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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