唯一对的组，每个组中成员出现一次 [英] Groups of unique pairs where members appear once per group

问题描述

我有此代码:

from itertools import groupby
from itertools import combinations


teams = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
combo = list(combinations(teams, 2))

输出是45个元组的列表.

The output is a list of 45 tuples.

[(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10), (2, 3), (2, 4), (2, 5), (2, 6), (2, 7), (2, 8), (2, 9), (2, 10), (3, 4), (3, 5), (3, 6), (3, 7), (3, 8), (3, 9), (3, 10), (4, 5), (4, 6), (4, 7), (4, 8), (4, 9), (4, 10), (5, 6), (5, 7), (5, 8), (5, 9), (5, 10), (6, 7), (6, 8), (6, 9), (6, 10), (7, 8), (7, 9), (7, 10), (8, 9), (8, 10), (9, 10)]

我想将45个元组分为9组，每组5个元组，每个组都是唯一的项.例如这样的

I would like to divide 45 tuples into 9 groups of 5 tuples, each of which are unique items. For example like so:

list_1 = [(1, 2), (3, 4), (5, 6), (7, 8), (9, 10)]
list_2 = [(1, 3), (2, 4), (5, 7), (6, 9), (8, 10)]
list_3 = 
list_4 = 
list_5 = 

因此，每个列表都包含元组，其中的项从1到10，没有重复.

So each list contains tuples with items from 1 to 10 without repetition.

推荐答案

这是一种基于循环锦标赛调度算法.基本上，此方法将列表分为两半，并将列表的前半部分与列表后半部分的反向版本配对.然后，对于每个阶段，它都会轮换"除列表中的第一支球队之外的所有球队(基于阶段或轮次的循环和列表串联正在模拟轮换).

Here is a fairly straightforward approach based on a round robin tournament scheduling algorithm. Basically, this approach splits the list in half and pairs the first half of the list with a reversed version of the second half of the list. Then, for each stage it "rotates" all teams except for the first team in the list (the loop and list concatenation based on the stage or round number is simulating the rotation).

# even number of teams required
teams = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
n = int(len(teams) / 2)

stages = []
for i in range(len(teams) - 1):
    t = teams[:1] + teams[-i:] + teams[1:-i] if i else teams
    stages.append(list(zip(t[:n], reversed(t[n:]))))

print(stages)
# [
#     [(1, 10), (2, 9), (3, 8), (4, 7), (5, 6)],
#     [(1, 9), (10, 8), (2, 7), (3, 6), (4, 5)],
#     [(1, 8), (9, 7), (10, 6), (2, 5), (3, 4)],
#     [(1, 7), (8, 6), (9, 5), (10, 4), (2, 3)],
#     [(1, 6), (7, 5), (8, 4), (9, 3), (10, 2)],
#     [(1, 5), (6, 4), (7, 3), (8, 2), (9, 10)],
#     [(1, 4), (5, 3), (6, 2), (7, 10), (8, 9)],
#     [(1, 3), (4, 2), (5, 10), (6, 9), (7, 8)],
#     [(1, 2), (3, 10), (4, 9), (5, 8), (6, 7)]
# ]

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