Question

我有此代码：

DATETIME2

这段代码正在生成三对，如下所示： from collections import Counter def groups(d, l, c = []): if l == len(c): yield c else: for i in d: if i not in c: _c = Counter([j for k in [*c, i] for j in k]) if all(j < 3 for j in _c.values()): yield from groups(d, l, c+[i]) data = [(1,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,4)] result = list(groups(data, 3))

问题是，有这样的重复项：

[[(1, 2), (2, 3), (3, 1)], [(1, 2), (2, 3), (3, 4)], [(1, 2), (2, 4), (3, 1)],1[(1, 2), (2, 4), (3, 4)], [(1, 2), (2, 5), (3, 1)], [(1, 2), (2, 5), (3, 4)] ...

有没有一种方法可以避免它们在生成过程中发生？

Answer 1

您正在重新发明轮子。只需使用itertools.combinations：

from itertools import combinations

data = [(1, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 4)]
print(list(combinations(data, 3)))
# [((1, 2), (2, 3), (2, 4)), ((1, 2), (2, 3), (2, 5)), ...

您可以通过检查返回列表的长度（即56）来确认没有重复，这正是您所期望的（8选择3等于56）

如果您需要应用自定义逻辑，您仍然可以这样做：

from itertools import combinations

data = [(1, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 4)]

wanted_combinations = []
for combination in combinations(data, 3):
    # apply logic
    if condition:
        wanted_combinations.append(combination)

生成组合时跳过重复项

1 个答案: