包括嵌套列表的排列

Question

I have the following function to get the number of permutations (without repeating elements) of a list:

import itertools

def permutations_without_repetition(samples, size):
    return len(list(itertools.permutations(samples, size)))

Which works fine for me as long as the list I provide doesn't contain nested lists. Itertools treats nested elements just as one whole element, and doesn't bother also generating permutations containing differently-ordered nested lists.

If I run permutations_without_repetition([[1, 2], 3], 2) , the only results I get are:

[([1, 2], 3), (3, [1, 2])]

I know this is the expected behaviour, but I would like the result to be:

[([1, 2], 3), (3, [1, 2]), ([2, 1], 3), (3, [2, 1])]

What's the easiest way to also return permutations that contain permutations of nested lists, to produce the result above?

Answer 1

You can use a recursive generator function:

def combos(d, c = []):
   if not isinstance(d, list):
      yield d
   elif not d:
      yield c
   else:
      for i, a in enumerate(d):
         for k in combos(a, c = []):
             yield from combos(d[:i]+d[i+1:], c+[k])

print(list(combos([[1, 2], 3])))

Output:

[[[1, 2], 3], [[2, 1], 3], [3, [1, 2]], [3, [2, 1]]]

Shorter solution using itertools :

import itertools as it
def combos(d):
   if not isinstance(d, list):
      yield d
   else:
      for i in it.permutations(d):
         yield from map(list, it.product(*[combos(j) for j in i]))

print(list(combos([[1, 2], 3])))

Output:

[[[1, 2], 3], [[2, 1], 3], [3, [1, 2]], [3, [2, 1]]]