能否快速获取python中一个列表中最内层子列表的最小长度？

Question

This展示如何获取一级内表的最小长度

我怎样才能 return 4 这个列表 b=[[1,0,1,2,1,1,1,3111111,[1,1,6,7]],[31,1,4,51,1,1,1],[1,1,6,7,8]] 因为 [1,1,6,7] 只有 4 个元素。

我可以运行一个 for 循环来得到它。但是可以更简单吗？

我的意思是"the length of the shortest list that is at the same depth as the deepest list"。

Answer 1

v1，任意深度的最小长度：

>>> b = [[1,0,1,2,1,1,1,3111111,[1,1,6,7]],[31,1,4,51,1,1,1],[1,1,6,7,8]]
>>> def lists_in(L):
...     for element in L:
...         if isinstance(element, list):
...             yield element
...             yield from lists_in(element)
...             
>>> min(lists_in(b), key=len)
[1, 1, 6, 7]
>>> len(min(lists_in(b), key=len))
4

v2，新要求 "the length of the shortest list that is at the same depth as the deepest list":

>>> def depths_and_lengths(L, depth=0):
...     for element in L:
...         if isinstance(element, list):
...             yield (depth, len(element))
...             yield from depths_and_lengths(element, depth-1)
...             
...             
>>> min(depths_and_lengths(b))[1]
4
>>> min(depths_and_lengths([[[1, 2]]]))[1]    # Stefan Pochmann example
2

Answer 2

这是一个错误的答案：

def mpr(x, inital=[]):
    if isinstance(x, list):
        inital.append(len(x))
        map(lambda x: mpr(x, inital), x)
    return inital

mpr(b) -> [3, 9, 4, 7, 5]

能否快速获取python中一个列表中最内层子列表的最小长度？

Can I quickly get the minimum length of the most inner sub-lists within a list in python?

python

python-3.x

list

variable-length