Get all combinations of a list in Python

Question

I would like to get all combinations of a list:

L = ["a","b","c"]
combinations(L,length=2)
# [("a","a"),("a","b"),("a","c"),("b","a"),("b","b"),("b","c"),("c","a"),("c","b"),("c","c")]

I've tried

itertools.combinations()

but this returned

[('a', 'b'), ('a', 'c'), ('b', 'c')]

When I use itertools.permutations(), it just returns the combinations with the length of the iteration, which is also not what I want.

Any librarys / function that I can use, without writing my own?

You mean `permutations` not `combinations`. Try that function instead. — michalwa, Jun 27 '20 at 13:26
Does this answer your question? [Get all combination with n -length from a list](https://stackoverflow.com/questions/24536725/get-all-combination-with-n-length-from-a-list) — ywbaek, Jun 27 '20 at 13:37

score 4 · Answer 1 · answered Jun 27 '20 at 13:30

4

You can use itertools.product with repeat=2 like so:

from itertools import product

L = ["a","b","c"]
print(list(product(L, repeat=2)))
#[('a', 'a'), ('a', 'b'), ('a', 'c'), ('b', 'a'), ('b', 'b'), ('b', 'c'), ('c', 'a'), ('c', 'b'), ('c', 'c')]

answered Jun 27 '20 at 13:30

Anwarvic

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This should be the accepted answer, using permutations to generate a product is a misuse of the itertools module. – Thierry Lathuille Jun 30 '20 at 13:24

score 3 · Answer 2 · answered Jun 27 '20 at 13:24

3

A simple list comprehesion can do the job too.

L = ["a","b","c"]
print([(a,b) for a in L for b in L])
#[('a', 'a'), ('a', 'b'), ('a', 'c'), ('b', 'a'), ('b', 'b'), ('b', 'c'), ('c', 'a'), ('c', 'b'), ('c', 'c')]

answered Jun 27 '20 at 13:24

Koralp Catalsakal

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Is there a way to put it in a function, so I can do `combinations(["a","b","c"], length=xyz)`? – CMinusMinus Jun 27 '20 at 13:30
1

As other answers suggest, I guess for variable length using `itertools.product` might be easier. – Koralp Catalsakal Jun 27 '20 at 13:33

taha · Answer 3 · 2020-06-27T13:37:30.293

2

itertools module has a function called product which is what you are looking for.

>>> L = ["a", "b", "c"]
>>> list(itertools.product(L, repeat=2))
[('a', 'a'), ('a', 'b'), ('a', 'c'), ('b', 'a'), ('b', 'b'), ('b', 'c'), ('c', 'a'), ('c', 'b'), ('c', 'c')]

edited Jun 27 '20 at 13:37

answered Jun 27 '20 at 13:31

taha

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score 2 · Answer 4 · answered Jun 27 '20 at 13:31

The product function from itertools offers a solution.

In [17]: from itertools import product

In [18]: L = ["a","b","c"]

In [19]: list(product(L, L))
Out[19]:
[('a', 'a'),
 ('a', 'b'),
 ('a', 'c'),
 ('b', 'a'),
 ('b', 'b'),
 ('b', 'c'),
 ('c', 'a'),
 ('c', 'b'),
 ('c', 'c')]

Red · Accepted Answer · 2020-06-27T13:38:01.103

You can use the second parameter of itertools.permutations():

from itertools import permutations

L = ["a","b","c"]

print([n for n in permutations(L,2)]+[(i,i) for i in L])

Output:

[('a', 'b'), ('a', 'c'), ('b', 'a'), ('b', 'c'), ('c', 'a'), ('c', 'b'), ('a', 'a'), ('b', 'b'), ('c', 'c')]

From the documentation:

itertools.permutations(iterable, r=None)

Return successive r length permutations of elements in the iterable. If r is not specified or is None, then r defaults to the length of the iterable and all possible full-length permutations are generated.

Get all combinations of a list in Python

5 Answers5