从最深层次开始的块嵌套列表

Question

我想要一个函数，根据这个方案将我的列表分成一个子列表，然后将这些子列表分成更多的子列表。嵌套数和块大小会有所不同。请注意lst2中的[4]和[9]，这是必不可少的。

block_sizes = [5, 2, 1]
my_lst = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]

# desired return
lst1 = [[0, 1, 2, 3, 4], [5, 6, 7, 8, 9], [10, 11]] # chunked into sublists of size 5
lst2 = [[0, 1], [2, 3], [4], [5, 6], [7, 8], [9], [10, 11]] # chunked into sublists of size 2; be aware of [4] and [9]
lst3 = [[0], [1], [2], [3], [4], [5], ... [11]]

我已经尝试过了，但是分块没有按预期工作：

block_sizes = [5, 2, 1]

def chunk_into_list(lst, chunk_size):
    return [lst[i:i + chunk_size] for i in range(0, len(lst), chunk_size)]

for i, block_size in enumerate(block_sizes):

    for step in range(len(block_sizes[:(i + 1)])):
        my_lst = chunk_in_sizes(my_lst, block_sizes[step])

    print(my_lst)

但这会导致：

[[0, 1, 2, 3, 4], [5, 6, 7, 8, 9], [10, 11]]
[[[0, 1, 2, 3, 4], [5, 6, 7, 8, 9]], [[10, 11]]]
[[[[0, 1, 2, 3, 4], [5, 6, 7, 8, 9]]], [[[10, 11]]]]

我将使用它进行模拟，所以如果可能的话，效率很重要。

Answer 1

你的意思是：

lst = [0,1,2,3,4,5,6,7,8,9,10,11]
for block_size in block_sizes:
    print(chunk_in_sizes(lst, block_size))

Answer 2

我猜这不是最理想的方式，但据我所知，它会做你想做的事。

block_sizes = [5, 2, 1]
my_lst = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]

def chunk_into_list(lst, chunk_size):
    if lst and isinstance(lst[0], list):
        lst_new = []
        for nested_lst in lst:
            lst_new += (chunk_into_list(nested_lst, chunk_size))
        return lst_new
    else:
        return [lst[i:i + chunk_size] for i in range(0, len(lst), chunk_size)]


for i, block_size in enumerate(block_sizes):
    for step in range(len(block_sizes[:(i + 1)])):
        my_lst = chunk_into_list(my_lst, block_sizes[step])

    print(my_lst)

#Output
[[0, 1, 2, 3, 4], [5, 6, 7, 8, 9], [10, 11]]
[[0, 1], [2, 3], [4], [5, 6], [7, 8], [9], [10, 11]]
[[0], [1], [2], [3], [4], [5], [6], [7], [8], [9], [10], [11]]