Question

我需要一个可以将序列拆分成对的函数，然后将它们组合起来，使得组合中的所有元素都是唯一的。我已经尝试了许多使用python的itertools的方法，但还没有找到解决方案。

为了说明我想要一个能够采用这个序列的函数： [1,2,3,4]

并将其分为以下3种组合：

[[1, 2], [3, 4]]
[[1, 3], [2, 4]]
[[1, 4], [2, 3]]

它也适用于较长的序列，但不必处理奇数长度的序列。例如

[1,2,3,4,5,6]

分为以下15种组合：

[[1, 2], [3, 4], [5, 6]]
[[1, 2], [3, 5], [4, 6]]
[[1, 2], [3, 6], [4, 5]]
[[1, 3], [2, 4], [5, 6]]
[[1, 3], [2, 5], [4, 6]]
[[1, 3], [2, 6], [4, 5]]
[[1, 4], [2, 3], [5, 6]]
[[1, 4], [2, 5], [3, 6]]
[[1, 4], [2, 6], [3, 5]]
[[1, 5], [2, 3], [4, 6]]
[[1, 5], [2, 4], [3, 6]]
[[1, 5], [2, 6], [3, 4]]
[[1, 6], [2, 3], [4, 5]]
[[1, 6], [2, 4], [3, 5]]
[[1, 6], [2, 5], [3, 4]]

......等等。

名为Maple的CAS具有以setpartition名称实现的此功能。

编辑：修复了由wks指出的一个严重的深夜打字错误，并澄清了输出。

Answer 1

itertools确实是你的朋友：

from itertools import permutations

def group(iterable, n=2):
    return zip(*([iter(iterable)] * n))

for each in permutations([1, 2, 3, 4, 5, 6]):
    print map(list, group(each))

结果：

[[1, 2], [3, 4], [5, 6]]
[[1, 2], [3, 4], [6, 5]]
[[1, 2], [3, 5], [4, 6]]
[[1, 2], [3, 5], [6, 4]]
[[1, 2], [3, 6], [4, 5]]
[[1, 2], [3, 6], [5, 4]]
[[1, 2], [4, 3], [5, 6]]
[[1, 2], [4, 3], [6, 5]]
[[1, 2], [4, 5], [3, 6]]
...

<强> [编辑] @FrederikNS：在您澄清问题and found an answer yourself之后，这是我的解决方案：

from itertools import combinations

def setpartition(iterable, n=2):
    iterable = list(iterable)
    partitions = combinations(combinations(iterable, r=n), r=len(iterable) / n)
    for partition in partitions:
        seen = set()
        for group in partition:
            if seen.intersection(group):
                break
            seen.update(group)
        else:
            yield partition

for each in setpartition([1, 2, 3, 4]):
    print each
print
for each in setpartition([1, 2, 3, 4, 5, 6]):
    print each

结果：

((1, 2), (3, 4))
((1, 3), (2, 4))
((1, 4), (2, 3))

((1, 2), (3, 4), (5, 6))
((1, 2), (3, 5), (4, 6))
((1, 2), (3, 6), (4, 5))
((1, 3), (2, 4), (5, 6))
((1, 3), (2, 5), (4, 6))
((1, 3), (2, 6), (4, 5))
((1, 4), (2, 3), (5, 6))
((1, 4), (2, 5), (3, 6))
((1, 4), (2, 6), (3, 5))
((1, 5), (2, 3), (4, 6))
((1, 5), (2, 4), (3, 6))
((1, 5), (2, 6), (3, 4))
((1, 6), (2, 3), (4, 5))
((1, 6), (2, 4), (3, 5))
((1, 6), (2, 5), (3, 4))

Answer 2

我终于得到了我的自我（pillmuncher的回答确实给了我正确方向的推动，并且小组功能完全是他的）：

def group(iterable, n=2):
    return zip(*([iter(iterable)] * n))

def set_partition(iterable, n=2):
    set_partitions = set()

    for permutation in itertools.permutations(iterable):
        grouped = group(list(permutation), n)
        sorted_group = tuple(sorted([tuple(sorted(partition)) for partition in grouped]))
        set_partitions.add(sorted_group)

    return set_partitions

partitions = set_partition([1,2,3,4], 2)
for part in partitions:
    print(part)

打印：

((1, 4), (2, 3))
((1, 3), (2, 4))
((1, 2), (3, 4))

Answer 3

试试这个：

def function(list):
    combinations = []
    for i in list:
        for i2 in list:
            if not [i2, i] in combinations:
                 combinations.append([i, i2])
    return combinations

这会返回所有可能的组合。

希望这有帮助！

将序列划分为多组唯一对

3 个答案: