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服務器之家 - 腳本之家 - Python - Python編程之黑板上排列組合,你舍得解開嗎

Python編程之黑板上排列組合,你舍得解開嗎

2020-12-14 00:38Inside_Zhang Python

這篇文章主要介紹了Python排列組合算法的相關內容,具有一定參考價值,需要的朋友可以了解下。

考慮這樣一個問題,給定一個矩陣(多維數組,numpy.ndarray()),如何shuffle這個矩陣(也就是對其行進行全排列),如何隨機地選擇其中的k行,這叫組合,實現一種某一維度空間的切片。例如五列中選三列(全部三列的排列數),便從原有的五維空間中降維到三維空間,因為是全部的排列數,故不會漏掉任何一種可能性。

涉及的函數主要有:

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

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# 1. 對0-5之間的數進行一次全排列
>>>np.random.permutation(6)
array([3, 1, 5, 4, 0, 2])
 
# 2. 創建待排矩陣
>>>A = np.array([[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]])
 
# 3. shuffle矩陣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的實現

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>>>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的實現

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>>>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
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# 5. 任取其中的k(k=2)行
>>>c = [c for c in combinations(range(A.shape[0]), 2)]
>>>A[c[0], :]   # 一種排列
array([[1, 2, 3, 4],
  [5, 6, 7, 8]])

下面再介紹一個列表數據任意組合,主要是利用自帶的庫

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#_*_ 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)
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[[(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)]]

排列的實現

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#_*_ 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)

運行結果:

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[[(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)]]

可以根據你需要隨意組合

python實現排列組合公式C(m,n)求值

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# -*- coding:utf-8 -*-
# 用python實現排列組合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

運行結果:

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

總結

以上就是本文關于Python排列組合算法的全部內容,希望對大家有所幫助。有什么問題可以隨時留言,小編會及時回復大家的。

原文鏈接:http://blog.csdn.net/lanchunhui/article/details/49494265

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