Python編程之黑板上排列組合，你舍得解開(kāi)嗎
更新時(shí)間：2017年10月30日 15:12:19 作者：Inside_Zhang
這篇文章主要介紹了Python排列組合算法的相關(guān)內(nèi)容，具有一定參考價(jià)值，需要的朋友可以了解下。
考慮這樣一個(gè)問(wèn)題，給定一個(gè)矩陣（多維數(shù)組，numpy.ndarray()），如何shuffle這個(gè)矩陣（也就是對(duì)其行進(jìn)行全排列），如何隨機(jī)地選擇其中的k行，這叫組合，實(shí)現(xiàn)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(shù)），便從原有的五維空間中降維到三維空間，因?yàn)槭侨康呐帕袛?shù)，故不會(huì)漏掉任何一種可能性。
涉及的函數(shù)主要有：
np.random.permutation()
itertools.combinations()
itertools.permutations()
# 1. 對(duì)0-5之間的數(shù)進(jìn)行一次全排列 >>>np.random.permutation(6) array([3, 1, 5, 4, 0, 2]) # 2. 創(chuàng)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(shí)現(xiàn)
>>>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(shí)現(xiàn)
>>>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], :] # 一種排列 array([[1, 2, 3, 4], [5, 6, 7, 8]])
下面再介紹一個(gè)列表數(shù)據(jù)任意組合，主要是利用自帶的庫(kù)
#_*_ 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)]]
排列的實(shí)現(xiàn)
#_*_ 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(yùn)行結(jié)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)]]
可以根據(jù)你需要隨意組合
python實(shí)現(xiàn)排列組合公式C(m,n)求值
# -*- coding:utf-8 -*- # 用python實(shí)現(xiàn)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(yùn)行結(jié)果：
n:5 value:120 n:3 value:6 n:2 value:2 value: 10
總結(jié)
以上就是本文關(guān)于Python排列組合算法的全部?jī)?nèi)容，希望對(duì)大家有所幫助。感興趣的朋友可以繼續(xù)參閱本站：Python數(shù)據(jù)結(jié)構(gòu)與算法之列表（鏈表，linked list）簡(jiǎn)單實(shí)現(xiàn)、Python算法之求n個(gè)節(jié)點(diǎn)不同二叉樹(shù)個(gè)數(shù)等，有什么問(wèn)題可以隨時(shí)留言，小編會(huì)及時(shí)回復(fù)大家的。
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