Input: [1, 2, 3] [a, b]
Expected Output: [(1,a),(1,b),(2,a),(2,b),(3,a),(3,b)]
This works, but is there a better way without an if statement?
[(x,y) for (x,y) in list(combinations(chain(a,b), 2)) if x in a and y in b]
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7`[(x,y) for x in a for y in b]`? – Wondercricket Oct 24 '17 at 19:12
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You can use [itertools.product()](https://docs.python.org/3/library/itertools.html#itertools.product): `list(x for x in itertools.product([1, 2, 3], ['a', 'b']))` – John Szakmeister Oct 24 '17 at 19:14
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1@jszakmeister Or just `list(itertools.product([1, 2, 3], ['a', 'b']))`, ;) – user2390182 Oct 24 '17 at 19:16
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Thanks Wondercricket, that's what I was looking for. @schwobaseggl does product work for n lists? Like (a,b,c) for [a][b][c], (a,b,c,d) for [a][b][c][d], etc? – Kyle K Oct 24 '17 at 19:38
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@schwobaseggl haha, yes! I have no idea what I was thinking. :-) – John Szakmeister Oct 24 '17 at 19:40
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1Sigh. I don't think the referenced question is actually similar enough to this one. :-( `zip()` is the wrong solution here and `map()` isn't necessarily a better solution either. – John Szakmeister Oct 24 '17 at 21:51
1 Answers
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Use itertools.product, your handy library tool for a cartesian product:
from itertools import product
l1, l2 = [1, 2, 3], ['a', 'b']
output = list(product(l1, l2))
# [(1, 'a'), (1, 'b'), (2, 'a'), (2, 'b'), (3, 'a'), (3, 'b')]
user2390182
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