我有一个[1,2,3]数组

我想使用数组的所有元素进行所有可能的组合:

结果:

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

最佳答案

既然这个好问题已经重生,那么这里就有一个新的答案。

该问题以递归方式解决:如果您已经有n-1个元素的分区,如何使用它来划分n个元素?将第n个元素放在现有子集中的一个子集中,或将其添加为新的单例子集。仅此而已;没有itertools,没有集合,没有重复的输出,并且总共只有n次调用partition():

def partition(collection):
    if len(collection) == 1:
        yield [ collection ]
        return

    first = collection[0]
    for smaller in partition(collection[1:]):
        # insert `first` in each of the subpartition's subsets
        for n, subset in enumerate(smaller):
            yield smaller[:n] + [[ first ] + subset]  + smaller[n+1:]
        # put `first` in its own subset
        yield [ [ first ] ] + smaller


something = list(range(1,5))

for n, p in enumerate(partition(something), 1):
    print(n, sorted(p))

输出:
1 [[1, 2, 3, 4]]
2 [[1], [2, 3, 4]]
3 [[1, 2], [3, 4]]
4 [[1, 3, 4], [2]]
5 [[1], [2], [3, 4]]
6 [[1, 2, 3], [4]]
7 [[1, 4], [2, 3]]
8 [[1], [2, 3], [4]]
9 [[1, 3], [2, 4]]
10 [[1, 2, 4], [3]]
11 [[1], [2, 4], [3]]
12 [[1, 2], [3], [4]]
13 [[1, 3], [2], [4]]
14 [[1, 4], [2], [3]]
15 [[1], [2], [3], [4]]

10-07 19:52
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