How to get all mappings between two lists?

Question

We have two lists, A and B:

A = ['a','b','c']
B = [1, 2]

Is there a pythonic way to build the set of all maps between A and B containing 2^n (here 2^3=8)? That is:

[(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)]

Using itertools.product, it's possible to get all the tuples:

import itertools as it
P = it.product(A, B)
[p for p in P]

Which gives:

Out[3]: [('a', 1), ('a', 2), ('b', 1), ('b', 2), ('c', 1), ('c', 2)]

Answer 1

You can do this with itertools.product and zip

from itertools import product
print [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.

Answer 2

import itertools as it

A = ['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.