Python全排列操作实例分析

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 -*-

#!python3

def 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 -*-

#!python3

def 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 itertools

t = 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 itertools

t = 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。

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