[python] 가능한 모든 방법으로 목록을 쌍으로 나누는 방법

내 상사는 아마도이 재미있는 문제에 약간의 시간을 보냈을 때 기뻐하지 않을 것입니다.하지만 여기에 재귀가 필요하지 않고 itertools.product. 독 스트링에 설명되어 있습니다. :). 결과는 괜찮아 보이지만 너무 많이 테스트하지는 않았습니다.

import itertools


def all_pairs(lst):
    """Generate all sets of unique pairs from a list `lst`.

    This is equivalent to all _partitions_ of `lst` (considered as an indexed
    set) which have 2 elements in each partition.

    Recall how we compute the total number of such partitions. Starting with
    a list

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

    one takes off the first element, and chooses its pair [from any of the
    remaining 5].  For example, we might choose our first pair to be (1, 4).
    Then, we take off the next element, 2, and choose which element it is
    paired to (say, 3). So, there are 5 * 3 * 1 = 15 such partitions.

    That sounds like a lot of nested loops (i.e. recursion), because 1 could
    pick 2, in which case our next element is 3. But, if one abstracts "what
    the next element is", and instead just thinks of what index it is in the
    remaining list, our choices are static and can be aided by the
    itertools.product() function.
    """
    N = len(lst)
    choice_indices = itertools.product(*[
        xrange(k) for k in reversed(xrange(1, N, 2)) ])

    for choice in choice_indices:
        # calculate the list corresponding to the choices
        tmp = lst[:]
        result = []
        for index in choice:
            result.append( (tmp.pop(0), tmp.pop(index)) )
        yield result

건배!