Python 3.x 如何使用循环按节拆分多行列表

如果一个_列表是一个包含多行整数的列表，那么假设使用三个循环将该列表的拆分版本打印为三部分：第一行加上后面两行，中间一行加上前后两行，最后一行加上前面两行

Print entire list a_list if its length is less than or equal to
3 times section_size (an int).

Otherwise, print the first
section_size elements of a_list, followed by an ellipsis ("...")

followed by the middle section_size elements, followed by another
ellipsis, followed by the final section_size elements.

Note: This function assumes its parameters' values are valid, and
does no exception handling.

这应该起作用：

def contract(lst, size=3):
    if len(lst) <= size * 3:
        return lst

    mid = (len(lst) - 1) // 2
    mid_size = (size - 1) // 2

    new_lst = lst[:size]
    new_lst += ["..."] + lst[mid-mid_size:mid+mid_size+1]
    new_lst += ["..."] + lst[-size:]

    return new_lst

lst = [[0, 0, 1, 2, 3, 4, 5, 6, 7, 8],
 [1, 0, 1, 2, 3, 4, 5, 6, 7, 8],
 [2, 0, 1, 2, 3, 4, 5, 6, 7, 8],
 [3, 0, 1, 2, 3, 4, 5, 6, 7, 8],
 [4, 0, 1, 2, 3, 4, 5, 6, 7, 8],
 [5, 0, 1, 2, 3, 4, 5, 6, 7, 8],
 [6, 0, 1, 2, 3, 4, 5, 6, 7, 8],
 [7, 0, 1, 2, 3, 4, 5, 6, 7, 8],
 [8, 0, 1, 2, 3, 4, 5, 6, 7, 8],
 [9, 0, 1, 2, 3, 4, 5, 6, 7, 8],
 [10, 0, 1, 2, 3, 4, 5, 6, 7, 8],
 [11, 0, 1, 2, 3, 4, 5, 6, 7, 8],
 [12, 0, 1, 2, 3, 4, 5, 6, 7, 8],
 [13, 0, 1, 2, 3, 4, 5, 6, 7, 8],
 [14, 0, 1, 2, 3, 4, 5, 6, 7, 8]]

for row in contract(lst, 3):
    print(row)

[0, 0, 1, 2, 3, 4, 5, 6, 7, 8]
[1, 0, 1, 2, 3, 4, 5, 6, 7, 8]
[2, 0, 1, 2, 3, 4, 5, 6, 7, 8]
...
[6, 0, 1, 2, 3, 4, 5, 6, 7, 8]
[7, 0, 1, 2, 3, 4, 5, 6, 7, 8]
[8, 0, 1, 2, 3, 4, 5, 6, 7, 8]
...
[12, 0, 1, 2, 3, 4, 5, 6, 7, 8]
[13, 0, 1, 2, 3, 4, 5, 6, 7, 8]
[14, 0, 1, 2, 3, 4, 5, 6, 7, 8]