Question

我有一个清单：

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，我想为2对2游戏创建所有玩家组合。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玩）。游戏允许重复播放，所以每一对元组应该重复，但顺序不同（元组的顺序不同，元组中的项目顺序不同）。

当列表包含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']

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

所以一场比赛总共有4名球员。其他玩家只是等待，并没有参与特定游戏。

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将每对与其补码相关联。所以输出是

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

编辑：如果列表中有超过4个元素，则应该有效：

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

[[('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

以下内容如何：

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为我们提供了所有长度为2的对（我们将其转换为list，因此我们不会在迭代时将其耗尽）。 set(x).intersection(set(y))会发现x和y之间是否存在任何共同的项目，如果情况并非如此，我们希望保留该组合。

打印：

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

Answer 3

这应该适合你：

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]

答案：

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

Answer 4

尝试使用permutations()

的itertools

import itertools
my_list = ['john','jeff','george','peter']
paired = []
for pair in itertools.permutations(my_list, 2):
    paired.append(pair)
print paired

[（＆＃39; john＆＃39;，＆＃39; jeff＆＃39;），（＆＃39; john＆＃39;，＆＃39; george＆＃39;），（＆＃39; john＆＃39;，＆＃39; peter＆＃39;），（＆＃39; jeff＆＃39;，＆＃39; john＆＃39;），（＆＃39; jeff＆＃39;，＆＃39; george＆＃39;），（＆＃39; jeff＆＃39;，＆＃39; peter＆＃39;），（＆＃39;乔治＆＃39;，＆＃39;约翰＆＃39;），（＆＃39; george＆＃39;，＆＃39; jeff＆＃39;），（＆＃39;乔治＆＃39;，＃pe;＆＃39;），（＆＃39; peter＆＃39;，＆＃39; john＆＃39;），（＆＃39; peter＆＃39;，＆＃39; jeff＆＃39;），（＆＃39; peter＆＃39;，＆＃39; george＆＃39;）]

python中两个列表的组合

4 个答案: