In the case of two sublists
`[[A,B],[1,2]]`
I would like to have the output
[[A,B],[1,2]
[A,B],[2,1]
[B,A],[1,2]
[B,A],[2,1]]
In the case of three sublists
`[[A,B],[1,2], [c, d]]`
I would like to have the output
`[[A,B],[1,2], [c, d],
[A,B],[1,2], [d, c],
[A,B],[2,1], [c, d],
[A,B],[2,1], [d, c],
[B,A],[1,2], [c, d],
[B,A],[1,2], [d, c],
[B,A],[2,1], [c, d]
[B,A],[2,1], [d, c]]`
Using itertools is the right way of doing this (it's part of the standard library so why not?). For example:
from itertools import permutations, product
values = [['A', 'B'], [1, 2]]
sub_permutations = ([list(p) for p in permutations(l)] for l in values)
values_perms = [list(v) for v in product(*sub_permutations)]
However, it is of course possible to implement a permutation function and a product function:
def permutations(values):
if not values:
return
for i in range(len(values)):
for p in all_perms(values[1:]):
yield [*p[:i], values[0], *p[i:]]
def product(lists):
result = [[]]
for pool in lists:
result = [r + [p] for r in result for p in pool]
return result
Which we can then use to achieve the same goal:
values = [['A', 'B'], [1, 2]]
values_perms = product(list(permutations(l)) for l in values)
But this will be slower, prone to bugs and it means more code to maintain. The only interesting thing here I think is to implement it yourself and see what you end up with, then simply use itertools.
