本文实例讲述了Python全排列操作。分享给大家供大家参考，具体如下：

step 1: 列表的全排列:

这个版本比较low

# -*-coding:utf-8 -*-#!python3def permutation(li,index): for i in range(index,len(li)): if index == len(li)-1: print(li) return tmp = li[index] li[index] = li[i] li[i] = tmp permutation(li,index+1) tmp = li[index] li[index] = li[i] li[i] = tmp

调用：

permutation([1,2,3,4],0)

运行结果：

[1, 2, 3, 4]
[1, 2, 4, 3]
[1, 3, 2, 4]
[1, 3, 4, 2]
[1, 4, 3, 2]
[1, 4, 2, 3]
[2, 1, 3, 4]
[2, 1, 4, 3]
[2, 3, 1, 4]
[2, 3, 4, 1]
[2, 4, 3, 1]
[2, 4, 1, 3]
[3, 2, 1, 4]
[3, 2, 4, 1]
[3, 1, 2, 4]
[3, 1, 4, 2]
[3, 4, 1, 2]
[3, 4, 2, 1]
[4, 2, 3, 1]
[4, 2, 1, 3]
[4, 3, 2, 1]
[4, 3, 1, 2]
[4, 1, 3, 2]
[4, 1, 2, 3]

step2: 字符串的全排列:

# -*-coding:utf-8 -*-#!python3def permutation(str): li = list(str) cnt = 0 #记录全排列的总数 def permutation_list(index): if index == len(li) -1: nonlocal cnt cnt += 1 print(li) for i in range(index,len(li)): li[index],li[i] = li[i],li[index] permutation_list(index+1) li[index], li[i] = li[i], li[index] ret = permutation_list(0) print("共有%d中全排列" % cnt) return ret

备注:

在闭包中，内部函数依然维持了外部函数中自由变量的引用—单元。内部函数不能修改单元对象的值(但是可以引用)。若尝试修改，则解释器会认为它是局部变量。这类似于全局变量和局部变量的关系。如果在函数内部修改全局变量，必须加上global声明，但是对于自由变量，尚没有类似的机制。所以，只能使用列表。(python3中引入了关键字：nonlocal)

测试:

permutation('abcd')

运行结果：

['a', 'b', 'c', 'd']
['a', 'b', 'd', 'c']
['a', 'c', 'b', 'd']
['a', 'c', 'd', 'b']
['a', 'd', 'c', 'b']
['a', 'd', 'b', 'c']
['b', 'a', 'c', 'd']
['b', 'a', 'd', 'c']
['b', 'c', 'a', 'd']
['b', 'c', 'd', 'a']
['b', 'd', 'c', 'a']
['b', 'd', 'a', 'c']
['c', 'b', 'a', 'd']
['c', 'b', 'd', 'a']
['c', 'a', 'b', 'd']
['c', 'a', 'd', 'b']
['c', 'd', 'a', 'b']
['c', 'd', 'b', 'a']
['d', 'b', 'c', 'a']
['d', 'b', 'a', 'c']
['d', 'c', 'b', 'a']
['d', 'c', 'a', 'b']
['d', 'a', 'c', 'b']
['d', 'a', 'b', 'c']
共有24中全排列

step3 : 使用python标准库

import itertoolst = list(itertools.permutations([1,2,3,4]))print(t)

运行结果：

[(1, 2, 3, 4), (1, 2, 4, 3), (1, 3, 2, 4), (1, 3, 4, 2), (1, 4, 2, 3), (1, 4, 3, 2), (2, 1, 3, 4), (2, 1, 4, 3), (2, 3, 1, 4), (2, 3, 4, 1), (2, 4, 1, 3), (2, 4, 3, 1), (3, 1, 2, 4), (3, 1, 4, 2), (3, 2, 1, 4), (3, 2, 4, 1), (3, 4, 1, 2), (3, 4, 2, 1), (4, 1, 2, 3), (4, 1, 3, 2), (4, 2, 1, 3), (4, 2, 3, 1), (4, 3, 1, 2), (4, 3, 2, 1)]

可以指定排列的位数:

import itertoolst = itertools.permutations([1,2,3,4],3) #只排列3位print(list(t))

运行结果：

[(1, 2, 3), (1, 2, 4), (1, 3, 2), (1, 3, 4), (1, 4, 2), (1, 4, 3), (2, 1, 3), (2, 1, 4), (2, 3, 1), (2, 3, 4), (2, 4, 1), (2, 4, 3), (3, 1, 2), (3, 1, 4), (3, 2, 1), (3, 2, 4), (3, 4, 1), (3, 4, 2), (4, 1, 2), (4, 1, 3), (4, 2, 1), (4, 2, 3), (4, 3, 1), (4, 3, 2)]

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希望本文所述对大家Python程序设计有所帮助。