如何解决Python-如何生成一个列表的所有排列？？

从Python 2.6（如果你使用的是Python 3）开始，你可以使用标准库工具：itertools.permutations。

import itertools
list(itertools.permutations([1, 2, 3]))

如果你出于某种原因使用旧版Python（<2.6），或者只是想知道它的工作原理，那么这是一种不错的方法，取自 http://code.activestate.com/recipes/252178/：

def all_perms(elements):
    if len(elements) <=1:
        yield elements
    else:
        for perm in all_perms(elements[1:]):
            for i in range(len(elements)):
                # nb elements[0:1] works in both string and list contexts
                yield perm[:i] + elements[0:1] + perm[i:]

的文档中列出了几种替代方法itertools.permutations。这是一个：

def permutations(iterable, r=None):
    # permutations(''ABCD'', 2) --> AB AC AD BA BC BD CA CB CD DA DB DC
    # permutations(range(3)) --> 012 021 102 120 201 210
    pool = tuple(iterable)
    n = len(pool)
    r = n if r is None else r
    if r > n:
        return
    indices = range(n)
    cycles = range(n, n-r, -1)
    yield tuple(pool[i] for i in indices[:r])
    while n:
        for i in reversed(range(r)):
            cycles[i] -= 1
            if cycles[i] == 0:
                indices[i:] = indices[i+1:] + indices[i:i+1]
                cycles[i] = n - i
            else:
                j = cycles[i]
                indices[i], indices[-j] = indices[-j], indices[i]
                yield tuple(pool[i] for i in indices[:r])
                break
        else:
            return

另一个基于itertools.product：

def permutations(iterable, r=None):
    pool = tuple(iterable)
    n = len(pool)
    r = n if r is None else r
    for indices in product(range(n), repeat=r):
        if len(set(indices)) == r:
            yield tuple(pool[i] for i in indices)

解决方法

permutations([])
[]

permutations([1])
[1]

permutations([1,2])
[1,2]
[2,1]

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

[[1,2,3],[4,5,6],[7,8,9,10]] -> [[1,4,7],[1,4,8],...,[3,6,10]]

>>> import itertools>>> a = [[1,2,3],[4,5,6],[7,8,9,10]]>>> list(itertools.product(*a))[(1, 4, 7), (1, 4, 8), (1, 4, 9), (1, 4, 10), (1, 5, 7), (1, 5, 8), (1, 5, 9), (1, 5, 10), (1, 6, 7), (1, 6, 8), (1, 6, 9), (1, 6, 10), (2, 4, 7), (2, 4, 8), (2, 4, 9), (2, 4, 10), (2, 5, 7), (2, 5, 8), (2, 5, 9), (2, 5, 10), (2, 6, 7), (2, 6, 8), (2, 6, 9), (2, 6, 10), (3, 4, 7), (3, 4, 8), (3, 4, 9), (3, 4, 10), (3, 5, 7), (3, 5, 8), (3, 5, 9), (3, 5, 10), (3, 6, 7), (3, 6, 8), (3, 6, 9), (3, 6, 10)]