侯体宗的博客
  • 首页
  • 人生(杂谈)
  • 技术
  • 关于我
  • 更多分类
    • 文件下载
    • 文字修仙
    • 中国象棋ai
    • 群聊
    • 九宫格抽奖
    • 拼图
    • 消消乐
    • 相册

Python编程之黑板上排列组合,你舍得解开吗

Python  /  管理员 发布于 7年前   166

考虑这样一个问题,给定一个矩阵(多维数组,numpy.ndarray()),如何shuffle这个矩阵(也就是对其行进行全排列),如何随机地选择其中的k行,这叫组合,实现一种某一维度空间的切片。例如五列中选三列(全部三列的排列数),便从原有的五维空间中降维到三维空间,因为是全部的排列数,故不会漏掉任何一种可能性。

涉及的函数主要有:

np.random.permutation()
itertools.combinations()
itertools.permutations()

# 1. 对0-5之间的数进行一次全排列>>>np.random.permutation(6)array([3, 1, 5, 4, 0, 2])# 2. 创建待排矩阵>>>A = np.array([[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]])# 3. shuffle矩阵A>>>p = np.random.permutation(A.shape[0])>>>parray([1, 2, 0])>>>A[p, :]     array([[ 5, 6, 7, 8],  [ 9, 10, 11, 12],  [ 1, 2, 3, 4]])

C52的实现

>>>from itertools import combinations>>>combins = [c for c in combinations(range(5), 2)]>>>len(combins)10>>>combins    # 而且是按序排列[(0, 1), (0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)]

A52的实现

>>>from itertools import permutations>>>pertumations(range(5), 2)<itertools.permutations object at 0x0233E360>>>>perms = permutations(range(5), 2)>>>perms[(0, 1), (0, 2), (0, 3), (0, 4), (1, 0), (1, 2), (1, 3), (1, 4), (2, 0), (2, 1), (2, 3), (2, 4), (3, 0), (3, 1), (3, 2), (3, 4), (4, 0), (4, 1), (4, 2), (4, 3)]>>>len(perms)20
# 5. 任取其中的k(k=2)行>>>c = [c for c in combinations(range(A.shape[0]), 2)]>>>A[c[0], :]   # 一种排列array([[1, 2, 3, 4],  [5, 6, 7, 8]])

下面再介绍一个列表数据任意组合,主要是利用自带的库

#_*_ coding:utf-8 _*_#__author__='dragon'import itertoolslist1 = [1,2,3,4,5]list2 = []for i in range(1,len(list1)+1): iter = itertools.combinations(list1,i) list2.append(list(iter))print(list2)
[[(1,), (2,), (3,), (4,), (5,)], [(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)], [(1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 3, 4), (1, 3, 5), (1, 4, 5), (2, 3, 4), (2, 3, 5), (2, 4, 5), (3, 4, 5)], [(1, 2, 3, 4), (1, 2, 3, 5), (1, 2, 4, 5), (1, 3, 4, 5), (2, 3, 4, 5)], [(1, 2, 3, 4, 5)]]

排列的实现

#_*_ coding:utf-8 _*_#__author__='dragon'import itertoolslist1 = [1,2,3,4,5]list2 = []for i in range(1,len(list1)+1): iter = itertools.permutations(list1,i) list2.append(list(iter))print(list2)

运行结果:

[[(1,), (2,), (3,), (4,), (5,)], [(1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4)], [(1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 3, 2), (1, 3, 4), (1, 3, 5), (1, 4, 2), (1, 4, 3), (1, 4, 5), (1, 5, 2), (1, 5, 3), (1, 5, 4), (2, 1, 3), (2, 1, 4), (2, 1, 5), (2, 3, 1), (2, 3, 4), (2, 3, 5), (2, 4, 1), (2, 4, 3), (2, 4, 5), (2, 5, 1), (2, 5, 3), (2, 5, 4), (3, 1, 2), (3, 1, 4), (3, 1, 5), (3, 2, 1), (3, 2, 4), (3, 2, 5), (3, 4, 1), (3, 4, 2), (3, 4, 5), (3, 5, 1), (3, 5, 2), (3, 5, 4), (4, 1, 2), (4, 1, 3), (4, 1, 5), (4, 2, 1), (4, 2, 3), (4, 2, 5), (4, 3, 1), (4, 3, 2), (4, 3, 5), (4, 5, 1), (4, 5, 2), (4, 5, 3), (5, 1, 2), (5, 1, 3), (5, 1, 4), (5, 2, 1), (5, 2, 3), (5, 2, 4), (5, 3, 1), (5, 3, 2), (5, 3, 4), (5, 4, 1), (5, 4, 2), (5, 4, 3)], [(1, 2, 3, 4), (1, 2, 3, 5), (1, 2, 4, 3), (1, 2, 4, 5), (1, 2, 5, 3), (1, 2, 5, 4), (1, 3, 2, 4), (1, 3, 2, 5), (1, 3, 4, 2), (1, 3, 4, 5), (1, 3, 5, 2), (1, 3, 5, 4), (1, 4, 2, 3), (1, 4, 2, 5), (1, 4, 3, 2), (1, 4, 3, 5), (1, 4, 5, 2), (1, 4, 5, 3), (1, 5, 2, 3), (1, 5, 2, 4), (1, 5, 3, 2), (1, 5, 3, 4), (1, 5, 4, 2), (1, 5, 4, 3), (2, 1, 3, 4), (2, 1, 3, 5), (2, 1, 4, 3), (2, 1, 4, 5), (2, 1, 5, 3), (2, 1, 5, 4), (2, 3, 1, 4), (2, 3, 1, 5), (2, 3, 4, 1), (2, 3, 4, 5), (2, 3, 5, 1), (2, 3, 5, 4), (2, 4, 1, 3), (2, 4, 1, 5), (2, 4, 3, 1), (2, 4, 3, 5), (2, 4, 5, 1), (2, 4, 5, 3), (2, 5, 1, 3), (2, 5, 1, 4), (2, 5, 3, 1), (2, 5, 3, 4), (2, 5, 4, 1), (2, 5, 4, 3), (3, 1, 2, 4), (3, 1, 2, 5), (3, 1, 4, 2), (3, 1, 4, 5), (3, 1, 5, 2), (3, 1, 5, 4), (3, 2, 1, 4), (3, 2, 1, 5), (3, 2, 4, 1), (3, 2, 4, 5), (3, 2, 5, 1), (3, 2, 5, 4), (3, 4, 1, 2), (3, 4, 1, 5), (3, 4, 2, 1), (3, 4, 2, 5), (3, 4, 5, 1), (3, 4, 5, 2), (3, 5, 1, 2), (3, 5, 1, 4), (3, 5, 2, 1), (3, 5, 2, 4), (3, 5, 4, 1), (3, 5, 4, 2), (4, 1, 2, 3), (4, 1, 2, 5), (4, 1, 3, 2), (4, 1, 3, 5), (4, 1, 5, 2), (4, 1, 5, 3), (4, 2, 1, 3), (4, 2, 1, 5), (4, 2, 3, 1), (4, 2, 3, 5), (4, 2, 5, 1), (4, 2, 5, 3), (4, 3, 1, 2), (4, 3, 1, 5), (4, 3, 2, 1), (4, 3, 2, 5), (4, 3, 5, 1), (4, 3, 5, 2), (4, 5, 1, 2), (4, 5, 1, 3), (4, 5, 2, 1), (4, 5, 2, 3), (4, 5, 3, 1), (4, 5, 3, 2), (5, 1, 2, 3), (5, 1, 2, 4), (5, 1, 3, 2), (5, 1, 3, 4), (5, 1, 4, 2), (5, 1, 4, 3), (5, 2, 1, 3), (5, 2, 1, 4), (5, 2, 3, 1), (5, 2, 3, 4), (5, 2, 4, 1), (5, 2, 4, 3), (5, 3, 1, 2), (5, 3, 1, 4), (5, 3, 2, 1), (5, 3, 2, 4), (5, 3, 4, 1), (5, 3, 4, 2), (5, 4, 1, 2), (5, 4, 1, 3), (5, 4, 2, 1), (5, 4, 2, 3), (5, 4, 3, 1), (5, 4, 3, 2)], [(1, 2, 3, 4, 5), (1, 2, 3, 5, 4), (1, 2, 4, 3, 5), (1, 2, 4, 5, 3), (1, 2, 5, 3, 4), (1, 2, 5, 4, 3), (1, 3, 2, 4, 5), (1, 3, 2, 5, 4), (1, 3, 4, 2, 5), (1, 3, 4, 5, 2), (1, 3, 5, 2, 4), (1, 3, 5, 4, 2), (1, 4, 2, 3, 5), (1, 4, 2, 5, 3), (1, 4, 3, 2, 5), (1, 4, 3, 5, 2), (1, 4, 5, 2, 3), (1, 4, 5, 3, 2), (1, 5, 2, 3, 4), (1, 5, 2, 4, 3), (1, 5, 3, 2, 4), (1, 5, 3, 4, 2), (1, 5, 4, 2, 3), (1, 5, 4, 3, 2), (2, 1, 3, 4, 5), (2, 1, 3, 5, 4), (2, 1, 4, 3, 5), (2, 1, 4, 5, 3), (2, 1, 5, 3, 4), (2, 1, 5, 4, 3), (2, 3, 1, 4, 5), (2, 3, 1, 5, 4), (2, 3, 4, 1, 5), (2, 3, 4, 5, 1), (2, 3, 5, 1, 4), (2, 3, 5, 4, 1), (2, 4, 1, 3, 5), (2, 4, 1, 5, 3), (2, 4, 3, 1, 5), (2, 4, 3, 5, 1), (2, 4, 5, 1, 3), (2, 4, 5, 3, 1), (2, 5, 1, 3, 4), (2, 5, 1, 4, 3), (2, 5, 3, 1, 4), (2, 5, 3, 4, 1), (2, 5, 4, 1, 3), (2, 5, 4, 3, 1), (3, 1, 2, 4, 5), (3, 1, 2, 5, 4), (3, 1, 4, 2, 5), (3, 1, 4, 5, 2), (3, 1, 5, 2, 4), (3, 1, 5, 4, 2), (3, 2, 1, 4, 5), (3, 2, 1, 5, 4), (3, 2, 4, 1, 5), (3, 2, 4, 5, 1), (3, 2, 5, 1, 4), (3, 2, 5, 4, 1), (3, 4, 1, 2, 5), (3, 4, 1, 5, 2), (3, 4, 2, 1, 5), (3, 4, 2, 5, 1), (3, 4, 5, 1, 2), (3, 4, 5, 2, 1), (3, 5, 1, 2, 4), (3, 5, 1, 4, 2), (3, 5, 2, 1, 4), (3, 5, 2, 4, 1), (3, 5, 4, 1, 2), (3, 5, 4, 2, 1), (4, 1, 2, 3, 5), (4, 1, 2, 5, 3), (4, 1, 3, 2, 5), (4, 1, 3, 5, 2), (4, 1, 5, 2, 3), (4, 1, 5, 3, 2), (4, 2, 1, 3, 5), (4, 2, 1, 5, 3), (4, 2, 3, 1, 5), (4, 2, 3, 5, 1), (4, 2, 5, 1, 3), (4, 2, 5, 3, 1), (4, 3, 1, 2, 5), (4, 3, 1, 5, 2), (4, 3, 2, 1, 5), (4, 3, 2, 5, 1), (4, 3, 5, 1, 2), (4, 3, 5, 2, 1), (4, 5, 1, 2, 3), (4, 5, 1, 3, 2), (4, 5, 2, 1, 3), (4, 5, 2, 3, 1), (4, 5, 3, 1, 2), (4, 5, 3, 2, 1), (5, 1, 2, 3, 4), (5, 1, 2, 4, 3), (5, 1, 3, 2, 4), (5, 1, 3, 4, 2), (5, 1, 4, 2, 3), (5, 1, 4, 3, 2), (5, 2, 1, 3, 4), (5, 2, 1, 4, 3), (5, 2, 3, 1, 4), (5, 2, 3, 4, 1), (5, 2, 4, 1, 3), (5, 2, 4, 3, 1), (5, 3, 1, 2, 4), (5, 3, 1, 4, 2), (5, 3, 2, 1, 4), (5, 3, 2, 4, 1), (5, 3, 4, 1, 2), (5, 3, 4, 2, 1), (5, 4, 1, 2, 3), (5, 4, 1, 3, 2), (5, 4, 2, 1, 3), (5, 4, 2, 3, 1), (5, 4, 3, 1, 2), (5, 4, 3, 2, 1)]]

可以根据你需要随意组合

python实现排列组合公式C(m,n)求值

# -*- coding:utf-8 -*- # 用python实现排列组合C(n,m) = n!/m!*(n-m)! def get_value(n):  if n==1:   return n  else:   return n * get_value(n-1) def gen_last_value(n,m):   first = get_value(n)   print "n:%s  value:%s"%(n, first)   second = get_value(m)   print "n:%s  value:%s"%(m, second)   third = get_value((n-m))   print "n:%s  value:%s"%((n-m), third)   return first/(second * third)    if __name__ == "__main__":  # C(12,5)  rest = gen_last_value(5,3)  print "value:", rest 

运行结果:

n:5  value:120n:3  value:6n:2  value:2value: 10

总结

以上就是本文关于Python排列组合算法的全部内容,希望对大家有所帮助。感兴趣的朋友可以继续参阅本站:Python数据结构与算法之列表(链表,linked list)简单实现、Python算法之求n个节点不同二叉树个数等,有什么问题可以随时留言,小编会及时回复大家的。


  • 上一条:
    python中requests库session对象的妙用详解
    下一条:
    Python数据结构与算法之列表(链表,linked list)简单实现
  • 昵称:

    邮箱:

    0条评论 (评论内容有缓存机制,请悉知!)
    最新最热
    • 分类目录
    • 人生(杂谈)
    • 技术
    • linux
    • Java
    • php
    • 框架(架构)
    • 前端
    • ThinkPHP
    • 数据库
    • 微信(小程序)
    • Laravel
    • Redis
    • Docker
    • Go
    • swoole
    • Windows
    • Python
    • 苹果(mac/ios)
    • 相关文章
    • 在python语言中Flask框架的学习及简单功能示例(0个评论)
    • 在Python语言中实现GUI全屏倒计时代码示例(0个评论)
    • Python + zipfile库实现zip文件解压自动化脚本示例(0个评论)
    • python爬虫BeautifulSoup快速抓取网站图片(1个评论)
    • vscode 配置 python3开发环境的方法(0个评论)
    • 近期文章
    • 在windows10中升级go版本至1.24后LiteIDE的Ctrl+左击无法跳转问题解决方案(0个评论)
    • 智能合约Solidity学习CryptoZombie第四课:僵尸作战系统(0个评论)
    • 智能合约Solidity学习CryptoZombie第三课:组建僵尸军队(高级Solidity理论)(0个评论)
    • 智能合约Solidity学习CryptoZombie第二课:让你的僵尸猎食(0个评论)
    • 智能合约Solidity学习CryptoZombie第一课:生成一只你的僵尸(0个评论)
    • 在go中实现一个常用的先进先出的缓存淘汰算法示例代码(0个评论)
    • 在go+gin中使用"github.com/skip2/go-qrcode"实现url转二维码功能(0个评论)
    • 在go语言中使用api.geonames.org接口实现根据国际邮政编码获取地址信息功能(1个评论)
    • 在go语言中使用github.com/signintech/gopdf实现生成pdf分页文件功能(95个评论)
    • gmail发邮件报错:534 5.7.9 Application-specific password required...解决方案(0个评论)
    • 近期评论
    • 122 在

      学历:一种延缓就业设计,生活需求下的权衡之选中评论 工作几年后,报名考研了,到现在还没认真学习备考,迷茫中。作为一名北漂互联网打工人..
    • 123 在

      Clash for Windows作者删库跑路了,github已404中评论 按理说只要你在国内,所有的流量进出都在监控范围内,不管你怎么隐藏也没用,想搞你分..
    • 原梓番博客 在

      在Laravel框架中使用模型Model分表最简单的方法中评论 好久好久都没看友情链接申请了,今天刚看,已经添加。..
    • 博主 在

      佛跳墙vpn软件不会用?上不了网?佛跳墙vpn常见问题以及解决办法中评论 @1111老铁这个不行了,可以看看近期评论的其他文章..
    • 1111 在

      佛跳墙vpn软件不会用?上不了网?佛跳墙vpn常见问题以及解决办法中评论 网站不能打开,博主百忙中能否发个APP下载链接,佛跳墙或极光..
    • 2016-10
    • 2016-11
    • 2018-04
    • 2020-03
    • 2020-04
    • 2020-05
    • 2020-06
    • 2022-01
    • 2023-07
    • 2023-10
    Top

    Copyright·© 2019 侯体宗版权所有· 粤ICP备20027696号 PHP交流群

    侯体宗的博客