python中两个列表的组合

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。在一条输出线内，一个名字只能显示一次（一名球员一次只能在一支球队中比赛）。游戏允许进行复赛，因此每对元组都应该重复，但顺序不同（元组顺序不同，元组中项目的顺序不同）。

当列表有超过4个元素时，例如5个，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')]
  ...
]

所以一场比赛总是有 4 名玩家。其余玩家只是等待，不参与特定游戏。

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 将每对与它的补码相关联。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，这应该有效：

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，因此在迭代它时不会耗尽它）。 set(x).intersection(set(y)) 查找x 和y 之间是否有任何共同项，如果不是，我们希望保留组合。

打印出来：

[[('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

这应该对你有用：

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

用permutations()试试itertools

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

[('约翰', '杰夫'), （“约翰”，“乔治”）， （'约翰'，'彼得'）， （'杰夫'，'约翰'）， （'杰夫'，'乔治'）， （'杰夫'，'彼得'）， （“乔治”，“约翰”）， （“乔治”，“杰夫”）， （“乔治”，“彼得”）， （“彼得”，“约翰”）， （“彼得”，“杰夫”）， ('彼得', '乔治')]