CODEWITHSUNDEEP
pythonpython-itertools
Bryant
Bryant

Reputation: 1

How do you return every possible combination of 2 lists in python?

I want to create a list of every possible combination between 2 lists of elements regardless of either list's size (they may or may not be identical lengths).

I've looked through itertools examples and searched stackoverflow but haven't found an exact example of what I'm looking for.

INPUT:

l1 = ['a', 'b', 'c']
l2 = [1, 2]

OUTPUT:

['a1-b1-c1', 'a1-b1-c2', 'a1-b2-c1', 'a1-b2-c2', 'a2-b1-c1', 'a2-b1-c2', 'a2-b2-c1', 'a2-b2-c2']

but again, l1 can be any size and l2 can be any size. Don't care at all about formatting, just want the complete output.

Thanks!

Upvotes: 0

Views: 209

Answers (4)

cs95
cs95

Reputation: 402483

You can use itertools.product to generate all possible 3-combinations of l2, and then join each combination with l1.

from itertools import product

combs = product(map(str, l2), repeat=3)
['-'.join([x + y for x, y in zip(l1, c)]) for c in combs]
# ['a1-b1-c1', 'a1-b1-c2', 'a1-b2-c1', 'a1-b2-c2', 'a2-b1-c1', 'a2-b1-c2', 'a2-b2-c1', 'a2-b2-c2']

Upvotes: 2

John La Rooy
John La Rooy

Reputation: 304147

This solution creates a template from l1: eg. 'a{}-b{}-c{}'

from itertools import product
l1 = ['a', 'b', 'c']
l2 = [1, 2]
template = "{}-".join(l1) + "{}"
[template.format(*c) for c in product(l2, repeat=len(l1))]

['a1-b1-c1', 'a1-b1-c2', 'a1-b2-c1', 'a1-b2-c2', 'a2-b1-c1', 'a2-b1-c2', 'a2-b2-c1', 'a2-b2-c2']

Upvotes: 1

blhsing
blhsing

Reputation: 106543

You can use itertools.product:

from itertools import product
['-'.join(map(''.join, zip(l1, c))) for c in product(map(str, l2), repeat=len(l1))]

This returns:

['a1-b1-c1', 'a1-b1-c2', 'a1-b2-c1', 'a1-b2-c2', 'a2-b1-c1', 'a2-b1-c2', 'a2-b2-c1', 'a2-b2-c2']

Upvotes: 0

Dillon Davis
Dillon Davis

Reputation: 7740

You can use itertools.product and list comprehensions to get the result you are after:

from itertools import product
result = ["-".join(y+str(z) for y, z in zip(l1, x)) for x in product(*[l2] * 3)]

I would like to note that approach does not produce any extra intermediate combinations.

Output:

['a1-b1-c1', 'a1-b1-c2', 'a1-b2-c1', 'a1-b2-c2', 'a2-b1-c1', 'a2-b1-c2', 'a2-b2-c1', 'a2-b2-c2']

Upvotes: 0

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