sort the implicit input by lambda N:
_ reversed
{ uniquified
S sorted
' string representation [of N]
_ then reverse the result.
Try it here.
Pyth, 7 6 bytes
-1 byte by @Sok
_o_{S`
Pyth, which uses only printable ASCII, is at a bit of a disadvantage here. Optimally encoded this would be 6*log(95)/log(256) = 4.927
bytes, beating 05AB1E.
Explained:
o Sort the implicit input by lambda N:
_ reversed
{ uniquified
S sorted
' string representation [of N]
_ then reverse the result.
Try it here.
edited Nov 15 at 13:00
answered Nov 15 at 10:36
lirtosiast
15.3k436104
15.3k436104
2
The trailingN
can be left out to save 1 byte - all lambda-type functions infer the presence of the principle lambda variable if any arguments are missing from the end. Examples includem
inferringd
,f
inferringT
,u
inferringG
...
– Sok
Nov 15 at 12:08
add a comment |
2
The trailingN
can be left out to save 1 byte - all lambda-type functions infer the presence of the principle lambda variable if any arguments are missing from the end. Examples includem
inferringd
,f
inferringT
,u
inferringG
...
– Sok
Nov 15 at 12:08
2
2
The trailing
N
can be left out to save 1 byte - all lambda-type functions infer the presence of the principle lambda variable if any arguments are missing from the end. Examples include m
inferring d
, f
inferring T
, u
inferring G
...– Sok
Nov 15 at 12:08
The trailing
N
can be left out to save 1 byte - all lambda-type functions infer the presence of the principle lambda variable if any arguments are missing from the end. Examples include m
inferring d
, f
inferring T
, u
inferring G
...– Sok
Nov 15 at 12:08
add a comment |
up vote
4
down vote
Jelly, 8 bytes
ADṢUQµÞU
Try it online!
How it works
ADṢUQµÞU Main link (monad). Input: integer list
µÞU Sort by (reversed):
AD Absolute value converted to decimal digits
ṢUQ Sort, reverse, take unique values
2
I just implemented this then found your post. I went with normal reverses,Ṛ
, rather than the upends,U
. Note, however, that you do not need theD
since sort,Ṣ
, is implemented with aniterable(z, make_digits=True)
call inside. So that wasAṢQṚµÞṚ
for 7.
– Jonathan Allan
Nov 15 at 18:16
add a comment |
up vote
4
down vote
Jelly, 8 bytes
ADṢUQµÞU
Try it online!
How it works
ADṢUQµÞU Main link (monad). Input: integer list
µÞU Sort by (reversed):
AD Absolute value converted to decimal digits
ṢUQ Sort, reverse, take unique values
2
I just implemented this then found your post. I went with normal reverses,Ṛ
, rather than the upends,U
. Note, however, that you do not need theD
since sort,Ṣ
, is implemented with aniterable(z, make_digits=True)
call inside. So that wasAṢQṚµÞṚ
for 7.
– Jonathan Allan
Nov 15 at 18:16
add a comment |
up vote
4
down vote
up vote
4
down vote
Jelly, 8 bytes
ADṢUQµÞU
Try it online!
How it works
ADṢUQµÞU Main link (monad). Input: integer list
µÞU Sort by (reversed):
AD Absolute value converted to decimal digits
ṢUQ Sort, reverse, take unique values
Jelly, 8 bytes
ADṢUQµÞU
Try it online!
How it works
ADṢUQµÞU Main link (monad). Input: integer list
µÞU Sort by (reversed):
AD Absolute value converted to decimal digits
ṢUQ Sort, reverse, take unique values
answered Nov 15 at 10:10
Bubbler
5,514754
5,514754
2
I just implemented this then found your post. I went with normal reverses,Ṛ
, rather than the upends,U
. Note, however, that you do not need theD
since sort,Ṣ
, is implemented with aniterable(z, make_digits=True)
call inside. So that wasAṢQṚµÞṚ
for 7.
– Jonathan Allan
Nov 15 at 18:16
add a comment |
2
I just implemented this then found your post. I went with normal reverses,Ṛ
, rather than the upends,U
. Note, however, that you do not need theD
since sort,Ṣ
, is implemented with aniterable(z, make_digits=True)
call inside. So that wasAṢQṚµÞṚ
for 7.
– Jonathan Allan
Nov 15 at 18:16
2
2
I just implemented this then found your post. I went with normal reverses,
Ṛ
, rather than the upends, U
. Note, however, that you do not need the D
since sort, Ṣ
, is implemented with an iterable(z, make_digits=True)
call inside. So that was AṢQṚµÞṚ
for 7.– Jonathan Allan
Nov 15 at 18:16
I just implemented this then found your post. I went with normal reverses,
Ṛ
, rather than the upends, U
. Note, however, that you do not need the D
since sort, Ṣ
, is implemented with an iterable(z, make_digits=True)
call inside. So that was AṢQṚµÞṚ
for 7.– Jonathan Allan
Nov 15 at 18:16
add a comment |
up vote
3
down vote
MathGolf, 7 6 bytes
áÉ░▀zx
Try it online! or as a test suite.
Explanation
After looking at Emigna's 05AB1E solution, I found that I didn't need the absolute operator (and my previous answer was actually incorrect because of that operator). Now the main difference is that I convert to string and get unique characters instead of using the 1-byte operator in 05AB1E.
áÉ Sort by the value generated from mapping each element using the next 3 instructions
░ Convert to string
▀ Get unique characters
z Sort reversed (last instruction of block)
x Reverse list (needed because I don't have a sort-reversed by mapping)
add a comment |
up vote
3
down vote
MathGolf, 7 6 bytes
áÉ░▀zx
Try it online! or as a test suite.
Explanation
After looking at Emigna's 05AB1E solution, I found that I didn't need the absolute operator (and my previous answer was actually incorrect because of that operator). Now the main difference is that I convert to string and get unique characters instead of using the 1-byte operator in 05AB1E.
áÉ Sort by the value generated from mapping each element using the next 3 instructions
░ Convert to string
▀ Get unique characters
z Sort reversed (last instruction of block)
x Reverse list (needed because I don't have a sort-reversed by mapping)
add a comment |
up vote
3
down vote
up vote
3
down vote
MathGolf, 7 6 bytes
áÉ░▀zx
Try it online! or as a test suite.
Explanation
After looking at Emigna's 05AB1E solution, I found that I didn't need the absolute operator (and my previous answer was actually incorrect because of that operator). Now the main difference is that I convert to string and get unique characters instead of using the 1-byte operator in 05AB1E.
áÉ Sort by the value generated from mapping each element using the next 3 instructions
░ Convert to string
▀ Get unique characters
z Sort reversed (last instruction of block)
x Reverse list (needed because I don't have a sort-reversed by mapping)
MathGolf, 7 6 bytes
áÉ░▀zx
Try it online! or as a test suite.
Explanation
After looking at Emigna's 05AB1E solution, I found that I didn't need the absolute operator (and my previous answer was actually incorrect because of that operator). Now the main difference is that I convert to string and get unique characters instead of using the 1-byte operator in 05AB1E.
áÉ Sort by the value generated from mapping each element using the next 3 instructions
░ Convert to string
▀ Get unique characters
z Sort reversed (last instruction of block)
x Reverse list (needed because I don't have a sort-reversed by mapping)
edited Nov 15 at 11:46
answered Nov 15 at 10:19
maxb
2,0181923
2,0181923
add a comment |
add a comment |
up vote
3
down vote
C (gcc), 114 111 bytes
a;v(n){n=n<0?-n:n;for(a=0;n;n/=10)a|=1<<n%10;n=a;}c(int*a,int*b){a=v(*a)<v(*b);}f(a,n)int*a,n;{qsort(a,n,4,c);}
Try it online!
Explanation:
f() uses qsort() to sort provided array in place. Using comparison function c() to compare numbers which evaluates numbers using v().
v() calculates a higher number if bigger digits are present in parameter.
[Edit 1]
Improved by 3 bytes. 2 byte credits to Kevin. Thanks
You can golfn>0
ton
I think in your loop of your methodv
.
– Kevin Cruijssen
Nov 15 at 11:55
Suggestfor(a=0;n=abs(n);
instead ofn=n<?-n:n;for(a=0;n;
– ceilingcat
2 days ago
f()'s argument listint*a,n
can be shortened toint*a
.
– gastropner
26 mins ago
add a comment |
up vote
3
down vote
C (gcc), 114 111 bytes
a;v(n){n=n<0?-n:n;for(a=0;n;n/=10)a|=1<<n%10;n=a;}c(int*a,int*b){a=v(*a)<v(*b);}f(a,n)int*a,n;{qsort(a,n,4,c);}
Try it online!
Explanation:
f() uses qsort() to sort provided array in place. Using comparison function c() to compare numbers which evaluates numbers using v().
v() calculates a higher number if bigger digits are present in parameter.
[Edit 1]
Improved by 3 bytes. 2 byte credits to Kevin. Thanks
You can golfn>0
ton
I think in your loop of your methodv
.
– Kevin Cruijssen
Nov 15 at 11:55
Suggestfor(a=0;n=abs(n);
instead ofn=n<?-n:n;for(a=0;n;
– ceilingcat
2 days ago
f()'s argument listint*a,n
can be shortened toint*a
.
– gastropner
26 mins ago
add a comment |
up vote
3
down vote
up vote
3
down vote
C (gcc), 114 111 bytes
a;v(n){n=n<0?-n:n;for(a=0;n;n/=10)a|=1<<n%10;n=a;}c(int*a,int*b){a=v(*a)<v(*b);}f(a,n)int*a,n;{qsort(a,n,4,c);}
Try it online!
Explanation:
f() uses qsort() to sort provided array in place. Using comparison function c() to compare numbers which evaluates numbers using v().
v() calculates a higher number if bigger digits are present in parameter.
[Edit 1]
Improved by 3 bytes. 2 byte credits to Kevin. Thanks
C (gcc), 114 111 bytes
a;v(n){n=n<0?-n:n;for(a=0;n;n/=10)a|=1<<n%10;n=a;}c(int*a,int*b){a=v(*a)<v(*b);}f(a,n)int*a,n;{qsort(a,n,4,c);}
Try it online!
Explanation:
f() uses qsort() to sort provided array in place. Using comparison function c() to compare numbers which evaluates numbers using v().
v() calculates a higher number if bigger digits are present in parameter.
[Edit 1]
Improved by 3 bytes. 2 byte credits to Kevin. Thanks
edited Nov 15 at 14:08
answered Nov 15 at 11:35
GPS
33115
33115
You can golfn>0
ton
I think in your loop of your methodv
.
– Kevin Cruijssen
Nov 15 at 11:55
Suggestfor(a=0;n=abs(n);
instead ofn=n<?-n:n;for(a=0;n;
– ceilingcat
2 days ago
f()'s argument listint*a,n
can be shortened toint*a
.
– gastropner
26 mins ago
add a comment |
You can golfn>0
ton
I think in your loop of your methodv
.
– Kevin Cruijssen
Nov 15 at 11:55
Suggestfor(a=0;n=abs(n);
instead ofn=n<?-n:n;for(a=0;n;
– ceilingcat
2 days ago
f()'s argument listint*a,n
can be shortened toint*a
.
– gastropner
26 mins ago
You can golf
n>0
to n
I think in your loop of your method v
.– Kevin Cruijssen
Nov 15 at 11:55
You can golf
n>0
to n
I think in your loop of your method v
.– Kevin Cruijssen
Nov 15 at 11:55
Suggest
for(a=0;n=abs(n);
instead of n=n<?-n:n;for(a=0;n;
– ceilingcat
2 days ago
Suggest
for(a=0;n=abs(n);
instead of n=n<?-n:n;for(a=0;n;
– ceilingcat
2 days ago
f()'s argument list
int*a,n
can be shortened to int*a
.– gastropner
26 mins ago
f()'s argument list
int*a,n
can be shortened to int*a
.– gastropner
26 mins ago
add a comment |
up vote
3
down vote
Stax, 6 7 bytes
èó≥ü≤♥¥
Run and debug it
This seems to give incorrect results. For example, in the test case of your TIO it outputs-904 8491 478 62778 6458 -7738 -73 373 123 3120 0
instead of the intended8491 -904 62778 478 -7738 6458 373 -73 3120 123 0
or8491 -904 62778 478 -7738 6458 -73 373 3120 123 0
. This test case is also used in the example, and to explain the rules, so I would take a look at that to understand it better. It seems you are only sorting by the single largest digit once, without any of the other rules?
– Kevin Cruijssen
2 days ago
@KevinCruijssen: Yes, my apologies. I mis-read the problem statement. I've adjusted the program to handle the stated requirements. This program is accepting the input integers as quoted strings. That's usually acceptable, but if not I might need to add another byte.
– recursive
2 days ago
Looks good now, +1 from me. And yes, inputting as strings is completely fine.
– Kevin Cruijssen
2 days ago
add a comment |
up vote
3
down vote
Stax, 6 7 bytes
èó≥ü≤♥¥
Run and debug it
This seems to give incorrect results. For example, in the test case of your TIO it outputs-904 8491 478 62778 6458 -7738 -73 373 123 3120 0
instead of the intended8491 -904 62778 478 -7738 6458 373 -73 3120 123 0
or8491 -904 62778 478 -7738 6458 -73 373 3120 123 0
. This test case is also used in the example, and to explain the rules, so I would take a look at that to understand it better. It seems you are only sorting by the single largest digit once, without any of the other rules?
– Kevin Cruijssen
2 days ago
@KevinCruijssen: Yes, my apologies. I mis-read the problem statement. I've adjusted the program to handle the stated requirements. This program is accepting the input integers as quoted strings. That's usually acceptable, but if not I might need to add another byte.
– recursive
2 days ago
Looks good now, +1 from me. And yes, inputting as strings is completely fine.
– Kevin Cruijssen
2 days ago
add a comment |
up vote
3
down vote
up vote
3
down vote
Stax, 6 7 bytes
èó≥ü≤♥¥
Run and debug it
Stax, 6 7 bytes
èó≥ü≤♥¥
Run and debug it
edited 2 days ago
answered Nov 15 at 23:28
recursive
4,9641221
4,9641221
This seems to give incorrect results. For example, in the test case of your TIO it outputs-904 8491 478 62778 6458 -7738 -73 373 123 3120 0
instead of the intended8491 -904 62778 478 -7738 6458 373 -73 3120 123 0
or8491 -904 62778 478 -7738 6458 -73 373 3120 123 0
. This test case is also used in the example, and to explain the rules, so I would take a look at that to understand it better. It seems you are only sorting by the single largest digit once, without any of the other rules?
– Kevin Cruijssen
2 days ago
@KevinCruijssen: Yes, my apologies. I mis-read the problem statement. I've adjusted the program to handle the stated requirements. This program is accepting the input integers as quoted strings. That's usually acceptable, but if not I might need to add another byte.
– recursive
2 days ago
Looks good now, +1 from me. And yes, inputting as strings is completely fine.
– Kevin Cruijssen
2 days ago
add a comment |
This seems to give incorrect results. For example, in the test case of your TIO it outputs-904 8491 478 62778 6458 -7738 -73 373 123 3120 0
instead of the intended8491 -904 62778 478 -7738 6458 373 -73 3120 123 0
or8491 -904 62778 478 -7738 6458 -73 373 3120 123 0
. This test case is also used in the example, and to explain the rules, so I would take a look at that to understand it better. It seems you are only sorting by the single largest digit once, without any of the other rules?
– Kevin Cruijssen
2 days ago
@KevinCruijssen: Yes, my apologies. I mis-read the problem statement. I've adjusted the program to handle the stated requirements. This program is accepting the input integers as quoted strings. That's usually acceptable, but if not I might need to add another byte.
– recursive
2 days ago
Looks good now, +1 from me. And yes, inputting as strings is completely fine.
– Kevin Cruijssen
2 days ago
This seems to give incorrect results. For example, in the test case of your TIO it outputs
-904 8491 478 62778 6458 -7738 -73 373 123 3120 0
instead of the intended 8491 -904 62778 478 -7738 6458 373 -73 3120 123 0
or 8491 -904 62778 478 -7738 6458 -73 373 3120 123 0
. This test case is also used in the example, and to explain the rules, so I would take a look at that to understand it better. It seems you are only sorting by the single largest digit once, without any of the other rules?– Kevin Cruijssen
2 days ago
This seems to give incorrect results. For example, in the test case of your TIO it outputs
-904 8491 478 62778 6458 -7738 -73 373 123 3120 0
instead of the intended 8491 -904 62778 478 -7738 6458 373 -73 3120 123 0
or 8491 -904 62778 478 -7738 6458 -73 373 3120 123 0
. This test case is also used in the example, and to explain the rules, so I would take a look at that to understand it better. It seems you are only sorting by the single largest digit once, without any of the other rules?– Kevin Cruijssen
2 days ago
@KevinCruijssen: Yes, my apologies. I mis-read the problem statement. I've adjusted the program to handle the stated requirements. This program is accepting the input integers as quoted strings. That's usually acceptable, but if not I might need to add another byte.
– recursive
2 days ago
@KevinCruijssen: Yes, my apologies. I mis-read the problem statement. I've adjusted the program to handle the stated requirements. This program is accepting the input integers as quoted strings. That's usually acceptable, but if not I might need to add another byte.
– recursive
2 days ago
Looks good now, +1 from me. And yes, inputting as strings is completely fine.
– Kevin Cruijssen
2 days ago
Looks good now, +1 from me. And yes, inputting as strings is completely fine.
– Kevin Cruijssen
2 days ago
add a comment |
up vote
2
down vote
J, 17 bytes
{~[::~.@:~@":@|
Try it online!
Explanation:
@| - find the absolute value and
@": - convert to string and
@:~ - sort down and
~. - keep only the unique symbols
: - grade down the entire list of strings
[: - function composition
{~ - use the graded-down list to index into the input
add a comment |
up vote
2
down vote
J, 17 bytes
{~[::~.@:~@":@|
Try it online!
Explanation:
@| - find the absolute value and
@": - convert to string and
@:~ - sort down and
~. - keep only the unique symbols
: - grade down the entire list of strings
[: - function composition
{~ - use the graded-down list to index into the input
add a comment |
up vote
2
down vote
up vote
2
down vote
J, 17 bytes
{~[::~.@:~@":@|
Try it online!
Explanation:
@| - find the absolute value and
@": - convert to string and
@:~ - sort down and
~. - keep only the unique symbols
: - grade down the entire list of strings
[: - function composition
{~ - use the graded-down list to index into the input
J, 17 bytes
{~[::~.@:~@":@|
Try it online!
Explanation:
@| - find the absolute value and
@": - convert to string and
@:~ - sort down and
~. - keep only the unique symbols
: - grade down the entire list of strings
[: - function composition
{~ - use the graded-down list to index into the input
edited Nov 15 at 12:20
answered Nov 15 at 12:09
Galen Ivanov
5,89711032
5,89711032
add a comment |
add a comment |
up vote
2
down vote
JavaScript (SpiderMonkey), 68 bytes
Thanks for @Arnauld for reminding me again that SpiderMonkey uses stable sort, so -4 bytes for removing ||-1
.
A=>A.sort((x,y,F=n=>[...new Set(""+n)].sort().reverse())=>F(x)<F(y))
Try it online!
JavaScript (Node.js), 72 bytes
A=>A.sort((x,y,F=n=>[...new Set(""+n)].sort().reverse())=>F(x)<F(y)||-1)
Try it online!
Or 68 bytes with SpiderMonkey.
– Arnauld
Nov 15 at 11:04
1
@Arnauld oh stable sort again ;P
– Shieru Asakoto
Nov 15 at 12:17
Actually, I think V8 is using at least 2 different sort algorithms. It seems like it's stable if the size of the array is less than or equal to $10$.
– Arnauld
Nov 15 at 12:37
1
@Arnauld V8 use quick sort before Chrome 70. The quick sort algorithm perform an insertion sort when array size is small enough. And latest Chrome had changed to stable sort for matching other browsers' (IE/Firefox/Safari) behavior.
– tsh
2 days ago
add a comment |
up vote
2
down vote
JavaScript (SpiderMonkey), 68 bytes
Thanks for @Arnauld for reminding me again that SpiderMonkey uses stable sort, so -4 bytes for removing ||-1
.
A=>A.sort((x,y,F=n=>[...new Set(""+n)].sort().reverse())=>F(x)<F(y))
Try it online!
JavaScript (Node.js), 72 bytes
A=>A.sort((x,y,F=n=>[...new Set(""+n)].sort().reverse())=>F(x)<F(y)||-1)
Try it online!
Or 68 bytes with SpiderMonkey.
– Arnauld
Nov 15 at 11:04
1
@Arnauld oh stable sort again ;P
– Shieru Asakoto
Nov 15 at 12:17
Actually, I think V8 is using at least 2 different sort algorithms. It seems like it's stable if the size of the array is less than or equal to $10$.
– Arnauld
Nov 15 at 12:37
1
@Arnauld V8 use quick sort before Chrome 70. The quick sort algorithm perform an insertion sort when array size is small enough. And latest Chrome had changed to stable sort for matching other browsers' (IE/Firefox/Safari) behavior.
– tsh
2 days ago
add a comment |
up vote
2
down vote
up vote
2
down vote
JavaScript (SpiderMonkey), 68 bytes
Thanks for @Arnauld for reminding me again that SpiderMonkey uses stable sort, so -4 bytes for removing ||-1
.
A=>A.sort((x,y,F=n=>[...new Set(""+n)].sort().reverse())=>F(x)<F(y))
Try it online!
JavaScript (Node.js), 72 bytes
A=>A.sort((x,y,F=n=>[...new Set(""+n)].sort().reverse())=>F(x)<F(y)||-1)
Try it online!
JavaScript (SpiderMonkey), 68 bytes
Thanks for @Arnauld for reminding me again that SpiderMonkey uses stable sort, so -4 bytes for removing ||-1
.
A=>A.sort((x,y,F=n=>[...new Set(""+n)].sort().reverse())=>F(x)<F(y))
Try it online!
JavaScript (Node.js), 72 bytes
A=>A.sort((x,y,F=n=>[...new Set(""+n)].sort().reverse())=>F(x)<F(y)||-1)
Try it online!
edited Nov 15 at 12:40
answered Nov 15 at 10:13
Shieru Asakoto
2,220314
2,220314
Or 68 bytes with SpiderMonkey.
– Arnauld
Nov 15 at 11:04
1
@Arnauld oh stable sort again ;P
– Shieru Asakoto
Nov 15 at 12:17
Actually, I think V8 is using at least 2 different sort algorithms. It seems like it's stable if the size of the array is less than or equal to $10$.
– Arnauld
Nov 15 at 12:37
1
@Arnauld V8 use quick sort before Chrome 70. The quick sort algorithm perform an insertion sort when array size is small enough. And latest Chrome had changed to stable sort for matching other browsers' (IE/Firefox/Safari) behavior.
– tsh
2 days ago
add a comment |
Or 68 bytes with SpiderMonkey.
– Arnauld
Nov 15 at 11:04
1
@Arnauld oh stable sort again ;P
– Shieru Asakoto
Nov 15 at 12:17
Actually, I think V8 is using at least 2 different sort algorithms. It seems like it's stable if the size of the array is less than or equal to $10$.
– Arnauld
Nov 15 at 12:37
1
@Arnauld V8 use quick sort before Chrome 70. The quick sort algorithm perform an insertion sort when array size is small enough. And latest Chrome had changed to stable sort for matching other browsers' (IE/Firefox/Safari) behavior.
– tsh
2 days ago
Or 68 bytes with SpiderMonkey.
– Arnauld
Nov 15 at 11:04
Or 68 bytes with SpiderMonkey.
– Arnauld
Nov 15 at 11:04
1
1
@Arnauld oh stable sort again ;P
– Shieru Asakoto
Nov 15 at 12:17
@Arnauld oh stable sort again ;P
– Shieru Asakoto
Nov 15 at 12:17
Actually, I think V8 is using at least 2 different sort algorithms. It seems like it's stable if the size of the array is less than or equal to $10$.
– Arnauld
Nov 15 at 12:37
Actually, I think V8 is using at least 2 different sort algorithms. It seems like it's stable if the size of the array is less than or equal to $10$.
– Arnauld
Nov 15 at 12:37
1
1
@Arnauld V8 use quick sort before Chrome 70. The quick sort algorithm perform an insertion sort when array size is small enough. And latest Chrome had changed to stable sort for matching other browsers' (IE/Firefox/Safari) behavior.
– tsh
2 days ago
@Arnauld V8 use quick sort before Chrome 70. The quick sort algorithm perform an insertion sort when array size is small enough. And latest Chrome had changed to stable sort for matching other browsers' (IE/Firefox/Safari) behavior.
– tsh
2 days ago
add a comment |
up vote
2
down vote
Java (JDK), 98 bytes
l->l.sort((a,b)->{int r=0,i=58;for(;r==0&i-->48;)r=(b.indexOf(i)>>9)-(a.indexOf(i)>>9);return r;})
Try it online!
Explanation
l-> // Consumer<List<String>>
l.sort( // Use the incorporated sort method which uses a...
(a,b)->{ // Comparator of Strings
int r=0, // define r as the result, initiated to 0
i=58; // i as the codepoint to test for.
for(;r==0&i-->48;) // for each digit codepoint from '9' to '0',
// and while no difference was found.
r= // set r as the difference between
(b.indexOf(i)>>9)- // was the digit found in b? then 0 else -1 using the bit-shift operator
(a.indexOf(i)>>9); // and was the digit found in a? then 0 else -1.
return r; // return the comparison result.
}
)
Note:
I needed a way to map numbers to either 0/1
or 0/-1
.
indexOf
has the nice property that it's consistently returning -1
for characters not found. -1
right-shifted by any number is always -1
. Any positive number right-shifted by a big enough number will always produce 0
.
So here we are:
input input.indexOf('9') input.indexOf('9')>>9
"999" 0 0
"111119" 5 0
"123456" -1 -1
1
Ah, yeah, that's what I mean. ;p Nice golf of using>>9
instead of>>32
due to the limited range of the numbers.
– Kevin Cruijssen
Nov 15 at 12:53
add a comment |
up vote
2
down vote
Java (JDK), 98 bytes
l->l.sort((a,b)->{int r=0,i=58;for(;r==0&i-->48;)r=(b.indexOf(i)>>9)-(a.indexOf(i)>>9);return r;})
Try it online!
Explanation
l-> // Consumer<List<String>>
l.sort( // Use the incorporated sort method which uses a...
(a,b)->{ // Comparator of Strings
int r=0, // define r as the result, initiated to 0
i=58; // i as the codepoint to test for.
for(;r==0&i-->48;) // for each digit codepoint from '9' to '0',
// and while no difference was found.
r= // set r as the difference between
(b.indexOf(i)>>9)- // was the digit found in b? then 0 else -1 using the bit-shift operator
(a.indexOf(i)>>9); // and was the digit found in a? then 0 else -1.
return r; // return the comparison result.
}
)
Note:
I needed a way to map numbers to either 0/1
or 0/-1
.
indexOf
has the nice property that it's consistently returning -1
for characters not found. -1
right-shifted by any number is always -1
. Any positive number right-shifted by a big enough number will always produce 0
.
So here we are:
input input.indexOf('9') input.indexOf('9')>>9
"999" 0 0
"111119" 5 0
"123456" -1 -1
1
Ah, yeah, that's what I mean. ;p Nice golf of using>>9
instead of>>32
due to the limited range of the numbers.
– Kevin Cruijssen
Nov 15 at 12:53
add a comment |
up vote
2
down vote
up vote
2
down vote
Java (JDK), 98 bytes
l->l.sort((a,b)->{int r=0,i=58;for(;r==0&i-->48;)r=(b.indexOf(i)>>9)-(a.indexOf(i)>>9);return r;})
Try it online!
Explanation
l-> // Consumer<List<String>>
l.sort( // Use the incorporated sort method which uses a...
(a,b)->{ // Comparator of Strings
int r=0, // define r as the result, initiated to 0
i=58; // i as the codepoint to test for.
for(;r==0&i-->48;) // for each digit codepoint from '9' to '0',
// and while no difference was found.
r= // set r as the difference between
(b.indexOf(i)>>9)- // was the digit found in b? then 0 else -1 using the bit-shift operator
(a.indexOf(i)>>9); // and was the digit found in a? then 0 else -1.
return r; // return the comparison result.
}
)
Note:
I needed a way to map numbers to either 0/1
or 0/-1
.
indexOf
has the nice property that it's consistently returning -1
for characters not found. -1
right-shifted by any number is always -1
. Any positive number right-shifted by a big enough number will always produce 0
.
So here we are:
input input.indexOf('9') input.indexOf('9')>>9
"999" 0 0
"111119" 5 0
"123456" -1 -1
Java (JDK), 98 bytes
l->l.sort((a,b)->{int r=0,i=58;for(;r==0&i-->48;)r=(b.indexOf(i)>>9)-(a.indexOf(i)>>9);return r;})
Try it online!
Explanation
l-> // Consumer<List<String>>
l.sort( // Use the incorporated sort method which uses a...
(a,b)->{ // Comparator of Strings
int r=0, // define r as the result, initiated to 0
i=58; // i as the codepoint to test for.
for(;r==0&i-->48;) // for each digit codepoint from '9' to '0',
// and while no difference was found.
r= // set r as the difference between
(b.indexOf(i)>>9)- // was the digit found in b? then 0 else -1 using the bit-shift operator
(a.indexOf(i)>>9); // and was the digit found in a? then 0 else -1.
return r; // return the comparison result.
}
)
Note:
I needed a way to map numbers to either 0/1
or 0/-1
.
indexOf
has the nice property that it's consistently returning -1
for characters not found. -1
right-shifted by any number is always -1
. Any positive number right-shifted by a big enough number will always produce 0
.
So here we are:
input input.indexOf('9') input.indexOf('9')>>9
"999" 0 0
"111119" 5 0
"123456" -1 -1
edited Nov 15 at 13:01
answered Nov 15 at 12:11
Olivier Grégoire
8,25711842
8,25711842
1
Ah, yeah, that's what I mean. ;p Nice golf of using>>9
instead of>>32
due to the limited range of the numbers.
– Kevin Cruijssen
Nov 15 at 12:53
add a comment |
1
Ah, yeah, that's what I mean. ;p Nice golf of using>>9
instead of>>32
due to the limited range of the numbers.
– Kevin Cruijssen
Nov 15 at 12:53
1
1
Ah, yeah, that's what I mean. ;p Nice golf of using
>>9
instead of >>32
due to the limited range of the numbers.– Kevin Cruijssen
Nov 15 at 12:53
Ah, yeah, that's what I mean. ;p Nice golf of using
>>9
instead of >>32
due to the limited range of the numbers.– Kevin Cruijssen
Nov 15 at 12:53
add a comment |
up vote
2
down vote
Japt, 12 bytes
ñ_a ì â ñnÃw
All test cases
Explanation:
ñ_ Ãw :Sort Descending by:
a : Get the absolute value
ì : Get the digits
â : Remove duplicates
ñn : Sort the digits in descending order
add a comment |
up vote
2
down vote
Japt, 12 bytes
ñ_a ì â ñnÃw
All test cases
Explanation:
ñ_ Ãw :Sort Descending by:
a : Get the absolute value
ì : Get the digits
â : Remove duplicates
ñn : Sort the digits in descending order
add a comment |
up vote
2
down vote
up vote
2
down vote
Japt, 12 bytes
ñ_a ì â ñnÃw
All test cases
Explanation:
ñ_ Ãw :Sort Descending by:
a : Get the absolute value
ì : Get the digits
â : Remove duplicates
ñn : Sort the digits in descending order
Japt, 12 bytes
ñ_a ì â ñnÃw
All test cases
Explanation:
ñ_ Ãw :Sort Descending by:
a : Get the absolute value
ì : Get the digits
â : Remove duplicates
ñn : Sort the digits in descending order
edited Nov 15 at 18:01
answered Nov 15 at 14:59
Kamil Drakari
2,551416
2,551416
add a comment |
add a comment |
up vote
2
down vote
Haskell, 54 52 bytes
import Data.List
f=r.sortOn(r.sort.nub.show);r=reverse
Try it online!
Definingr=reverse
saves two bytes. We also allow anonymous functions, so thef=
does not need to ve counted.
– Laikoni
Nov 15 at 23:33
I moved the import and f= to the TIO header. Is that OK?
– Martin Lütke
Nov 15 at 23:34
Same byte count, but maybe of some interest:f=r$r id.nub.show;r=(reverse.).sortOn
.
– Laikoni
Nov 15 at 23:43
1
The import actually needs to be counted.
– Laikoni
Nov 15 at 23:45
2
You may want to have a look at our Guide to golfing rules in Haskell.
– Laikoni
Nov 15 at 23:52
|
show 1 more comment
up vote
2
down vote
Haskell, 54 52 bytes
import Data.List
f=r.sortOn(r.sort.nub.show);r=reverse
Try it online!
Definingr=reverse
saves two bytes. We also allow anonymous functions, so thef=
does not need to ve counted.
– Laikoni
Nov 15 at 23:33
I moved the import and f= to the TIO header. Is that OK?
– Martin Lütke
Nov 15 at 23:34
Same byte count, but maybe of some interest:f=r$r id.nub.show;r=(reverse.).sortOn
.
– Laikoni
Nov 15 at 23:43
1
The import actually needs to be counted.
– Laikoni
Nov 15 at 23:45
2
You may want to have a look at our Guide to golfing rules in Haskell.
– Laikoni
Nov 15 at 23:52
|
show 1 more comment
up vote
2
down vote
up vote
2
down vote
Haskell, 54 52 bytes
import Data.List
f=r.sortOn(r.sort.nub.show);r=reverse
Try it online!
Haskell, 54 52 bytes
import Data.List
f=r.sortOn(r.sort.nub.show);r=reverse
Try it online!
edited Nov 16 at 0:09
answered Nov 15 at 23:21
Martin Lütke
37125
37125
Definingr=reverse
saves two bytes. We also allow anonymous functions, so thef=
does not need to ve counted.
– Laikoni
Nov 15 at 23:33
I moved the import and f= to the TIO header. Is that OK?
– Martin Lütke
Nov 15 at 23:34
Same byte count, but maybe of some interest:f=r$r id.nub.show;r=(reverse.).sortOn
.
– Laikoni
Nov 15 at 23:43
1
The import actually needs to be counted.
– Laikoni
Nov 15 at 23:45
2
You may want to have a look at our Guide to golfing rules in Haskell.
– Laikoni
Nov 15 at 23:52
|
show 1 more comment
Definingr=reverse
saves two bytes. We also allow anonymous functions, so thef=
does not need to ve counted.
– Laikoni
Nov 15 at 23:33
I moved the import and f= to the TIO header. Is that OK?
– Martin Lütke
Nov 15 at 23:34
Same byte count, but maybe of some interest:f=r$r id.nub.show;r=(reverse.).sortOn
.
– Laikoni
Nov 15 at 23:43
1
The import actually needs to be counted.
– Laikoni
Nov 15 at 23:45
2
You may want to have a look at our Guide to golfing rules in Haskell.
– Laikoni
Nov 15 at 23:52
Defining
r=reverse
saves two bytes. We also allow anonymous functions, so the f=
does not need to ve counted.– Laikoni
Nov 15 at 23:33
Defining
r=reverse
saves two bytes. We also allow anonymous functions, so the f=
does not need to ve counted.– Laikoni
Nov 15 at 23:33
I moved the import and f= to the TIO header. Is that OK?
– Martin Lütke
Nov 15 at 23:34
I moved the import and f= to the TIO header. Is that OK?
– Martin Lütke
Nov 15 at 23:34
Same byte count, but maybe of some interest:
f=r$r id.nub.show;r=(reverse.).sortOn
.– Laikoni
Nov 15 at 23:43
Same byte count, but maybe of some interest:
f=r$r id.nub.show;r=(reverse.).sortOn
.– Laikoni
Nov 15 at 23:43
1
1
The import actually needs to be counted.
– Laikoni
Nov 15 at 23:45
The import actually needs to be counted.
– Laikoni
Nov 15 at 23:45
2
2
You may want to have a look at our Guide to golfing rules in Haskell.
– Laikoni
Nov 15 at 23:52
You may want to have a look at our Guide to golfing rules in Haskell.
– Laikoni
Nov 15 at 23:52
|
show 1 more comment
up vote
2
down vote
APL (Dyalog Extended), 19 bytes
{⍵[⍒∪¨(∨'¯'~⍨⍕)¨⍵]}
Try it online!
Fixed at a cost of +2 bytes thanks to OP.
I think you are missing an 'uniquify' somewhere? If I try the example test case in your TIO for example,¯7738
is placed before478
, but it should be after it: digits[8,7,4]
come before digits[8,7,3]
.
– Kevin Cruijssen
2 days ago
Thanks, @KevinCruijssen
– Zacharý
2 days ago
add a comment |
up vote
2
down vote
APL (Dyalog Extended), 19 bytes
{⍵[⍒∪¨(∨'¯'~⍨⍕)¨⍵]}
Try it online!
Fixed at a cost of +2 bytes thanks to OP.
I think you are missing an 'uniquify' somewhere? If I try the example test case in your TIO for example,¯7738
is placed before478
, but it should be after it: digits[8,7,4]
come before digits[8,7,3]
.
– Kevin Cruijssen
2 days ago
Thanks, @KevinCruijssen
– Zacharý
2 days ago
add a comment |
up vote
2
down vote
up vote
2
down vote
APL (Dyalog Extended), 19 bytes
{⍵[⍒∪¨(∨'¯'~⍨⍕)¨⍵]}
Try it online!
Fixed at a cost of +2 bytes thanks to OP.
APL (Dyalog Extended), 19 bytes
{⍵[⍒∪¨(∨'¯'~⍨⍕)¨⍵]}
Try it online!
Fixed at a cost of +2 bytes thanks to OP.
edited 2 days ago
answered 2 days ago
Zacharý
5,10511035
5,10511035
I think you are missing an 'uniquify' somewhere? If I try the example test case in your TIO for example,¯7738
is placed before478
, but it should be after it: digits[8,7,4]
come before digits[8,7,3]
.
– Kevin Cruijssen
2 days ago
Thanks, @KevinCruijssen
– Zacharý
2 days ago
add a comment |
I think you are missing an 'uniquify' somewhere? If I try the example test case in your TIO for example,¯7738
is placed before478
, but it should be after it: digits[8,7,4]
come before digits[8,7,3]
.
– Kevin Cruijssen
2 days ago
Thanks, @KevinCruijssen
– Zacharý
2 days ago
I think you are missing an 'uniquify' somewhere? If I try the example test case in your TIO for example,
¯7738
is placed before 478
, but it should be after it: digits [8,7,4]
come before digits [8,7,3]
.– Kevin Cruijssen
2 days ago
I think you are missing an 'uniquify' somewhere? If I try the example test case in your TIO for example,
¯7738
is placed before 478
, but it should be after it: digits [8,7,4]
come before digits [8,7,3]
.– Kevin Cruijssen
2 days ago
Thanks, @KevinCruijssen
– Zacharý
2 days ago
Thanks, @KevinCruijssen
– Zacharý
2 days ago
add a comment |
up vote
1
down vote
Ruby, 55 bytes
->a{a.sort_by{|x|x.abs.digits.sort.reverse|}.reverse}
Try it online!
add a comment |
up vote
1
down vote
Ruby, 55 bytes
->a{a.sort_by{|x|x.abs.digits.sort.reverse|}.reverse}
Try it online!
add a comment |
up vote
1
down vote
up vote
1
down vote
Ruby, 55 bytes
->a{a.sort_by{|x|x.abs.digits.sort.reverse|}.reverse}
Try it online!
Ruby, 55 bytes
->a{a.sort_by{|x|x.abs.digits.sort.reverse|}.reverse}
Try it online!
answered Nov 15 at 14:29
Kirill L.
3,2161118
3,2161118
add a comment |
add a comment |
up vote
1
down vote
Perl 5 -nl
, 68 bytes
$a{eval"9876543210=~y/$_//dcr"}=$_}{say$a{$_}for reverse sort keys%a
Try it online!
add a comment |
up vote
1
down vote
Perl 5 -nl
, 68 bytes
$a{eval"9876543210=~y/$_//dcr"}=$_}{say$a{$_}for reverse sort keys%a
Try it online!
add a comment |
up vote
1
down vote
up vote
1
down vote
Perl 5 -nl
, 68 bytes
$a{eval"9876543210=~y/$_//dcr"}=$_}{say$a{$_}for reverse sort keys%a
Try it online!
Perl 5 -nl
, 68 bytes
$a{eval"9876543210=~y/$_//dcr"}=$_}{say$a{$_}for reverse sort keys%a
Try it online!
answered Nov 15 at 19:22
Xcali
4,990520
4,990520
add a comment |
add a comment |
up vote
1
down vote
Julia 1.0, 50 bytes
x->sort(x,by=y->sort(unique([digits(-abs(y));1])))
Try it online!
add a comment |
up vote
1
down vote
Julia 1.0, 50 bytes
x->sort(x,by=y->sort(unique([digits(-abs(y));1])))
Try it online!
add a comment |
up vote
1
down vote
up vote
1
down vote
Julia 1.0, 50 bytes
x->sort(x,by=y->sort(unique([digits(-abs(y));1])))
Try it online!
Julia 1.0, 50 bytes
x->sort(x,by=y->sort(unique([digits(-abs(y));1])))
Try it online!
answered 2 days ago
Kirill L.
3,2161118
3,2161118
add a comment |
add a comment |
up vote
1
down vote
APL(NARS), 366 chars, 732 bytes
_gb←⍬
∇a _s w;t
t←_gb[a]⋄_gb[a]←_gb[w]⋄_gb[w]←t
∇
∇(_f _q)w;l;r;ls;i
(l r)←w⋄→0×⍳l≥r⋄l _s⌊2÷⍨l+r⋄ls←i←l⋄→3
→3×⍳∼0<_gb[i]_f _gb[l]⋄ls+←1⋄ls _s i
→2×⍳r≥i+←1
l _s ls⋄_f _q l(ls-1)⋄_f _q(ls+1)r
∇
∇r←(a qsort)w
r←¯1⋄→0×⍳1≠⍴⍴w⋄_gb←w⋄a _q 1(↑⍴w)⋄r←_gb
∇
f←{∪t[⍒t←⍎¨⍕∣⍵]}
∇r←a c b;x;y;i;m
x←f a⋄y←f b⋄r←i←0⋄m←(↑⍴x)⌊(↑⍴y)⋄→3
→0×⍳x[i]<y[i]⋄→3×⍳∼x[i]>y[i]⋄r←1⋄→0
→2×⍳m≥i+←1⋄r←(↑⍴x)>(↑⍴y)
∇
For the qsort operator, it is one traslation in APL of algo page 139 K&R Linguaggio C.
I think in it there is using data as C with pointers...
Test
c qsort 123, 478, ¯904, 62778, 0, ¯73, 8491, 3120, 6458, ¯7738, 373
8491 ¯904 62778 478 ¯7738 6458 ¯73 373 3120 123 0
c qsort 11, ¯312, 902, 23, 321, 2132, 34202, ¯34, ¯382
902 ¯382 34202 ¯34 321 ¯312 2132 23 11
c qsort 9, 44, 2212, 4, 6, 6, 1, 2, 192, 21, 29384, 0
29384 192 9 6 6 4 44 2212 21 2 1 0
c qsort 44, ¯88, 9, 233, ¯3, 14, 101, 77, 555, 67
9 ¯88 67 77 555 14 44 233 ¯3 101
add a comment |
up vote
1
down vote
APL(NARS), 366 chars, 732 bytes
_gb←⍬
∇a _s w;t
t←_gb[a]⋄_gb[a]←_gb[w]⋄_gb[w]←t
∇
∇(_f _q)w;l;r;ls;i
(l r)←w⋄→0×⍳l≥r⋄l _s⌊2÷⍨l+r⋄ls←i←l⋄→3
→3×⍳∼0<_gb[i]_f _gb[l]⋄ls+←1⋄ls _s i
→2×⍳r≥i+←1
l _s ls⋄_f _q l(ls-1)⋄_f _q(ls+1)r
∇
∇r←(a qsort)w
r←¯1⋄→0×⍳1≠⍴⍴w⋄_gb←w⋄a _q 1(↑⍴w)⋄r←_gb
∇
f←{∪t[⍒t←⍎¨⍕∣⍵]}
∇r←a c b;x;y;i;m
x←f a⋄y←f b⋄r←i←0⋄m←(↑⍴x)⌊(↑⍴y)⋄→3
→0×⍳x[i]<y[i]⋄→3×⍳∼x[i]>y[i]⋄r←1⋄→0
→2×⍳m≥i+←1⋄r←(↑⍴x)>(↑⍴y)
∇
For the qsort operator, it is one traslation in APL of algo page 139 K&R Linguaggio C.
I think in it there is using data as C with pointers...
Test
c qsort 123, 478, ¯904, 62778, 0, ¯73, 8491, 3120, 6458, ¯7738, 373
8491 ¯904 62778 478 ¯7738 6458 ¯73 373 3120 123 0
c qsort 11, ¯312, 902, 23, 321, 2132, 34202, ¯34, ¯382
902 ¯382 34202 ¯34 321 ¯312 2132 23 11
c qsort 9, 44, 2212, 4, 6, 6, 1, 2, 192, 21, 29384, 0
29384 192 9 6 6 4 44 2212 21 2 1 0
c qsort 44, ¯88, 9, 233, ¯3, 14, 101, 77, 555, 67
9 ¯88 67 77 555 14 44 233 ¯3 101
add a comment |
up vote
1
down vote
up vote
1
down vote
APL(NARS), 366 chars, 732 bytes
_gb←⍬
∇a _s w;t
t←_gb[a]⋄_gb[a]←_gb[w]⋄_gb[w]←t
∇
∇(_f _q)w;l;r;ls;i
(l r)←w⋄→0×⍳l≥r⋄l _s⌊2÷⍨l+r⋄ls←i←l⋄→3
→3×⍳∼0<_gb[i]_f _gb[l]⋄ls+←1⋄ls _s i
→2×⍳r≥i+←1
l _s ls⋄_f _q l(ls-1)⋄_f _q(ls+1)r
∇
∇r←(a qsort)w
r←¯1⋄→0×⍳1≠⍴⍴w⋄_gb←w⋄a _q 1(↑⍴w)⋄r←_gb
∇
f←{∪t[⍒t←⍎¨⍕∣⍵]}
∇r←a c b;x;y;i;m
x←f a⋄y←f b⋄r←i←0⋄m←(↑⍴x)⌊(↑⍴y)⋄→3
→0×⍳x[i]<y[i]⋄→3×⍳∼x[i]>y[i]⋄r←1⋄→0
→2×⍳m≥i+←1⋄r←(↑⍴x)>(↑⍴y)
∇
For the qsort operator, it is one traslation in APL of algo page 139 K&R Linguaggio C.
I think in it there is using data as C with pointers...
Test
c qsort 123, 478, ¯904, 62778, 0, ¯73, 8491, 3120, 6458, ¯7738, 373
8491 ¯904 62778 478 ¯7738 6458 ¯73 373 3120 123 0
c qsort 11, ¯312, 902, 23, 321, 2132, 34202, ¯34, ¯382
902 ¯382 34202 ¯34 321 ¯312 2132 23 11
c qsort 9, 44, 2212, 4, 6, 6, 1, 2, 192, 21, 29384, 0
29384 192 9 6 6 4 44 2212 21 2 1 0
c qsort 44, ¯88, 9, 233, ¯3, 14, 101, 77, 555, 67
9 ¯88 67 77 555 14 44 233 ¯3 101
APL(NARS), 366 chars, 732 bytes
_gb←⍬
∇a _s w;t
t←_gb[a]⋄_gb[a]←_gb[w]⋄_gb[w]←t
∇
∇(_f _q)w;l;r;ls;i
(l r)←w⋄→0×⍳l≥r⋄l _s⌊2÷⍨l+r⋄ls←i←l⋄→3
→3×⍳∼0<_gb[i]_f _gb[l]⋄ls+←1⋄ls _s i
→2×⍳r≥i+←1
l _s ls⋄_f _q l(ls-1)⋄_f _q(ls+1)r
∇
∇r←(a qsort)w
r←¯1⋄→0×⍳1≠⍴⍴w⋄_gb←w⋄a _q 1(↑⍴w)⋄r←_gb
∇
f←{∪t[⍒t←⍎¨⍕∣⍵]}
∇r←a c b;x;y;i;m
x←f a⋄y←f b⋄r←i←0⋄m←(↑⍴x)⌊(↑⍴y)⋄→3
→0×⍳x[i]<y[i]⋄→3×⍳∼x[i]>y[i]⋄r←1⋄→0
→2×⍳m≥i+←1⋄r←(↑⍴x)>(↑⍴y)
∇
For the qsort operator, it is one traslation in APL of algo page 139 K&R Linguaggio C.
I think in it there is using data as C with pointers...
Test
c qsort 123, 478, ¯904, 62778, 0, ¯73, 8491, 3120, 6458, ¯7738, 373
8491 ¯904 62778 478 ¯7738 6458 ¯73 373 3120 123 0
c qsort 11, ¯312, 902, 23, 321, 2132, 34202, ¯34, ¯382
902 ¯382 34202 ¯34 321 ¯312 2132 23 11
c qsort 9, 44, 2212, 4, 6, 6, 1, 2, 192, 21, 29384, 0
29384 192 9 6 6 4 44 2212 21 2 1 0
c qsort 44, ¯88, 9, 233, ¯3, 14, 101, 77, 555, 67
9 ¯88 67 77 555 14 44 233 ¯3 101
answered yesterday
RosLuP
1,747514
1,747514
add a comment |
add a comment |
up vote
1
down vote
Powershell, 44 bytes
$args|sort{$_-split'(.)'-ne'-'|sort -u -d}-d
Test script:
$f = {
$args|sort{$_-split'(.)'-ne'-'|sort -u -d}-d
}
@(
,( (123, 478, -904, 62778, 0, -73, 8491, 3120, 6458, -7738, 373),
(8491, -904, 62778, 478, -7738, 6458, 373, -73, 3120, 123, 0),
(8491, -904, 62778, 478, -7738, 6458, -73, 373, 3120, 123, 0) )
,( (11, -312, 902, 23, 321, 2132, 34202, -34, -382),
(902, -382, 34202, -34, -312, 321, 2132, 23, 11),
(902, -382, 34202, -34, 2132, -312, 321, 23, 11) )
,( (9, 44, 2212, 4, 6, 6, 1, 2, 192, 21, 29384, 0),
(29384, 192, 9, 6, 6, 4, 44, 2212, 21, 2, 1, 0),
(29384, 192, 9, 6, 6, 44, 4, 2212, 21, 2, 1, 0),
(29384, 192, 9, 6, 6, 44, 4, 21, 2212, 2, 1, 0) )
,( (44, -88, 9, 233, -3, 14, 101, 77, 555, 67),
,(9, -88, 67, 77, 555, 14, 44, 233, -3, 101) )
) | % {
$a, $expected = $_
$result = &$f @a
$true-in($expected|%{"$result"-eq"$_"})
"$result"
}
Output:
True
8491 -904 62778 478 -7738 6458 -73 373 3120 123 0
True
902 -382 34202 -34 2132 -312 321 23 11
True
29384 192 9 6 6 44 4 21 2212 2 1 0
True
9 -88 67 77 555 14 44 233 -3 101
add a comment |
up vote
1
down vote
Powershell, 44 bytes
$args|sort{$_-split'(.)'-ne'-'|sort -u -d}-d
Test script:
$f = {
$args|sort{$_-split'(.)'-ne'-'|sort -u -d}-d
}
@(
,( (123, 478, -904, 62778, 0, -73, 8491, 3120, 6458, -7738, 373),
(8491, -904, 62778, 478, -7738, 6458, 373, -73, 3120, 123, 0),
(8491, -904, 62778, 478, -7738, 6458, -73, 373, 3120, 123, 0) )
,( (11, -312, 902, 23, 321, 2132, 34202, -34, -382),
(902, -382, 34202, -34, -312, 321, 2132, 23, 11),
(902, -382, 34202, -34, 2132, -312, 321, 23, 11) )
,( (9, 44, 2212, 4, 6, 6, 1, 2, 192, 21, 29384, 0),
(29384, 192, 9, 6, 6, 4, 44, 2212, 21, 2, 1, 0),
(29384, 192, 9, 6, 6, 44, 4, 2212, 21, 2, 1, 0),
(29384, 192, 9, 6, 6, 44, 4, 21, 2212, 2, 1, 0) )
,( (44, -88, 9, 233, -3, 14, 101, 77, 555, 67),
,(9, -88, 67, 77, 555, 14, 44, 233, -3, 101) )
) | % {
$a, $expected = $_
$result = &$f @a
$true-in($expected|%{"$result"-eq"$_"})
"$result"
}
Output:
True
8491 -904 62778 478 -7738 6458 -73 373 3120 123 0
True
902 -382 34202 -34 2132 -312 321 23 11
True
29384 192 9 6 6 44 4 21 2212 2 1 0
True
9 -88 67 77 555 14 44 233 -3 101
add a comment |
up vote
1
down vote
up vote
1
down vote
Powershell, 44 bytes
$args|sort{$_-split'(.)'-ne'-'|sort -u -d}-d
Test script:
$f = {
$args|sort{$_-split'(.)'-ne'-'|sort -u -d}-d
}
@(
,( (123, 478, -904, 62778, 0, -73, 8491, 3120, 6458, -7738, 373),
(8491, -904, 62778, 478, -7738, 6458, 373, -73, 3120, 123, 0),
(8491, -904, 62778, 478, -7738, 6458, -73, 373, 3120, 123, 0) )
,( (11, -312, 902, 23, 321, 2132, 34202, -34, -382),
(902, -382, 34202, -34, -312, 321, 2132, 23, 11),
(902, -382, 34202, -34, 2132, -312, 321, 23, 11) )
,( (9, 44, 2212, 4, 6, 6, 1, 2, 192, 21, 29384, 0),
(29384, 192, 9, 6, 6, 4, 44, 2212, 21, 2, 1, 0),
(29384, 192, 9, 6, 6, 44, 4, 2212, 21, 2, 1, 0),
(29384, 192, 9, 6, 6, 44, 4, 21, 2212, 2, 1, 0) )
,( (44, -88, 9, 233, -3, 14, 101, 77, 555, 67),
,(9, -88, 67, 77, 555, 14, 44, 233, -3, 101) )
) | % {
$a, $expected = $_
$result = &$f @a
$true-in($expected|%{"$result"-eq"$_"})
"$result"
}
Output:
True
8491 -904 62778 478 -7738 6458 -73 373 3120 123 0
True
902 -382 34202 -34 2132 -312 321 23 11
True
29384 192 9 6 6 44 4 21 2212 2 1 0
True
9 -88 67 77 555 14 44 233 -3 101
Powershell, 44 bytes
$args|sort{$_-split'(.)'-ne'-'|sort -u -d}-d
Test script:
$f = {
$args|sort{$_-split'(.)'-ne'-'|sort -u -d}-d
}
@(
,( (123, 478, -904, 62778, 0, -73, 8491, 3120, 6458, -7738, 373),
(8491, -904, 62778, 478, -7738, 6458, 373, -73, 3120, 123, 0),
(8491, -904, 62778, 478, -7738, 6458, -73, 373, 3120, 123, 0) )
,( (11, -312, 902, 23, 321, 2132, 34202, -34, -382),
(902, -382, 34202, -34, -312, 321, 2132, 23, 11),
(902, -382, 34202, -34, 2132, -312, 321, 23, 11) )
,( (9, 44, 2212, 4, 6, 6, 1, 2, 192, 21, 29384, 0),
(29384, 192, 9, 6, 6, 4, 44, 2212, 21, 2, 1, 0),
(29384, 192, 9, 6, 6, 44, 4, 2212, 21, 2, 1, 0),
(29384, 192, 9, 6, 6, 44, 4, 21, 2212, 2, 1, 0) )
,( (44, -88, 9, 233, -3, 14, 101, 77, 555, 67),
,(9, -88, 67, 77, 555, 14, 44, 233, -3, 101) )
) | % {
$a, $expected = $_
$result = &$f @a
$true-in($expected|%{"$result"-eq"$_"})
"$result"
}
Output:
True
8491 -904 62778 478 -7738 6458 -73 373 3120 123 0
True
902 -382 34202 -34 2132 -312 321 23 11
True
29384 192 9 6 6 44 4 21 2212 2 1 0
True
9 -88 67 77 555 14 44 233 -3 101
answered yesterday
mazzy
1,747313
1,747313
add a comment |
add a comment |
up vote
1
down vote
PHP, 87 86 84 bytes
while(--$argc)$a[_.strrev(count_chars($n=$argv[++$i],3))]=$n;krsort($a);print_r($a);
Run with -nr
or try it online.
Replace ++$i
with $argc
(+1 byte) to suppress the Notice (and render -n
obosolete).
breakdown
while(--$argc) # loop through command line arguments
$a[ # key=
_. # 3. prepend non-numeric char for non-numeric sort
strrev( # 2. reverse =^= sort descending
count_chars($n=$argv[++$i],3) # 1. get characters used in argument
)
]=$n; # value=argument
krsort($a); # sort by key descending
print_r($a); # print
-
is "smaller" than the digits, so it has no affect on the sorting.
add a comment |
up vote
1
down vote
PHP, 87 86 84 bytes
while(--$argc)$a[_.strrev(count_chars($n=$argv[++$i],3))]=$n;krsort($a);print_r($a);
Run with -nr
or try it online.
Replace ++$i
with $argc
(+1 byte) to suppress the Notice (and render -n
obosolete).
breakdown
while(--$argc) # loop through command line arguments
$a[ # key=
_. # 3. prepend non-numeric char for non-numeric sort
strrev( # 2. reverse =^= sort descending
count_chars($n=$argv[++$i],3) # 1. get characters used in argument
)
]=$n; # value=argument
krsort($a); # sort by key descending
print_r($a); # print
-
is "smaller" than the digits, so it has no affect on the sorting.
add a comment |
up vote
1
down vote
up vote
1
down vote
PHP, 87 86 84 bytes
while(--$argc)$a[_.strrev(count_chars($n=$argv[++$i],3))]=$n;krsort($a);print_r($a);
Run with -nr
or try it online.
Replace ++$i
with $argc
(+1 byte) to suppress the Notice (and render -n
obosolete).
breakdown
while(--$argc) # loop through command line arguments
$a[ # key=
_. # 3. prepend non-numeric char for non-numeric sort
strrev( # 2. reverse =^= sort descending
count_chars($n=$argv[++$i],3) # 1. get characters used in argument
)
]=$n; # value=argument
krsort($a); # sort by key descending
print_r($a); # print
-
is "smaller" than the digits, so it has no affect on the sorting.
PHP, 87 86 84 bytes
while(--$argc)$a[_.strrev(count_chars($n=$argv[++$i],3))]=$n;krsort($a);print_r($a);
Run with -nr
or try it online.
Replace ++$i
with $argc
(+1 byte) to suppress the Notice (and render -n
obosolete).
breakdown
while(--$argc) # loop through command line arguments
$a[ # key=
_. # 3. prepend non-numeric char for non-numeric sort
strrev( # 2. reverse =^= sort descending
count_chars($n=$argv[++$i],3) # 1. get characters used in argument
)
]=$n; # value=argument
krsort($a); # sort by key descending
print_r($a); # print
-
is "smaller" than the digits, so it has no affect on the sorting.
edited yesterday
answered yesterday
Titus
12.8k11237
12.8k11237
add a comment |
add a comment |
up vote
1
down vote
Common Lisp, 88 bytes
(sort(read)'string> :key(lambda(x)(sort(remove-duplicates(format()"~d"(abs x)))'char>)))
Try it online!
Good old verbose Common Lisp!
Explanation:
(sort ; sort
(read) ; what to sort: a list of numbers, read on input stream
'string> ; comparison predicate (remember: this is a typed language!)
:key (lambda (x) ; how to get an element to sort; get a number
(sort (remove-duplicates ; then sort the unique digits (characters)
(format() "~d" (abs x))) ; from its string representation
'char>))) ; with the appropriate comparison operator for characters
add a comment |
up vote
1
down vote
Common Lisp, 88 bytes
(sort(read)'string> :key(lambda(x)(sort(remove-duplicates(format()"~d"(abs x)))'char>)))
Try it online!
Good old verbose Common Lisp!
Explanation:
(sort ; sort
(read) ; what to sort: a list of numbers, read on input stream
'string> ; comparison predicate (remember: this is a typed language!)
:key (lambda (x) ; how to get an element to sort; get a number
(sort (remove-duplicates ; then sort the unique digits (characters)
(format() "~d" (abs x))) ; from its string representation
'char>))) ; with the appropriate comparison operator for characters
add a comment |
up vote
1
down vote
up vote
1
down vote
Common Lisp, 88 bytes
(sort(read)'string> :key(lambda(x)(sort(remove-duplicates(format()"~d"(abs x)))'char>)))
Try it online!
Good old verbose Common Lisp!
Explanation:
(sort ; sort
(read) ; what to sort: a list of numbers, read on input stream
'string> ; comparison predicate (remember: this is a typed language!)
:key (lambda (x) ; how to get an element to sort; get a number
(sort (remove-duplicates ; then sort the unique digits (characters)
(format() "~d" (abs x))) ; from its string representation
'char>))) ; with the appropriate comparison operator for characters
Common Lisp, 88 bytes
(sort(read)'string> :key(lambda(x)(sort(remove-duplicates(format()"~d"(abs x)))'char>)))
Try it online!
Good old verbose Common Lisp!
Explanation:
(sort ; sort
(read) ; what to sort: a list of numbers, read on input stream
'string> ; comparison predicate (remember: this is a typed language!)
:key (lambda (x) ; how to get an element to sort; get a number
(sort (remove-duplicates ; then sort the unique digits (characters)
(format() "~d" (abs x))) ; from its string representation
'char>))) ; with the appropriate comparison operator for characters
edited 15 hours ago
answered 15 hours ago
Renzo
1,570516
1,570516
add a comment |
add a comment |
up vote
1
down vote
C# (Visual C# Interactive Compiler), 75 74 bytes
-1 thanks @ASCII-only
x=>x.OrderByDescending(y=>String.Concat((y+"").Distinct().OrderBy(z=>-z)))
Try it online!
In C#, strings are considered "enumerables" of characters. I use this to my advantage by first converting each number to a string. LINQ is then leveraged to get the unique characters (digits) sorted in reverse order. I convert each sorted character array back into a string and use that as the sort key to order the whole list.
Looks like you'll be able to get away with not adding-
, looks like order of those doesn't really matter?
– ASCII-only
21 hours ago
Without the-
, test case #2 returns... 321 2132 ...
which seems incorrect?
– dana
15 hours ago
nah, read the example more carefully
– ASCII-only
9 hours ago
OK - I think your right. Thanks for the tip!
– dana
9 hours ago
add a comment |
up vote
1
down vote
C# (Visual C# Interactive Compiler), 75 74 bytes
-1 thanks @ASCII-only
x=>x.OrderByDescending(y=>String.Concat((y+"").Distinct().OrderBy(z=>-z)))
Try it online!
In C#, strings are considered "enumerables" of characters. I use this to my advantage by first converting each number to a string. LINQ is then leveraged to get the unique characters (digits) sorted in reverse order. I convert each sorted character array back into a string and use that as the sort key to order the whole list.
Looks like you'll be able to get away with not adding-
, looks like order of those doesn't really matter?
– ASCII-only
21 hours ago
Without the-
, test case #2 returns... 321 2132 ...
which seems incorrect?
– dana
15 hours ago
nah, read the example more carefully
– ASCII-only
9 hours ago
OK - I think your right. Thanks for the tip!
– dana
9 hours ago
add a comment |
up vote
1
down vote
up vote
1
down vote
C# (Visual C# Interactive Compiler), 75 74 bytes
-1 thanks @ASCII-only
x=>x.OrderByDescending(y=>String.Concat((y+"").Distinct().OrderBy(z=>-z)))
Try it online!
In C#, strings are considered "enumerables" of characters. I use this to my advantage by first converting each number to a string. LINQ is then leveraged to get the unique characters (digits) sorted in reverse order. I convert each sorted character array back into a string and use that as the sort key to order the whole list.
C# (Visual C# Interactive Compiler), 75 74 bytes
-1 thanks @ASCII-only
x=>x.OrderByDescending(y=>String.Concat((y+"").Distinct().OrderBy(z=>-z)))
Try it online!
In C#, strings are considered "enumerables" of characters. I use this to my advantage by first converting each number to a string. LINQ is then leveraged to get the unique characters (digits) sorted in reverse order. I convert each sorted character array back into a string and use that as the sort key to order the whole list.
edited 8 hours ago
answered Nov 16 at 4:37
dana
1714
1714
Looks like you'll be able to get away with not adding-
, looks like order of those doesn't really matter?
– ASCII-only
21 hours ago
Without the-
, test case #2 returns... 321 2132 ...
which seems incorrect?
– dana
15 hours ago
nah, read the example more carefully
– ASCII-only
9 hours ago
OK - I think your right. Thanks for the tip!
– dana
9 hours ago
add a comment |
Looks like you'll be able to get away with not adding-
, looks like order of those doesn't really matter?
– ASCII-only
21 hours ago
Without the-
, test case #2 returns... 321 2132 ...
which seems incorrect?
– dana
15 hours ago
nah, read the example more carefully
– ASCII-only
9 hours ago
OK - I think your right. Thanks for the tip!
– dana
9 hours ago
Looks like you'll be able to get away with not adding
-
, looks like order of those doesn't really matter?– ASCII-only
21 hours ago
Looks like you'll be able to get away with not adding
-
, looks like order of those doesn't really matter?– ASCII-only
21 hours ago
Without the
-
, test case #2 returns ... 321 2132 ...
which seems incorrect?– dana
15 hours ago
Without the
-
, test case #2 returns ... 321 2132 ...
which seems incorrect?– dana
15 hours ago
nah, read the example more carefully
– ASCII-only
9 hours ago
nah, read the example more carefully
– ASCII-only
9 hours ago
OK - I think your right. Thanks for the tip!
– dana
9 hours ago
OK - I think your right. Thanks for the tip!
– dana
9 hours ago
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