How to get all mappings between two lists?

You can do this with itertools.product and zip

from itertools import productprint [zip(A, item) for item in product(B, repeat=len(A))]

Output

[[('a', 1), ('b', 1), ('c', 1)], [('a', 1), ('b', 1), ('c', 2)], [('a', 1), ('b', 2), ('c', 1)], [('a', 1), ('b', 2), ('c', 2)], [('a', 2), ('b', 1), ('c', 1)], [('a', 2), ('b', 1), ('c', 2)], [('a', 2), ('b', 2), ('c', 1)], [('a', 2), ('b', 2), ('c', 2)]]

product(B, repeat=len(A)) produces

[(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)]

Then we pick each element from the product and zip it with A, to get your desired output.

import itertools as itA = ['a','b','c']B = [1, 2]for i in it.product(*([B]*len(A))):    print(list(zip(A, i)))

outputs:

[('a', 1), ('b', 1), ('c', 1)][('a', 1), ('b', 1), ('c', 2)][('a', 1), ('b', 2), ('c', 1)][('a', 1), ('b', 2), ('c', 2)][('a', 2), ('b', 1), ('c', 1)][('a', 2), ('b', 1), ('c', 2)][('a', 2), ('b', 2), ('c', 1)][('a', 2), ('b', 2), ('c', 2)]

Not sure if it's very pythonic, it is if you look at it.product(*([B]*len(A))), because it uses multiple python-specific language features. But it's actually too cryptic to be pythonic... B is repeated n-times based on length of A and unpacked to the product-function.