排列组合,python排列组合

作者: 韦德国际1946手机版  发布:2019-09-03

python 排列组合,python

 

#coding:utf-8
__author__ = 'similarface'
'''
序列的排列组合
'''
def permute(list):
    '''
    序列的排列数: abc=abc,acb,bac,aca,cab,cba
    :param list:
    :return:
    '''
    #接受任何序列
    if not list:
        #返回空序列
        return [list]
    else:
        res=[]
        for i in range(len(list)):
            #删除当前节点
            rest=list[:i] list[i 1:]
            #排列其他的节点
            for x in permute(rest):
                #把当前节点添加到前面
                res.append(list[i:i 1] x)
        return res

def subset(list, size):
    '''
    子排列
    :param list:
    :param size:
    :return:
    '''
    if size == 0 or not list:
        return [list[:0]]
    else:
        result=[]
        for i in range(len(list)):
            pick = list[i:i 1]
            rest = list[:i]   list[i 1:]
            for x in subset(rest, size-1):
                result.append(pick   x)
        return result

def combo(list, size):
    '''
    组合数
    :param list:
    :param size:
    :return:
    '''
    if size == 0 or not list:
        return [list[:0]]
    else:
        result = []
        for i in range(0, (len(list) - size)   1):
            #重i出截断
            pick = list[i:i 1]
            rest = list[i 1:]
            #截断位置后的list继续组合
            for x in combo(rest, size - 1):
                result.append(pick   x)
        return result

if __name__=="__main__":
    print(permute([1,2,3]))
    print(permute('abc'))
    print(combo('abc', 2))
    print(subset([1, 2, 3], 2))
    '''
    [[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]
    ['abc', 'acb', 'bac', 'bca', 'cab', 'cba']
    ['ab', 'ac', 'bc']
    [[1, 2], [1, 3], [2, 1], [2, 3], [3, 1], [3, 2]]
    '''

 

排列组合,python # coding:utf-8 __author__ = ' similarface ' ''' 序列的排列组合 ''' def permute(list): ''' 系列的排列数: abc=abc,acb,bac,aca,ca...

 

python 2.6 引进了itertools模块,使得排列组合的贯彻非常轻易:

排列组合,python排列组合。Python编制程序之黑板上排列组合,你舍得解开吗,python排列组合

虚构这么二个主题材料,给定贰个矩阵(多维数组,numpy.ndarray()),怎么着shuffle这些矩阵(也等于对其行开展全排列),怎么样随机地挑选在那之中的k行,那叫组合,完结一种某一维度空间的切块。例如五列中选三列(全部三列的排列数),便从原始的五维空间中降维到三维空间,因为是整个的排列数,故不会孤陋寡闻任何一种或者。

关联的函数首要有:

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])
>>>p
array([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]])

上面再介绍三个列表数据任意组合,主纵然接纳自带的库

#_*_ coding:utf-8 _*_
#__author__='dragon'
import itertools
list1 = [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)]]

排列的落到实处

#_*_ coding:utf-8 _*_
#__author__='dragon'
import itertools
list1 = [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)

运作结果:

[[(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)]]

能够凭仗你供给自由组合

python完结排列组合公式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 

运维结果:

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

总结

如上就是本文关于Python排列组合算法的全体内容,希望对大家有着协助。感兴趣的朋友能够持续参照本站:Python数据结构与算法之列表(链表,linked list)轻松完成、Python算法之求n个节点差别二叉树个数等,有何难点得以每天留言,笔者会及时回复大家的。

思量那样二个难点,给定一个矩阵(多维数组,numpy.ndarray()),怎样shuffle那个...

#coding:utf-8
__author__ = 'similarface'
'''
序列的排列组合
'''
def permute(list):
    '''
    序列的排列数: abc=abc,acb,bac,aca,cab,cba
    :param list:
    :return:
    '''
    #接受任何序列
    if not list:
        #返回空序列
        return [list]
    else:
        res=[]
        for i in range(len(list)):
            #删除当前节点
            rest=list[:i] list[i 1:]
            #排列其他的节点
            for x in permute(rest):
                #把当前节点添加到前面
                res.append(list[i:i 1] x)
        return res

def subset(list, size):
    '''
    子排列
    :param list:
    :param size:
    :return:
    '''
    if size == 0 or not list:
        return [list[:0]]
    else:
        result=[]
        for i in range(len(list)):
            pick = list[i:i 1]
            rest = list[:i]   list[i 1:]
            for x in subset(rest, size-1):
                result.append(pick   x)
        return result

def combo(list, size):
    '''
    组合数
    :param list:
    :param size:
    :return:
    '''
    if size == 0 or not list:
        return [list[:0]]
    else:
        result = []
        for i in range(0, (len(list) - size)   1):
            #重i出截断
            pick = list[i:i 1]
            rest = list[i 1:]
            #截断位置后的list继续组合
            for x in combo(rest, size - 1):
                result.append(pick   x)
        return result

if __name__=="__main__":
    print(permute([1,2,3]))
    print(permute('abc'))
    print(combo('abc', 2))
    print(subset([1, 2, 3], 2))
    '''
    [[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]
    ['abc', 'acb', 'bac', 'bca', 'cab', 'cba']
    ['ab', 'ac', 'bc']
    [[1, 2], [1, 3], [2, 1], [2, 3], [3, 1], [3, 2]]
    '''

复制代码 代码如下:

 

import itertools 

本文由韦德国际1946发布于韦德国际1946手机版,转载请注明出处:排列组合,python排列组合

关键词: python 数据结构

上一篇:在大剧院,剧院惊魂第二天
下一篇:没有了