如何使用 python 函数将列表有效地划分为所有可能数量的块？

Question

我想根据以下条件拆分列表（浮点数或整数）：

将列表拆分为所有可能的子样本。
没有重复。
单位样本不能是子样本。

我有通过给定子样本数将列表分成相等大小的东西。

我手上的代码有效但没有给我想要的东西

import numpy as np
x = [1,2,3,4,5,6,7,8,9,10]
l = np.array_split(x,3)

输出

[[1, 2, 3, 4], [5, 6, 7], [8, 9, 10]]

我希望获得所有可能性的子样本列表，而不会重复。也就是说，当列表分为 2、3、4 等时，所有唯一可能性的组合（没有两个或更多子样本将具有相同的元素））。

我不指定什么块号，这样就不会被限制为这个数量。

这是我手动做的

来自以下系列[1,2,3,4,5,6,7,8,9,10]，我将其切成所有可能的块，如下所示：

拆分为2时

[1][2,3,4,5,6,7,8,9,10]
[1,2][3,4,5,6,7,8,9,10]
[1,2,3][4,5,6,7,8,9,10]
[1,2,3,4][5,6,7,8,9,10]
[1,2,3,4,5][6,7,8,9,10]
[1,2,3,4,5,6][7,8,9,10]
[1,2,3,4,5,6,7,8][9,10]
[1,2,3,4,5,6,7,8,9][10]

拆分成3时

[1][2][3,4,5,6,7,8,9,10]
[1][2,3][4,5,6,7,8,9,10]
[1][2,3,4][5,6,7,8,9,10]
[1][2,3,4,5][6,7,8,9,10]
[1][2,3,4,5,6][7,8,9,10]
[1][2,3,4,5,6,7][8,9,10]
[1][2,3,4,5,6,7,8][9,10]
[1][2,3,4,5,6,7,8,9][10]

[1,2][3][4,5,6,7,8,9,10]
[1,2][3,4][5,6,7,8,9,10]
[1,2][3,4,5][6,7,8,9,10]
[1,2][3,4,5,6][7,8,9,10]
[1,2][3,4,5,6,7][8,9,10]
[1,2][3,4,5,6,7,8][9,10]
[1,2][3,4,5,6,7,8,9][10]

[1,2,3][4][5,6,7,8,9,10]
[1,2,3][4,5][6,7,8,9,10]
[1,2,3][4,5,6][7,8,9,10]
[1,2,3][4,5,6,7][8,9,10]
[1,2,3][4,5,6,7,8][9,10]
[1,2,3][4,5,6,7,8,9][10]

[1,2,3,4][5][6,7,8,9,10]
[1,2,3,4][5,6][7,8,9,10]
[1,2,3,4][5,6,7][8,9,10]
[1,2,3,4][5,6,7,8][9,10]
[1,2,3,4][5,6,7,8,9][10]

[1,2,3,4,5][6][7,8,9,10]
[1,2,3,4,5][6,7][8,9,10]
[1,2,3,4,5][6,7,8][9,10]
[1,2,3,4,5][6,7,8,9][10]

[1,2,3,4,5,6][7][8,9,10]
[1,2,3,4,5,6][7,8][9,10]
[1,2,3,4,5,6][7,8,9][10]

拆分为4时

[1,2,3,4,5,6,7][8][9,10]
[1,2,3,4,5,6,7][8,9][10]

[1,2,3,4,5,6,7,8][9][10]

在所有可能的分块之后，我删除了所有的个位数块，也删除了所有重复的块。

[2,3,4,5,6,7,8,9,10]
[1,2,3,4,5,6,7,8,9]


[3,4,5,6,7,8,9,10]
[2,3]
[2,3,4]
[2,3,4,5]
[2,3,4,5,6]
[2,3,4,5,6,7]
[2,3,4,5,6,7,8]
[2,3,4,5,6,7,8,9]

[4,5,6,7,8,9,10]
[3,4]
[3,4,5]
[3,4,5,6]
[3,4,5,6,7]
[3,4,5,6,7,8]
[1,2][3,4,5,6,7,8,9]

[5,6,7,8,9,10]
[4,5]
[4,5,6]
[4,5,6,7]
[4,5,6,7,8]
[1,2,3][4,5,6,7,8,9]

[6,7,8,9,10]
[5,6]
[5,6,7]
[5,6,7,8]
[1,2,3,4][5,6,7,8,9]

[1,2,3,4,5][7,8,9,10]
[6,7]
[6,7,8]
[6,7,8,9]

[8,9,10]
[7,8]
[1,2,3,4,5,6][7,8,9]


[9,10]
[1,2,3,4,5,6,7][8,9]

[1,2,3,4,5,6,7,8]

在这里，我将所有可能的块收集在一起。

这就是我想要的输出。

[[2,3,4,5,6,7,8,9,10], [1,2,3,4,5,6,7,8,9], [3,4,5,6,7,8,9,10], [2,3], [2,3,4], [2,3,4,5], [2,3,4,5,6], [2,3,4,5,6,7], [2,3,4,5,6,7,8], [2,3,4,5,6,7,8,9], [4,5,6,7,8,9,10],[3,4], [3,4,5], [3,4,5,6], [3,4,5,6,7], [3,4,5,6,7,8], [1,2], [3,4,5,6,7,8,9], [5,6,7,8,9,10], [4,5], [4,5,6], [4,5,6,7], [4,5,6,7,8], [1,2,3], [4,5,6,7,8,9], [6,7,8,9,10], [5,6], [5,6,7], [5,6,7,8], [1,2,3,4], [5,6,7,8,9], [1,2,3,4,5], [7,8,9,10], [6,7], [6,7,8], [6,7,8,9], [8,9,10], [7,8], [1,2,3,4,5,6], [7,8,9], [9,10], [1,2,3,4,5,6,7], [8,9], [1,2,3,4,5,6,7,8]]

Answer 1

在我看来，问题简化为找到所有长度为 2 或更大且至少留下一个长度为 1 的片段的子串。换句话说，您不必枚举每个分区来找到它们。

def parts(thing):
    result = []
    for i in range(len(thing)):
        for j in range(i + 1, len(thing) + 1):
            if 1 < len(thing[i:j]) < len(thing):
                result.append(thing[i:j])
    return result

res = parts([*range(1,11)])
# res
# [[1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 6],
# [1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 8],
# [1, 2, 3, 4, 5, 6, 7, 8, 9], [2, 3], [2, 3, 4], [2, 3, 4, 5],
# [2, 3, 4, 5, 6], [2, 3, 4, 5, 6, 7], [2, 3, 4, 5, 6, 7, 8],
# [2, 3, 4, 5, 6, 7, 8, 9], [2, 3, 4, 5, 6, 7, 8, 9, 10], [3, 4], [3, 4, 5],
# [3, 4, 5, 6], [3, 4, 5, 6, 7], [3, 4, 5, 6, 7, 8], [3, 4, 5, 6, 7, 8, 9],
# [3, 4, 5, 6, 7, 8, 9, 10], [4, 5], [4, 5, 6], [4, 5, 6, 7], [4, 5, 6, 7, 8],
# [4, 5, 6, 7, 8, 9], [4, 5, 6, 7, 8, 9, 10], [5, 6], [5, 6, 7], [5, 6, 7, 8],
# [5, 6, 7, 8, 9], [5, 6, 7, 8, 9, 10], [6, 7], [6, 7, 8], [6, 7, 8, 9],
# [6, 7, 8, 9, 10], [7, 8], [7, 8, 9], [7, 8, 9, 10], [8, 9], [8, 9, 10],
# [9, 10]]

如何使用 python 函数将列表有效地划分为所有可能数量的块？

How can I have a python function that effectively divide a list to all possible number of chunks?

chunks

python-3.x