python - 成员每组出现一次的唯一对组

Question

我有这个代码:

from itertools import groupby
from itertools import combinations


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

输出是一个包含 45 个元组的列表。

[(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 个元组分成 9 组，每组 5 个元组，每个元组都是唯一项。例如像这样:

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 = 

所以每个列表都包含从 1 到 10 没有重复项的元组。

Answer 1

这是一个基于 round robin tournament scheduling algorithm 的相当简单的方法.基本上，这种方法将列表分成两半，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，它会“轮换”除列表中第一个团队之外的所有团队(基于阶段或轮数的循环和列表连接正在模拟轮换)。

# 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)]
# ]