Python-如何生成大小大于列表元素个数的排列

Question

Sorry if that title made no sense. What I mean is this: I have a list [A, B, C] and I want all possible permutations of those elements that will fill a list of length 10.

[A, B, C] => [A, A, A, A, A, A, A, A, A, A]
          => [A, A, A, A, A, A, A, A, A, B]
          ...
          => [C, C, C, C, C, C, C, C, C, C]

I've been reading through the itertools documentation but the permutations function wouldn't work in this case unless the output list length was less than or equal to 3. Thanks!

Answer 1

You are producing the product of the values, so use itertools.product() with a repeat set:

from itertools import product

for combo in product(['A', 'B', 'C'], repeat=10):

Demo:

>>> from itertools import product
>>> products = product(['A', 'B', 'C'], repeat=10)
>>> next(products)
('A', 'A', 'A', 'A', 'A', 'A', 'A', 'A', 'A', 'A')
>>> next(products)
('A', 'A', 'A', 'A', 'A', 'A', 'A', 'A', 'A', 'B')
>>> next(products)
('A', 'A', 'A', 'A', 'A', 'A', 'A', 'A', 'A', 'C')
>>> from itertools import islice
>>> skip_to_end = islice(products, (3 ** 10) - 6, None)
>>> next(skip_to_end)
('C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'A')
>>> next(skip_to_end)
('C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'B')
>>> next(skip_to_end)
('C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C', 'C')