一个集合或一个列表的所有可能子组的划分

Question

假设我有一个 [1,2,3,4] 的列表，我想生成这个集合的所有子集，一次覆盖所有成员，结果应该有 15 个列表，顺序不重要，相反 t 提供所有可能的子组：

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

这是集合分区问题或partitions of a set which is discussed here, but the response made me confused as it just suggests recalling permutations, but I don't know how! and another response also doesn't include [1,3] 同时我应该为高数字解决这个问题，结果应该提供 Bell Number 抱歉，我是 python 的新手，因此感到困惑。谁能帮我弄清楚？

Answer 1

与其进行所有排列并删除重复项，这是我最初的想法，不如使用我发现的这个递归函数 here and here:

def partitions(set_):
    if not set_:
        yield []
        return
    for i in range(int(2**len(set_)/2)):
        parts = [set(), set()]
        for item in set_:
            parts[i&1].add(item)
            i >>= 1
        for b in partitions(parts[1]):
            yield [parts[0]]+b

l = [1, 2, 3, 4]
for p in reversed(sorted(partitions(l))):
    print(p)
print('The Bell number is', len(list(partitions(l))))

它打印：

[{1, 2, 3, 4}]
[{1, 2, 4}, {3}]
[{1, 4}, {2, 3}]
[{1, 4}, {3}, {2}]
[{2, 4}, {1, 3}]
[{2, 4}, {3}, {1}]
[{1, 3, 4}, {2}]
[{2, 3, 4}, {1}]
[{3, 4}, {1, 2}]
[{3, 4}, {2}, {1}]
[{4}, {1, 2, 3}]
[{4}, {1, 3}, {2}]
[{4}, {2, 3}, {1}]
[{4}, {3}, {1, 2}]
[{4}, {3}, {2}, {1}]
The Bell number is 15

一个集合或一个列表的所有可能子组的划分

partition of a set or all possible subgroups of a list

arrays

partitioning

set

python-3.x