使嵌套列表的每个元素成为一个列表

Question

我有一个嵌套列表说：

lst = [[1,2,3,4], [2,3,4,5], [3,4,5,6]]

我希望输出为：

new_list = [[[1], [2], [3], [4]], [[2], [3], [4], [5]], [[3], [4], [5], [6]]]

这就是我的想法，但我输出的是一个平面列表。

new_lst = []
for i in lst:
  i = list(i)
  for el in i:
    new_list.append([el])
print(new_lst)

我想保留 lst 中每个预定义列表的长度

Answer 1

尝试列表理解 [[ [e] for e in l] for l in lst]

Answer 2

您可以使用列表理解并将每个元素附加到另一个列表。

lst = [[1,2,3,4], [2,3,4,5], [3,4,5,6]]
new = [[[numbers] for numbers in elements] for elements in lst]

上面的例子根据你想要的输出进行了调整。

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

Answer 3

你可以使用 numpy 的 reshape np.reshape

>>> import numpy as np

>>> lst = np.array([[1,2,3,4], [2,3,4,5], [3,4,5,6]])
>>> lst
array([[1, 2, 3, 4],
       [2, 3, 4, 5],
       [3, 4, 5, 6]])

>>> lst.shape
(3, 4)

>>> lst.reshape(3,4,1)
array([[[1],
        [2],
        [3],
        [4]],

       [[2],
        [3],
        [4],
        [5]],

       [[3],
        [4],
        [5],
        [6]]])

>>> lst.reshape(3,4,1).tolist()
[[[1], [2], [3], [4]], [[2], [3], [4], [5]], [[3], [4], [5], [6]]]

Answer 4

另一个版本，使用递归（你可以在你的输入中有更多的深度）：

lst = [[1, 2, 3, 4], [2, 3, 4, 5], [3, 4, 5, 6]]


def split(i):
    if isinstance(i, list):
        return [split(v) for v in i]
    else:
        return [i]


print(split(lst))

打印：

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

Answer 5

您以后可以使用辅助函数来更灵活地划分较小的列表。基于 here.

的回答

def split_list(alist, wanted_parts=1):
    length = len(alist)
    return [ alist[i*length // wanted_parts: (i+1)*length // wanted_parts]
             for i in range(wanted_parts) ]

lst = [[1,2,3,4], [2,3,4,5], [3,4,5,6]]

ret =[]
for mini_list in lst:
    ret.append(split_list(mini_list, len(mini_list)))

Answer 6

我认为您使用多个 for 循环的想法是正确的。这应该有效：

list = [[1,2,3,4], [2,3,4,5], [3,4,5,6]]

for i in range(0, 3):
     for j in range(0, 4):
          (list[i])[j] = [(list[i])[j]]

使嵌套列表的每个元素成为一个列表

Making each element of a nested list a list

python

for-loop

list-comprehension

list

nested-loops