递归之美：

>>> import copy
>>> def perm(prefix,rest):
...      for e in rest:
...              new_rest=copy.copy(rest)
...              new_prefix=copy.copy(prefix)
...              new_prefix.append(e)
...              new_rest.remove(e)
...              if len(new_rest) == 0:
...                      print new_prefix + new_rest
...                      continue
...              perm(new_prefix,new_rest)
... 
>>> perm([],['a','b','c','d'])
['a', 'b', 'c', 'd']
['a', 'b', 'd', 'c']
['a', 'c', 'b', 'd']
['a', 'c', 'd', 'b']
['a', 'd', 'b', 'c']
['a', 'd', 'c', 'b']
['b', 'a', 'c', 'd']
['b', 'a', 'd', 'c']
['b', 'c', 'a', 'd']
['b', 'c', 'd', 'a']
['b', 'd', 'a', 'c']
['b', 'd', 'c', 'a']
['c', 'a', 'b', 'd']
['c', 'a', 'd', 'b']
['c', 'b', 'a', 'd']
['c', 'b', 'd', 'a']
['c', 'd', 'a', 'b']
['c', 'd', 'b', 'a']
['d', 'a', 'b', 'c']
['d', 'a', 'c', 'b']
['d', 'b', 'a', 'c']
['d', 'b', 'c', 'a']
['d', 'c', 'a', 'b']
['d', 'c', 'b', 'a']

无论如何，我们可以使用sympy库，也支持多集合排列

import sympy
from sympy.utilities.iterables import multiset_permutations
t = [1,2,3]
p = list(multiset_permutations(t))
print(p)

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

答案的灵感来自获取numpy数组的所有排列

以下代码是给定列表的就地排列，作为生成器实现。由于它只返回对列表的引用，因此不应在生成器外部修改列表。该解决方案是非递归的，因此使用了低内存。还可以很好地处理输入列表中元素的多个副本。

def permute_in_place(a):
    a.sort()
    yield list(a)

    if len(a) <= 1:
        return

    first = 0
    last = len(a)
    while 1:
        i = last - 1

        while 1:
            i = i - 1
            if a[i] < a[i+1]:
                j = last - 1
                while not (a[i] < a[j]):
                    j = j - 1
                a[i], a[j] = a[j], a[i] # swap the values
                r = a[i+1:last]
                r.reverse()
                a[i+1:last] = r
                yield list(a)
                break
            if i == first:
                a.reverse()
                return

if __name__ == '__main__':
    for n in range(5):
        for a in permute_in_place(range(1, n+1)):
            print a
        print

    for a in permute_in_place([0, 0, 1, 1, 1]):
        print a
    print

使用标准库中的itertools.permutations：

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

从这里改编的是itertools.permutations如何实现的演示：

def permutations(elements):
    if len(elements) <= 1:
        yield elements
        return
    for perm in permutations(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:]

itertools.permutations文档中列出了两种替代方法

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)

对于Python 2.6及以上版本：

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

这将作为生成器返回。使用列表（排列（xs））作为列表返回。

此解决方案实现了一个生成器，以避免在内存中保留所有排列：

def permutations (orig_list):
    if not isinstance(orig_list, list):
        orig_list = list(orig_list)

    yield orig_list

    if len(orig_list) == 1:
        return

    for n in sorted(orig_list):
        new_list = orig_list[:]
        pos = new_list.index(n)
        del(new_list[pos])
        new_list.insert(0, n)
        for resto in permutations(new_list[1:]):
            if new_list[:1] + resto <> orig_list:
                yield new_list[:1] + resto