Basis enumeration of large lists
I want to combine two list in the following way without using Table or any loop structure.
genList[n_] := Table[RandomInteger[{1, 10}, 4], n] (* list generating function*)
The two lists that needs to be combined are:-
list1 = genList[3]
list2 = genList[4]
What I want to achieve is as follows:-
Partition[
Flatten[
Table[{list1[[i]], list2[[j]]}, {i, 1
Length[list1]},{j,1 Length[list2]}]
]
,8]
So I simply need to enumerate each of the elements of list 1 combined with all of the elements of list 2.
What would be an efficient way of doing this with large lists?
Let's say with Length[list1] = 100 and Length[list2] = 200.
Also Length[list2] > Length[list1]
list-manipulation
edited Dec 3 '18 at 8:01
Henrik Schumacher
48.8k467139
asked Dec 3 '18 at 7:20
Hubble07
2,887720
add a comment |
I want to combine two list in the following way without using Table or any loop structure.
genList[n_] := Table[RandomInteger[{1, 10}, 4], n] (* list generating function*)
The two lists that needs to be combined are:-
list1 = genList[3]
list2 = genList[4]
What I want to achieve is as follows:-
Partition[
Flatten[
Table[{list1[[i]], list2[[j]]}, {i, 1
Length[list1]},{j,1 Length[list2]}]
]
,8]
So I simply need to enumerate each of the elements of list 1 combined with all of the elements of list 2.
What would be an efficient way of doing this with large lists?
Let's say with Length[list1] = 100 and Length[list2] = 200.
Also Length[list2] > Length[list1]
list-manipulation
edited Dec 3 '18 at 8:01
Henrik Schumacher
48.8k467139
asked Dec 3 '18 at 7:20
Hubble07
2,887720
add a comment |
I want to combine two list in the following way without using Table or any loop structure.
genList[n_] := Table[RandomInteger[{1, 10}, 4], n] (* list generating function*)
The two lists that needs to be combined are:-
list1 = genList[3]
list2 = genList[4]
What I want to achieve is as follows:-
Partition[
Flatten[
Table[{list1[[i]], list2[[j]]}, {i, 1
Length[list1]},{j,1 Length[list2]}]
]
,8]
So I simply need to enumerate each of the elements of list 1 combined with all of the elements of list 2.
What would be an efficient way of doing this with large lists?
Let's say with Length[list1] = 100 and Length[list2] = 200.
Also Length[list2] > Length[list1]
list-manipulation
edited Dec 3 '18 at 8:01
Henrik Schumacher
48.8k467139
asked Dec 3 '18 at 7:20
Hubble07
2,887720
I want to combine two list in the following way without using Table or any loop structure.
genList[n_] := Table[RandomInteger[{1, 10}, 4], n] (* list generating function*)
The two lists that needs to be combined are:-
list1 = genList[3]
list2 = genList[4]
What I want to achieve is as follows:-
Partition[
Flatten[
Table[{list1[[i]], list2[[j]]}, {i, 1
Length[list1]},{j,1 Length[list2]}]
]
,8]
So I simply need to enumerate each of the elements of list 1 combined with all of the elements of list 2.
What would be an efficient way of doing this with large lists?
Let's say with Length[list1] = 100 and Length[list2] = 200.
Also Length[list2] > Length[list1]
list-manipulation
list-manipulation
edited Dec 3 '18 at 8:01
Henrik Schumacher
48.8k467139
asked Dec 3 '18 at 7:20
Hubble07
2,887720
edited Dec 3 '18 at 8:01
Henrik Schumacher
48.8k467139
asked Dec 3 '18 at 7:20
Hubble07
2,887720
edited Dec 3 '18 at 8:01
Henrik Schumacher
48.8k467139
edited Dec 3 '18 at 8:01
Henrik Schumacher
48.8k467139
edited Dec 3 '18 at 8:01
Henrik Schumacher
48.8k467139
48.8k467139
asked Dec 3 '18 at 7:20
Hubble07
2,887720
asked Dec 3 '18 at 7:20
Hubble07
2,887720
asked Dec 3 '18 at 7:20
Hubble07
2,887720
2,887720
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Join @@@ Tuples[{list1, list2}] (* or *)
Flatten /@ Tuples[{list1, list2}] (* or *)
Join @@ Outer[Join, list1, list2, 1]
{{5, 3, 5, 4, 9, 2, 10, 6}, {5, 3, 5, 4, 10, 6, 4, 7},
{5, 3, 5, 4, 2,
1, 7, 1}, {5, 3, 5, 4, 6, 6, 1, 9},
{2, 6, 9, 5, 9, 2, 10, 6}, {2,
6, 9, 5, 10, 6, 4, 7},
{2, 6, 9, 5, 2, 1, 7, 1}, {2, 6, 9, 5, 6, 6,
1, 9}, {10, 8, 9, 6, 9, 2, 10, 6},
{10, 8, 9, 6, 10, 6, 4, 7}, {10,
8, 9, 6, 2, 1, 7, 1}, {10, 8, 9, 6, 6, 6, 1, 9}}
answered Dec 3 '18 at 7:45
kglr
177k9198405
add a comment |
This is a variant of kglr's approach that employs ArrayFlatten instead of mapping Flatten. Notice also that I changed genList so that it produces packed arrays; this is crucial for performance.
genList[n_] := RandomInteger[{1, 10}, {n, 4}];
m = 100;
n = 100;
list1 = genList[m];
list2 = genList[n];
a = Partition[Flatten[Table[{list1[[i]], list2[[j]]}, {i, 1 Length[list1]}, {j, 1 Length[list2]}]], 8]; // RepeatedTiming // First
b = Flatten /@ Tuples[{list1, list2}]; // RepeatedTiming // First
c = ArrayReshape[
Tuples[{list1, list2}],
{Length[list1] Length[list2], Dimensions[list1][[2]] + Dimensions[list2][[2]]}
]; // RepeatedTiming // First
a == b == c
0.026
0.0012
0.000095
True
answered Dec 3 '18 at 7:59
Henrik Schumacher
48.8k467139
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Join @@@ Tuples[{list1, list2}] (* or *)
Flatten /@ Tuples[{list1, list2}] (* or *)
Join @@ Outer[Join, list1, list2, 1]
{{5, 3, 5, 4, 9, 2, 10, 6}, {5, 3, 5, 4, 10, 6, 4, 7},
{5, 3, 5, 4, 2,
1, 7, 1}, {5, 3, 5, 4, 6, 6, 1, 9},
{2, 6, 9, 5, 9, 2, 10, 6}, {2,
6, 9, 5, 10, 6, 4, 7},
{2, 6, 9, 5, 2, 1, 7, 1}, {2, 6, 9, 5, 6, 6,
1, 9}, {10, 8, 9, 6, 9, 2, 10, 6},
{10, 8, 9, 6, 10, 6, 4, 7}, {10,
8, 9, 6, 2, 1, 7, 1}, {10, 8, 9, 6, 6, 6, 1, 9}}
answered Dec 3 '18 at 7:45
kglr
177k9198405
add a comment |
Join @@@ Tuples[{list1, list2}] (* or *)
Flatten /@ Tuples[{list1, list2}] (* or *)
Join @@ Outer[Join, list1, list2, 1]
{{5, 3, 5, 4, 9, 2, 10, 6}, {5, 3, 5, 4, 10, 6, 4, 7},
{5, 3, 5, 4, 2,
1, 7, 1}, {5, 3, 5, 4, 6, 6, 1, 9},
{2, 6, 9, 5, 9, 2, 10, 6}, {2,
6, 9, 5, 10, 6, 4, 7},
{2, 6, 9, 5, 2, 1, 7, 1}, {2, 6, 9, 5, 6, 6,
1, 9}, {10, 8, 9, 6, 9, 2, 10, 6},
{10, 8, 9, 6, 10, 6, 4, 7}, {10,
8, 9, 6, 2, 1, 7, 1}, {10, 8, 9, 6, 6, 6, 1, 9}}
answered Dec 3 '18 at 7:45
kglr
177k9198405
add a comment |
Join @@@ Tuples[{list1, list2}] (* or *)
Flatten /@ Tuples[{list1, list2}] (* or *)
Join @@ Outer[Join, list1, list2, 1]
{{5, 3, 5, 4, 9, 2, 10, 6}, {5, 3, 5, 4, 10, 6, 4, 7},
{5, 3, 5, 4, 2,
1, 7, 1}, {5, 3, 5, 4, 6, 6, 1, 9},
{2, 6, 9, 5, 9, 2, 10, 6}, {2,
6, 9, 5, 10, 6, 4, 7},
{2, 6, 9, 5, 2, 1, 7, 1}, {2, 6, 9, 5, 6, 6,
1, 9}, {10, 8, 9, 6, 9, 2, 10, 6},
{10, 8, 9, 6, 10, 6, 4, 7}, {10,
8, 9, 6, 2, 1, 7, 1}, {10, 8, 9, 6, 6, 6, 1, 9}}
answered Dec 3 '18 at 7:45
kglr
177k9198405
Join @@@ Tuples[{list1, list2}] (* or *)
Flatten /@ Tuples[{list1, list2}] (* or *)
Join @@ Outer[Join, list1, list2, 1]
{{5, 3, 5, 4, 9, 2, 10, 6}, {5, 3, 5, 4, 10, 6, 4, 7},
{5, 3, 5, 4, 2,
1, 7, 1}, {5, 3, 5, 4, 6, 6, 1, 9},
{2, 6, 9, 5, 9, 2, 10, 6}, {2,
6, 9, 5, 10, 6, 4, 7},
{2, 6, 9, 5, 2, 1, 7, 1}, {2, 6, 9, 5, 6, 6,
1, 9}, {10, 8, 9, 6, 9, 2, 10, 6},
{10, 8, 9, 6, 10, 6, 4, 7}, {10,
8, 9, 6, 2, 1, 7, 1}, {10, 8, 9, 6, 6, 6, 1, 9}}
answered Dec 3 '18 at 7:45
kglr
177k9198405
edited Dec 3 '18 at 7:54
answered Dec 3 '18 at 7:45
kglr
177k9198405
answered Dec 3 '18 at 7:45
kglr
177k9198405
answered Dec 3 '18 at 7:45
kglr
177k9198405
177k9198405
add a comment |
add a comment |
This is a variant of kglr's approach that employs ArrayFlatten instead of mapping Flatten. Notice also that I changed genList so that it produces packed arrays; this is crucial for performance.
genList[n_] := RandomInteger[{1, 10}, {n, 4}];
m = 100;
n = 100;
list1 = genList[m];
list2 = genList[n];
a = Partition[Flatten[Table[{list1[[i]], list2[[j]]}, {i, 1 Length[list1]}, {j, 1 Length[list2]}]], 8]; // RepeatedTiming // First
b = Flatten /@ Tuples[{list1, list2}]; // RepeatedTiming // First
c = ArrayReshape[
Tuples[{list1, list2}],
{Length[list1] Length[list2], Dimensions[list1][[2]] + Dimensions[list2][[2]]}
]; // RepeatedTiming // First
a == b == c
0.026
0.0012
0.000095
True
answered Dec 3 '18 at 7:59
Henrik Schumacher
48.8k467139
add a comment |
This is a variant of kglr's approach that employs ArrayFlatten instead of mapping Flatten. Notice also that I changed genList so that it produces packed arrays; this is crucial for performance.
genList[n_] := RandomInteger[{1, 10}, {n, 4}];
m = 100;
n = 100;
list1 = genList[m];
list2 = genList[n];
a = Partition[Flatten[Table[{list1[[i]], list2[[j]]}, {i, 1 Length[list1]}, {j, 1 Length[list2]}]], 8]; // RepeatedTiming // First
b = Flatten /@ Tuples[{list1, list2}]; // RepeatedTiming // First
c = ArrayReshape[
Tuples[{list1, list2}],
{Length[list1] Length[list2], Dimensions[list1][[2]] + Dimensions[list2][[2]]}
]; // RepeatedTiming // First
a == b == c
0.026
0.0012
0.000095
True
answered Dec 3 '18 at 7:59
Henrik Schumacher
48.8k467139
add a comment |
This is a variant of kglr's approach that employs ArrayFlatten instead of mapping Flatten. Notice also that I changed genList so that it produces packed arrays; this is crucial for performance.
genList[n_] := RandomInteger[{1, 10}, {n, 4}];
m = 100;
n = 100;
list1 = genList[m];
list2 = genList[n];
a = Partition[Flatten[Table[{list1[[i]], list2[[j]]}, {i, 1 Length[list1]}, {j, 1 Length[list2]}]], 8]; // RepeatedTiming // First
b = Flatten /@ Tuples[{list1, list2}]; // RepeatedTiming // First
c = ArrayReshape[
Tuples[{list1, list2}],
{Length[list1] Length[list2], Dimensions[list1][[2]] + Dimensions[list2][[2]]}
]; // RepeatedTiming // First
a == b == c
0.026
0.0012
0.000095
True
answered Dec 3 '18 at 7:59
Henrik Schumacher
48.8k467139
This is a variant of kglr's approach that employs ArrayFlatten instead of mapping Flatten. Notice also that I changed genList so that it produces packed arrays; this is crucial for performance.
genList[n_] := RandomInteger[{1, 10}, {n, 4}];
m = 100;
n = 100;
list1 = genList[m];
list2 = genList[n];
a = Partition[Flatten[Table[{list1[[i]], list2[[j]]}, {i, 1 Length[list1]}, {j, 1 Length[list2]}]], 8]; // RepeatedTiming // First
b = Flatten /@ Tuples[{list1, list2}]; // RepeatedTiming // First
c = ArrayReshape[
Tuples[{list1, list2}],
{Length[list1] Length[list2], Dimensions[list1][[2]] + Dimensions[list2][[2]]}
]; // RepeatedTiming // First
a == b == c
0.026
0.0012
0.000095
True
answered Dec 3 '18 at 7:59
Henrik Schumacher
48.8k467139
edited Dec 3 '18 at 9:00
answered Dec 3 '18 at 7:59
Henrik Schumacher
48.8k467139
answered Dec 3 '18 at 7:59
Henrik Schumacher
48.8k467139
answered Dec 3 '18 at 7:59
Henrik Schumacher
48.8k467139
48.8k467139
add a comment |
add a comment |
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