如何生成列表的所有排列?例如:

permutations([])
[]

permutations([1])
[1]

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

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

当前回答

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。

其他回答

生成所有可能的排列

我正在使用python3.4:

def calcperm(arr, size):
    result = set([()])
    for dummy_idx in range(size):
        temp = set()
        for dummy_lst in result:
            for dummy_outcome in arr:
                if dummy_outcome not in dummy_lst:
                    new_seq = list(dummy_lst)
                    new_seq.append(dummy_outcome)
                    temp.add(tuple(new_seq))
        result = temp
    return result

测试用例:

lst = [1, 2, 3, 4]
#lst = ["yellow", "magenta", "white", "blue"]
seq = 2
final = calcperm(lst, seq)
print(len(final))
print(final)

如果有人喜欢这个丑陋的单行线(虽然只适用于字符串):

def p(a):
    return a if len(a) == 1 else [[a[i], *j] for i in range(len(a)) for j in p(a[:i] + a[i + 1:])]
#!/usr/bin/env python

def perm(a, k=0):
   if k == len(a):
      print a
   else:
      for i in xrange(k, len(a)):
         a[k], a[i] = a[i] ,a[k]
         perm(a, k+1)
         a[k], a[i] = a[i], a[k]

perm([1,2,3])

输出:

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

当我交换列表的内容时,需要一个可变的序列类型作为输入。例如,烫发(list(“ball”)会起作用,而烫发(“ball”)不会起作用,因为你不能更改字符串。

这种Python实现的灵感来自Horowitz、Sahni和Rajasekeran在《计算机算法》一书中提出的算法。

在我看来,一个很明显的方式可能是:

def permutList(l):
    if not l:
            return [[]]
    res = []
    for e in l:
            temp = l[:]
            temp.remove(e)
            res.extend([[e] + r for r in permutList(temp)])

    return res
list2Perm = [1, 2.0, 'three']
listPerm = [[a, b, c]
            for a in list2Perm
            for b in list2Perm
            for c in list2Perm
            if ( a != b and b != c and a != c )
            ]
print listPerm

输出:

[
    [1, 2.0, 'three'], 
    [1, 'three', 2.0], 
    [2.0, 1, 'three'], 
    [2.0, 'three', 1], 
    ['three', 1, 2.0], 
    ['three', 2.0, 1]
]