Drop 1st element of {#a, #b, #c} &
up vote
4
down vote
favorite
I pass the following to a function:
{#a, #b, #c} &
These are keys that are used in a list of associations.
There are instances where I want to pass only the last 2 elements, such as:
{#b, #c} &
Right now, I just rewrite it. But it seems like there should be a way to do this by getting rid of the first slot.
I looked at the full form and see that it's:
FullForm[{#a, #b, #c} &]
Function[List[Slot["a"], Slot["b"], Slot["c"]]]
I thought I could somehow get inside and drop Slot["a"], but can't seem to do it. I tried replacing Slot["a"] with a blank, but I'm left with a comma at the beginning of the list. I also tried Apply to change Function to List and doing it in the result. I couldn't.
Is there any way to get rid of the first element, #a, without rewriting the expression?
expression-manipulation slot
edited Nov 20 at 21:49
Kuba♦
103k12200511
asked Nov 20 at 21:07
Mitchell Kaplan
1,6421026
add a comment |
up vote
4
down vote
favorite
I pass the following to a function:
{#a, #b, #c} &
These are keys that are used in a list of associations.
There are instances where I want to pass only the last 2 elements, such as:
{#b, #c} &
Right now, I just rewrite it. But it seems like there should be a way to do this by getting rid of the first slot.
I looked at the full form and see that it's:
FullForm[{#a, #b, #c} &]
Function[List[Slot["a"], Slot["b"], Slot["c"]]]
I thought I could somehow get inside and drop Slot["a"], but can't seem to do it. I tried replacing Slot["a"] with a blank, but I'm left with a comma at the beginning of the list. I also tried Apply to change Function to List and doing it in the result. I couldn't.
Is there any way to get rid of the first element, #a, without rewriting the expression?
expression-manipulation slot
edited Nov 20 at 21:49
Kuba♦
103k12200511
asked Nov 20 at 21:07
Mitchell Kaplan
1,6421026
Instead of a blank you could replaceSlot["a"]->Nothing. You'll get thisNothingin the resulting function object, but once the function is applied,Nothingwill collapse.
– Ruslan
Nov 21 at 6:51
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I pass the following to a function:
{#a, #b, #c} &
These are keys that are used in a list of associations.
There are instances where I want to pass only the last 2 elements, such as:
{#b, #c} &
Right now, I just rewrite it. But it seems like there should be a way to do this by getting rid of the first slot.
I looked at the full form and see that it's:
FullForm[{#a, #b, #c} &]
Function[List[Slot["a"], Slot["b"], Slot["c"]]]
I thought I could somehow get inside and drop Slot["a"], but can't seem to do it. I tried replacing Slot["a"] with a blank, but I'm left with a comma at the beginning of the list. I also tried Apply to change Function to List and doing it in the result. I couldn't.
Is there any way to get rid of the first element, #a, without rewriting the expression?
expression-manipulation slot
edited Nov 20 at 21:49
Kuba♦
103k12200511
asked Nov 20 at 21:07
Mitchell Kaplan
1,6421026
I pass the following to a function:
{#a, #b, #c} &
These are keys that are used in a list of associations.
There are instances where I want to pass only the last 2 elements, such as:
{#b, #c} &
Right now, I just rewrite it. But it seems like there should be a way to do this by getting rid of the first slot.
I looked at the full form and see that it's:
FullForm[{#a, #b, #c} &]
Function[List[Slot["a"], Slot["b"], Slot["c"]]]
I thought I could somehow get inside and drop Slot["a"], but can't seem to do it. I tried replacing Slot["a"] with a blank, but I'm left with a comma at the beginning of the list. I also tried Apply to change Function to List and doing it in the result. I couldn't.
Is there any way to get rid of the first element, #a, without rewriting the expression?
expression-manipulation slot
expression-manipulation slot
edited Nov 20 at 21:49
Kuba♦
103k12200511
asked Nov 20 at 21:07
Mitchell Kaplan
1,6421026
edited Nov 20 at 21:49
Kuba♦
103k12200511
asked Nov 20 at 21:07
Mitchell Kaplan
1,6421026
edited Nov 20 at 21:49
Kuba♦
103k12200511
edited Nov 20 at 21:49
Kuba♦
103k12200511
edited Nov 20 at 21:49
Kuba♦
103k12200511
103k12200511
asked Nov 20 at 21:07
Mitchell Kaplan
1,6421026
asked Nov 20 at 21:07
Mitchell Kaplan
1,6421026
asked Nov 20 at 21:07
Mitchell Kaplan
1,6421026
1,6421026
Instead of a blank you could replaceSlot["a"]->Nothing. You'll get thisNothingin the resulting function object, but once the function is applied,Nothingwill collapse.
– Ruslan
Nov 21 at 6:51
add a comment |
Instead of a blank you could replaceSlot["a"]->Nothing. You'll get thisNothingin the resulting function object, but once the function is applied,Nothingwill collapse.
– Ruslan
Nov 21 at 6:51
Instead of a blank you could replace
Slot["a"]->Nothing. You'll get this Nothing in the resulting function object, but once the function is applied, Nothing will collapse.– Ruslan
Nov 21 at 6:51
Instead of a blank you could replace
Slot["a"]->Nothing. You'll get this Nothing in the resulting function object, but once the function is applied, Nothing will collapse.– Ruslan
Nov 21 at 6:51
add a comment |
2 Answers
2
active
oldest
votes
up vote
7
down vote
accepted
You could pass it to Delete:
Delete[#, {1, 1}] &[{#a, #b, #c} &]
{#b, #c} &
answered Nov 20 at 21:09
Coolwater
14.4k32452
add a comment |
up vote
5
down vote
foo = {#a, #b, #c} &;
foo[[{1}, 2 ;;]]
{#b, #c} &
or, but only because we know there won't be any side effects:
Evaluate /@ Rest /@ foo
{#b, #c} &
or
foo /. {_Slot, rest__Slot} :> {rest}
At the end consider using an operator form of a KeyDrop
bar = KeyTake[{"a", "b", "c"}];
(Rest /@ bar)@<|"a" -> 1, "b" -> 1, "c" -> 1|>
<|"b" -> 1, "c" -> 1|>
answered Nov 20 at 21:48
Kuba♦
103k12200511
I accepted Coolwater's answer because it solved my problem simply. I appreciate the extra insight that your answer provided. Actually, I'll have to spend more time to really understand it though.
– Mitchell Kaplan
Nov 20 at 21:51
@MitchellKaplan sure, btw, can you give a little insigth into how do you use it? Because it feels like a XY problem, maybe there is a better way anyway.
– Kuba♦
Nov 20 at 21:55
I use a lot of lists of associations and very often I have to sum things by one or more keys. Why I asked this question: I work in reinsurance. My list has, among other things: treaty, effective date, line of business, IBNR. I'm estimating IBNR by all 3, but it has to balance to the total IBNR for effective/line. I need to sum both by all 3, but also by only effective/line. I call a function that does this, and I want to adjust the keys I'm using to eliminate treaty when I'm summing over just effective/line. Left some detail out to fit answer into comment.
– Mitchell Kaplan
Nov 21 at 16:51
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
You could pass it to Delete:
Delete[#, {1, 1}] &[{#a, #b, #c} &]
{#b, #c} &
answered Nov 20 at 21:09
Coolwater
14.4k32452
add a comment |
up vote
7
down vote
accepted
You could pass it to Delete:
Delete[#, {1, 1}] &[{#a, #b, #c} &]
{#b, #c} &
answered Nov 20 at 21:09
Coolwater
14.4k32452
add a comment |
up vote
7
down vote
accepted
up vote
7
down vote
accepted
You could pass it to Delete:
Delete[#, {1, 1}] &[{#a, #b, #c} &]
{#b, #c} &
answered Nov 20 at 21:09
Coolwater
14.4k32452
You could pass it to Delete:
Delete[#, {1, 1}] &[{#a, #b, #c} &]
{#b, #c} &
answered Nov 20 at 21:09
Coolwater
14.4k32452
answered Nov 20 at 21:09
Coolwater
14.4k32452
answered Nov 20 at 21:09
Coolwater
14.4k32452
answered Nov 20 at 21:09
Coolwater
14.4k32452
14.4k32452
add a comment |
add a comment |
up vote
5
down vote
foo = {#a, #b, #c} &;
foo[[{1}, 2 ;;]]
{#b, #c} &
or, but only because we know there won't be any side effects:
Evaluate /@ Rest /@ foo
{#b, #c} &
or
foo /. {_Slot, rest__Slot} :> {rest}
At the end consider using an operator form of a KeyDrop
bar = KeyTake[{"a", "b", "c"}];
(Rest /@ bar)@<|"a" -> 1, "b" -> 1, "c" -> 1|>
<|"b" -> 1, "c" -> 1|>
answered Nov 20 at 21:48
Kuba♦
103k12200511
I accepted Coolwater's answer because it solved my problem simply. I appreciate the extra insight that your answer provided. Actually, I'll have to spend more time to really understand it though.
– Mitchell Kaplan
Nov 20 at 21:51
@MitchellKaplan sure, btw, can you give a little insigth into how do you use it? Because it feels like a XY problem, maybe there is a better way anyway.
– Kuba♦
Nov 20 at 21:55
I use a lot of lists of associations and very often I have to sum things by one or more keys. Why I asked this question: I work in reinsurance. My list has, among other things: treaty, effective date, line of business, IBNR. I'm estimating IBNR by all 3, but it has to balance to the total IBNR for effective/line. I need to sum both by all 3, but also by only effective/line. I call a function that does this, and I want to adjust the keys I'm using to eliminate treaty when I'm summing over just effective/line. Left some detail out to fit answer into comment.
– Mitchell Kaplan
Nov 21 at 16:51
add a comment |
up vote
5
down vote
foo = {#a, #b, #c} &;
foo[[{1}, 2 ;;]]
{#b, #c} &
or, but only because we know there won't be any side effects:
Evaluate /@ Rest /@ foo
{#b, #c} &
or
foo /. {_Slot, rest__Slot} :> {rest}
At the end consider using an operator form of a KeyDrop
bar = KeyTake[{"a", "b", "c"}];
(Rest /@ bar)@<|"a" -> 1, "b" -> 1, "c" -> 1|>
<|"b" -> 1, "c" -> 1|>
answered Nov 20 at 21:48
Kuba♦
103k12200511
I accepted Coolwater's answer because it solved my problem simply. I appreciate the extra insight that your answer provided. Actually, I'll have to spend more time to really understand it though.
– Mitchell Kaplan
Nov 20 at 21:51
@MitchellKaplan sure, btw, can you give a little insigth into how do you use it? Because it feels like a XY problem, maybe there is a better way anyway.
– Kuba♦
Nov 20 at 21:55
I use a lot of lists of associations and very often I have to sum things by one or more keys. Why I asked this question: I work in reinsurance. My list has, among other things: treaty, effective date, line of business, IBNR. I'm estimating IBNR by all 3, but it has to balance to the total IBNR for effective/line. I need to sum both by all 3, but also by only effective/line. I call a function that does this, and I want to adjust the keys I'm using to eliminate treaty when I'm summing over just effective/line. Left some detail out to fit answer into comment.
– Mitchell Kaplan
Nov 21 at 16:51
add a comment |
up vote
5
down vote
up vote
5
down vote
foo = {#a, #b, #c} &;
foo[[{1}, 2 ;;]]
{#b, #c} &
or, but only because we know there won't be any side effects:
Evaluate /@ Rest /@ foo
{#b, #c} &
or
foo /. {_Slot, rest__Slot} :> {rest}
At the end consider using an operator form of a KeyDrop
bar = KeyTake[{"a", "b", "c"}];
(Rest /@ bar)@<|"a" -> 1, "b" -> 1, "c" -> 1|>
<|"b" -> 1, "c" -> 1|>
answered Nov 20 at 21:48
Kuba♦
103k12200511
foo = {#a, #b, #c} &;
foo[[{1}, 2 ;;]]
{#b, #c} &
or, but only because we know there won't be any side effects:
Evaluate /@ Rest /@ foo
{#b, #c} &
or
foo /. {_Slot, rest__Slot} :> {rest}
At the end consider using an operator form of a KeyDrop
bar = KeyTake[{"a", "b", "c"}];
(Rest /@ bar)@<|"a" -> 1, "b" -> 1, "c" -> 1|>
<|"b" -> 1, "c" -> 1|>
answered Nov 20 at 21:48
Kuba♦
103k12200511
answered Nov 20 at 21:48
Kuba♦
103k12200511
answered Nov 20 at 21:48
Kuba♦
103k12200511
answered Nov 20 at 21:48
Kuba♦
103k12200511
103k12200511
I accepted Coolwater's answer because it solved my problem simply. I appreciate the extra insight that your answer provided. Actually, I'll have to spend more time to really understand it though.
– Mitchell Kaplan
Nov 20 at 21:51
@MitchellKaplan sure, btw, can you give a little insigth into how do you use it? Because it feels like a XY problem, maybe there is a better way anyway.
– Kuba♦
Nov 20 at 21:55
I use a lot of lists of associations and very often I have to sum things by one or more keys. Why I asked this question: I work in reinsurance. My list has, among other things: treaty, effective date, line of business, IBNR. I'm estimating IBNR by all 3, but it has to balance to the total IBNR for effective/line. I need to sum both by all 3, but also by only effective/line. I call a function that does this, and I want to adjust the keys I'm using to eliminate treaty when I'm summing over just effective/line. Left some detail out to fit answer into comment.
– Mitchell Kaplan
Nov 21 at 16:51
add a comment |
I accepted Coolwater's answer because it solved my problem simply. I appreciate the extra insight that your answer provided. Actually, I'll have to spend more time to really understand it though.
– Mitchell Kaplan
Nov 20 at 21:51
@MitchellKaplan sure, btw, can you give a little insigth into how do you use it? Because it feels like a XY problem, maybe there is a better way anyway.
– Kuba♦
Nov 20 at 21:55
I use a lot of lists of associations and very often I have to sum things by one or more keys. Why I asked this question: I work in reinsurance. My list has, among other things: treaty, effective date, line of business, IBNR. I'm estimating IBNR by all 3, but it has to balance to the total IBNR for effective/line. I need to sum both by all 3, but also by only effective/line. I call a function that does this, and I want to adjust the keys I'm using to eliminate treaty when I'm summing over just effective/line. Left some detail out to fit answer into comment.
– Mitchell Kaplan
Nov 21 at 16:51
I accepted Coolwater's answer because it solved my problem simply. I appreciate the extra insight that your answer provided. Actually, I'll have to spend more time to really understand it though.
– Mitchell Kaplan
Nov 20 at 21:51
I accepted Coolwater's answer because it solved my problem simply. I appreciate the extra insight that your answer provided. Actually, I'll have to spend more time to really understand it though.
– Mitchell Kaplan
Nov 20 at 21:51
@MitchellKaplan sure, btw, can you give a little insigth into how do you use it? Because it feels like a XY problem, maybe there is a better way anyway.
– Kuba♦
Nov 20 at 21:55
@MitchellKaplan sure, btw, can you give a little insigth into how do you use it? Because it feels like a XY problem, maybe there is a better way anyway.
– Kuba♦
Nov 20 at 21:55
I use a lot of lists of associations and very often I have to sum things by one or more keys. Why I asked this question: I work in reinsurance. My list has, among other things: treaty, effective date, line of business, IBNR. I'm estimating IBNR by all 3, but it has to balance to the total IBNR for effective/line. I need to sum both by all 3, but also by only effective/line. I call a function that does this, and I want to adjust the keys I'm using to eliminate treaty when I'm summing over just effective/line. Left some detail out to fit answer into comment.
– Mitchell Kaplan
Nov 21 at 16:51
I use a lot of lists of associations and very often I have to sum things by one or more keys. Why I asked this question: I work in reinsurance. My list has, among other things: treaty, effective date, line of business, IBNR. I'm estimating IBNR by all 3, but it has to balance to the total IBNR for effective/line. I need to sum both by all 3, but also by only effective/line. I call a function that does this, and I want to adjust the keys I'm using to eliminate treaty when I'm summing over just effective/line. Left some detail out to fit answer into comment.
– Mitchell Kaplan
Nov 21 at 16:51
add a comment |
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Instead of a blank you could replace
Slot["a"]->Nothing. You'll get thisNothingin the resulting function object, but once the function is applied,Nothingwill collapse.– Ruslan
Nov 21 at 6:51