python中两个列表的组合

Question

我有一份清单：

list = ['john','jeff','george','peter']

我想创建以下输出：

[
  [('john','jeff'),('george','peter')],
  [('john','george'),('jeff','peter')],
  [('john','peter'),('jeff','george')],
  [('george','peter'),('john','jeff')],
  [('jeff','peter'),('john','george')],
  [('jeff','george'),('john','peter')]
]

一般来说，我想创建2对2游戏的所有玩家组合。在一个输出行中，一个名称只能显示一次（一个玩家一次只能在一个团队中比赛）。游戏允许玩重赛，所以每对元组都应该重复，但顺序不同（元组的顺序不同，元组中的项目没有不同的顺序）。在

当列表有4个以上的元素时，例如5，输出应该是这样的：

list = ['john','jeff','george','peter','simon']

^{pr2}$

所以一场比赛总有4个队员。其他玩家只是等待，不参与特定的游戏。在

Answer 1

你可以这样做：

import itertools
l = set(['john','jeff','george','peter'])
m=list(itertools.combinations(l, 2))
res=[[i,tuple(l.symmetric_difference(i))] for i in m]

m是所有对的列表，res将每个对与其补码相关联。所以输出是

^{pr2}$

编辑：如果列表中有4个以上的元素，则可以这样做：

import itertools
l = set(['john','jeff','george','peter','a'])
four_tuples=list(itertools.combinations(l, 4))
pairs=[(set(i),list(itertools.combinations(i, 2))) for i in four_tuples]
pair_and_comp=[[[r,tuple(el[0].symmetric_difference(r))] for r in el[1:][0]] for el in pairs]
res=sum(pair_and_comp,[])
res

输出是

[[('john', 'jeff'), ('peter', 'george')],
 [('john', 'peter'), ('jeff', 'george')],
 [('john', 'george'), ('jeff', 'peter')],
 [('jeff', 'peter'), ('john', 'george')],
 [('jeff', 'george'), ('john', 'peter')],
 [('peter', 'george'), ('john', 'jeff')],
 [('john', 'jeff'), ('peter', 'a')],
 [('john', 'peter'), ('jeff', 'a')],
 [('john', 'a'), ('jeff', 'peter')],
 [('jeff', 'peter'), ('john', 'a')],
 [('jeff', 'a'), ('john', 'peter')],
 [('peter', 'a'), ('john', 'jeff')],
 [('john', 'jeff'), ('george', 'a')],
 [('john', 'george'), ('jeff', 'a')],
 [('john', 'a'), ('jeff', 'george')],
 [('jeff', 'george'), ('john', 'a')],
 [('jeff', 'a'), ('john', 'george')],
 [('george', 'a'), ('john', 'jeff')],
 [('john', 'peter'), ('george', 'a')],
 [('john', 'george'), ('peter', 'a')],
 [('john', 'a'), ('peter', 'george')],
 [('peter', 'george'), ('john', 'a')],
 [('peter', 'a'), ('john', 'george')],
 [('george', 'a'), ('john', 'peter')],
 [('jeff', 'peter'), ('george', 'a')],
 [('jeff', 'george'), ('peter', 'a')],
 [('jeff', 'a'), ('peter', 'george')],
 [('peter', 'george'), ('jeff', 'a')],
 [('peter', 'a'), ('jeff', 'george')],
 [('george', 'a'), ('jeff', 'peter')]]

Answer 2

以下情况如何：

from itertools import combinations
from pprint import pprint

names = ['john', 'jeff', 'george', 'peter', 'ringo']

combos = list(combinations(names, 2))
pairs = [[x, y] for x in combos for y in combos if not set(x).intersection(set(y))]

pprint(pairs)

combinations为我们提供了长度为2的所有对（我们将其转换为list以便在迭代时不会耗尽它）。set(x).intersection(set(y))查找x和{}之间是否有共同点，如果不是这样的话，我们希望保留这个组合。在

打印：

^{pr2}$

Answer 3

这应该能帮你做到：

from itertools import combinations
l = ['john','jeff','george','peter','beni']
x= list(combinations(l,2))
y=list(combinations(x,2))
remove_dup =lambda y: y if len(set(y[0])-set(y[1]))==2 else None
answer=[remove_dup(t) for t in y if remove_dup(t) is not None]

回答：

^{pr2}$