python自带的排列组合函数

需求： 在你的面前有一个n阶的台阶，你一步只能上1级或者2级，请计算出你可以采用多少种不同的方法爬完这个楼梯？输入一个正整数表示这个台阶的级数，输出一个正整数表示有多少种方法爬完这个楼梯。

 

分析：提炼出题干的意思：用1和2产生不同组合，使得他们的和等于台阶的级数，输出有多少种组合方式。

 

解决： 主要的问题就是如何利用1和2产生不同的组合，查阅了python关于排列组合相关的资料

　　最后发现了一个强大的python库 itertools

介绍一下常用的几个函数：

　　itertools.product(sequence，repeat) 　 #从sequence中拿出repeat个数做排列（repeat关键字传参） 有放回的拿出  repeat与sequence的长度无关。

demo: 输出为类型为元组，

In [2]: import itertools

In [3]: for i in itertools.product([1,2],repeat= 1):
   ...:     print(i)
   ...:     
   ...: 
(1,)
(2,)

In [4]: for i in itertools.product([1,2],repeat= 2):
   ...:     print(i)
   ...:     
   ...: 
(1, 1)
(1, 2)
(2, 1)
(2, 2)

In [5]: for i in itertools.product([1,2],repeat= 3):
    ...:     print(i)
    ...:     
    ...: 
(1, 1, 1)
(1, 1, 2)
(1, 2, 1)
(1, 2, 2)
(2, 1, 1)
(2, 1, 2)
(2, 2, 1)
(2, 2, 2)

 

 

  itertools.permutations(sequence，n)  #  从sequence中拿出n个数做排列， 不放回的拿出， n 只能小于等于sequence的长度 否则没有输出。

demo：

In [6]: for i in itertools.permutations([1,2], 1):
   ...:     print(i)
   ...:     
   ...: 
(1,)
(2,)

In [7]: for i in itertools.permutations([1,2], 2):
   ...:     print(i)
   ...:     
   ...: 
(1, 2)
(2, 1)

In [8]: for i in itertools.permutations([1,2],3):
   ...:     print(i)
   ...:     
   ...: 

In [9]: 

 

itertools.combinations(sequence, n )   # 从sequence中选n个数做组合，相当于不放回， 当n 大于sequence的长度自然没有数据。

demo：

In [10]: for i in itertools.combinations([1,2],1):
    ...:     print(i)
    ...:     
    ...: 
(1,)
(2,)

In [11]: for i in itertools.combinations([1,2],2):
    ...:     print(i)
    ...:     
    ...: 
(1, 2)

In [12]: for i in itertools.combinations([1,2],3):
    ...:     print(i)
    ...:     
    ...: 

In [13]: 

 

itertools.combinations_with_replacement(sequence, n )  # 从sequence中选n个数做组合，允许元素重复,repeat与sequence的长度无关。　

demo:

In [14]: for i in itertools.combinations_with_replacement([1,2],1):
    ...:     print(i)
    ...:     
    ...: 
(1,)
(2,)

In [15]: for i in itertools.combinations_with_replacement([1,2],2):
    ...:     print(i)
    ...:     
    ...: 
(1, 1)
(1, 2)
(2, 2)

In [16]: for i in itertools.combinations_with_replacement([1,2],3):
    ...:     print(i)
    ...:     
    ...: 
(1, 1, 1)
(1, 1, 2)
(1, 2, 2)
(2, 2, 2)

In [17]: for i in itertools.combinations_with_replacement([1,2],4):
    ...:     print(i)
    ...:     
    ...: 
(1, 1, 1, 1)
(1, 1, 1, 2)
(1, 1, 2, 2)
(1, 2, 2, 2)
(2, 2, 2, 2)

回到咱们的问题， 在这几个函数中，选择一个，很明显 itertools.product(sequence，repeat)  符合我们的要求：

code：

  1 import itertools
  2 n = int(input("输入台阶数:"))
  3 l = [1,2] # 或者"12"，序列即可
  4 m = 0  # 组合数
  5 for i in range(1,n+1):
  6     for j in itertools.product(l,repeat = i):
  7         asum = 0
  8         for k in j:
  9             asum += k # 累加步数 
 10         if asum == n: # 判断条件 步数等于台阶数
 11              m += 1  # 组合数加1
 12 print("总的组合数:{}".format(m)) 
 13 

 

bash：

 

kali@Deepin:~$ python3 demo.py 
输入台阶数:1
总的组合数:1
kali@Deepin:~$ python3 demo.py 
输入台阶数:2
总的组合数:2
kali@Deepin:~$ python3 demo.py 
输入台阶数:3
总的组合数:3
kali@Deepin:~$ python3 demo.py 
输入台阶数:4
总的组合数:5
kali@Deepin:~$ python3 demo.py 
输入台阶数:5
总的组合数:8
kali@Deepin:~$ python3 demo.py 
输入台阶数:6
总的组合数:13
kali@Deepin:~$ python3 demo.py 
输入台阶数:7
总的组合数:21
kali@Deepin:~$ 

 

功能倒是实现了， 但是效率确实不咋地， 台阶数上20就得花点时间了。

import requests
import itertools
import json
import threading
import time
class IDCMinterface():  
    def getassetsrecordpagelist(self,p):
        url=''
        header = {'Content-Type': 'application/json; charset=UTF-8',
                  'Accept': 'application/json;text/plain',
                  'Accept-Encoding':'gzip',
                  'Connection':'keep-alive',
                  'Content-Length':'400'}
        #arg={"AssetsRecordType":'',"AssetsID":'',"StartDate":'',"EndDate":'',"BillNO":'',"PageSize":10, "PageIndex":1, "clientType":"undefined"}
        dat=requests.post(url,headers=header,data=p)
        dat=dat.json()
        #print(dat['status'])
        if dat['status']!=105:
            print(dat)
            print(p)

if __name__=='__main__':
    #Requestss().gettoken()
    num=0
    A = [{"serial_number":num,"phrase":"junior"},{"serial_number":num,"phrase":"onion"},{"serial_number":num,"phrase":"detail"}]
    B = [{"serial_number": num, "phrase": "emerge"}, {"serial_number": num, "phrase": "weapon"}, {"serial_number": num, "phrase": "robust"}]
    C = [{"serial_number": num, "phrase": "isolate"}, {"serial_number": num, "phrase": "volcano"}, {"serial_number": num, "phrase": "rally"}]
    D = [{"serial_number": num, "phrase": "private"}, {"serial_number": num, "phrase": "detect"}, {"serial_number": num, "phrase": "become"}]
    hw=[A,B,C,D]
    helpword = ['license', 'agent','close', 'wine', 'field', 'wide', 'steak',  'tell', 'text', 'deal', 'pole','purity']
    print(time.strftime("%Y-%m-%d %H:%M:%S"))
    p=0
    for i in itertools.permutations(hw,4):
        # i=str(i)
        # print(i)
        # i=i.replace('], [',',').replace("([",'[').replace("])",']').replace('}, {','}, ,{').replace('[','').replace(']','')
        # print(type(i),i)
        # i=i.split(', ,')
        # print(type(i),i)
        num=0
        word=[]
        for m in range(0,4):
            for n in range(1,4):
                num=num+1
                #print(i[m][n-1]['serial_number'])
                #print(i[0])
                i[m][n-1]['serial_number']=num
                word.append(i[m][n-1])
                #print(i)
        #print(help)
        i=list(word)
        i=json.dumps(i) #或者post方法的data直接改为json
        #print(i[0])
        p = p + 1
        IDCMinterface().getassetsrecordpagelist(i)