如何将列表 S 的所有分区查找为 k 个子集（可以为空）？

Question

我有一个唯一元素列表，比如说 [1,2]，我想将它拆分为 k=2 个子列表。现在我想要所有可能的子列表：

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

我想拆分为 1

[ [1, 2] ]

如何使用 Python 3 做到这一点？

更新：我的目标是获取 N 个唯一编号列表的所有可能分区，其中每个分区将有 k 个子列表。我想展示比上面展示的更好的例子，我希望我不会错过任何东西。所以对于列表 [1, 2, 3] 和 k=2 我想要下一个列表：

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

更新 2：到目前为止，我已经结合了两个建议，并且很少修改得到下一个代码：

def sorted_k_partitions(seq, k):
    """Returns a list of all unique k-partitions of `seq`.

    Each partition is a list of parts, and each part is a tuple.

    The parts in each individual partition will be sorted in shortlex
    order (i.e., by length first, then lexicographically).

    The overall list of partitions will then be sorted by the length
    of their first part, the length of their second part, ...,
    the length of their last part, and then lexicographically.
    """
    n = len(seq)
    groups = []  # a list of lists, currently empty

    def generate_partitions(i):
        if i >= n:
            yield list(map(tuple, groups))
        else:
            if n - i > k - len(groups):
                for group in groups:
                    group.append(seq[i])
                    yield from generate_partitions(i + 1)
                    group.pop()

            if len(groups) < k:
                groups.append([seq[i]])
                yield from generate_partitions(i + 1)
                groups.pop()

    result = generate_partitions(0)

    # Sort the parts in each partition in shortlex order
    result = [sorted(ps, key = lambda p: (len(p), p)) for ps in result]
    # Sort partitions by the length of each part, then lexicographically.
    result = sorted(result, key = lambda ps: (*map(len, ps), ps))

    return result

有了这个功能，我接下来可以做：

import itertools as it
k=2
S = [1, 2, 3]
for i in (range(k)):
    for groups in sorted_k_partitions(S, k-i):
        for perm in it.permutations(groups+[tuple() for j in range(i)]):
            print(perm)

输出是：

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

我还不确定，这段代码是否给了我正确的解决方案，也许还有其他方法？

Answer 1

这是一个替代解决方案：

def partition_k(l, k):
    n = len(l)
    if k > n:
        raise ValueError("k = {0} should be no more than n = {1}".format(k, n))

    if n == 0:
        yield []
        return

    pos = [0] * n
    while True:
        # generate output for the value
        out = [[] for _ in range(k)]
        for i in range(n):
            out[pos[i]].append(l[i])
        yield out

        #increment the value
        pos[0] += 1
        for i in range(n):
            # should we carry from this digit to the next one?
            if pos[i] == k:
                # overflow of the whole value?
                if i == n - 1:
                    return
                pos[i] = 0
                pos[i + 1] += 1
            else:
                break

设n 为列表长度，k 为分区数。这段代码背后的想法是，输出的每一行都可以表示为 base-k 系统中的多个n 数字。每个“数字”显示对应位置的值在哪个桶中。例如行

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

可以编码为[1,0,0,2]，这意味着

• 1 去桶#1
• 2 去桶#0
• 3 去桶#0
• 4 去桶#2

显然，每个这样的n-digits base-k 数字都代表列表的一个有效分区，并且每个分区都由某个数字表示。因此，要生成所有分区，我们只需遍历所有这些数字并生成相应的分区。如果您使用数字列表来表示数字（在代码中为pos），则更容易做到。