Question

我需要执行一个迭代，该迭代输出一名球员参加其前队的所有实例。到目前为止，我有：

teams = ['team1', 'team2', 'team3', 'team4']

for previous_team, curr_team in zip(teams, teams[1:]):
    print (curr_team, previous_team)

打印：

'team2', 'team1'
'team3', 'team2'
'team4', 'team3'

但是我需要更大的范围，在这个范围内，玩家不仅玩了最后一支球队，而且还玩过更远的球队，考虑所有间隔回到索引0，最后得出：

'team2', 'team1'
'team3', 'team2'
'team4', 'team3'
'team3', 'team1'   <----
'team4', 'team2'   <----
'team4', 'team1    <----

这是逻辑，应该应用于n个项目的列表。

我该怎么做？

Answer 1

天真的方法是在您的团队中进行2次循环，然后在每个循环中用第一个元素减少第二个列表

所以：

for i in range(len(team)):
    for j in range(len(team)-i):
       print(team[i],team[j+i])

Answer 2

如果您想要的是所有组合

from itertools import combinations
teams = ['team1', 'team2', 'team3', 'team4']

for combo in combinations(teams, 2):  # 2 for pairs, 3 for triplets, etc
    print(combo)

（'team1'，'team2'）

（'team1'，'team3'）

（'team1'，'team4'）

（'team2'，'team3'）

（'team2'，'team4'）

（'team3'，'team4'）

Answer 3

您可以使用理解力尝试此操作，如果列表中出现两次，也可以删除相同的团队价值。

teams = ['team1', 'team2', 'team3', 'team4']

[(team1, team2) for team1 in teams for team2 in teams if team1 != team2]

结果：

[('team1', 'team2'), ('team1', 'team3'), ('team1', 'team4'), ('team2', 'team1'), ('team2', 'team3'), ('team2', 'team4'), ('team3', 'team1'), ('team3', 'team2'), ('team3', 'team4'), ('team4', 'team1'), ('team4', 'team2'), ('team4', 'team3')]

Answer 4

您可以使用product：

import itertools

teams = ["team1", "team2", "team3", "team4"]

matches = itertools.product(teams, repeat=2)

print(list(matches))

# [('team1', 'team2'), ('team1', 'team3'), ('team1', 'team4'), ('team2', 'team1'), ...]

然后您可以找到与您之前的球队的比赛：

previous_team = "team2"

current_team = "team3"

awkward = [previous_team in match and current_team in match for match in matches]

print(awkward)

以特定顺序遍历列表项

4 个答案: