All possible replacements of two lists?

Question

(I am aware that the title of the question might be misleading, but I could not find any other way to formulate it - feel free to edit it)

I have two lists, both of

Answer 1

Creating 3 lists of two elements would not over-complicate the code at all. zip can "flip the axes" of multiple lists trivially (making X sequences of Y elements into Y sequences of X elements), making it easy to use itertools.product:

import itertools

a = [1,2,3]
b = [4,5,6]

# Unpacking result of zip(a, b) means you automatically pass
# (1, 4), (2, 5), (3, 6)
# as the arguments to itertools.product
output = list(itertools.product(*zip(a, b)))

print(*output, sep="\n")

Which outputs:

(1, 2, 3)
(1, 2, 6)
(1, 5, 3)
(1, 5, 6)
(4, 2, 3)
(4, 2, 6)
(4, 5, 3)
(4, 5, 6)

Different ordering than your example output, but it's the same set of possible replacements.

Answer 2

Okay this is similar to the other answers, but taking a bit from both. You can model your problem as finding all possible bits of a sequence of given length, and replacing only when there is 1, and otherwise not.

from itertools import product

a = [1,2,3]
b = [4,5,6]

## All binary combinations of length of a (or b)
combinations = product([0,1], repeat=len(a))

for combination in combinations:
    y = []
    for l, i in zip(zip(a,b),combination):
         y.append(l[i])
    print y

All combinations of bits are:

(0, 0, 0)
(0, 0, 1)
(0, 1, 0)
(0, 1, 1)
(1, 0, 0)
(1, 0, 1)
(1, 1, 0)
(1, 1, 1)

Which results in:

[1, 2, 3]
[1, 2, 6]
[1, 5, 3]
[1, 5, 6]
[4, 2, 3]
[4, 2, 6]
[4, 5, 3]
[4, 5, 6]

Answer 3

Each item may independently be replaced or left alone. This can be modeled by a bit being 1 or 0. If you consider each item to be a separate bit, then iterating over all of the possibilities can be mapped to iterating over all of the combinations of n bits.

In other words, iterate from 0 to 2ⁿ-1 and look at the bit patterns.

n = len(a)
for i in range(2**n):
    yield [a[j] if i & (1 << j) != 0 else b[j] for j in range(n)]

Breaking this down, i & (1 << j) != 0 checks if the jth bit of i is set. If it is, use a[j], otherwise b[j].

Result:

[1, 2, 3]
[4, 2, 3]
[1, 5, 3]
[4, 5, 3]
[1, 2, 6]
[4, 2, 6]
[1, 5, 6]
[4, 5, 6]