计算嵌套列表进入的深度或最深层次

本文介绍了计算嵌套列表进入的深度或最深层次的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

一个作业确实有一个问题(和令人头疼)...

A have a real problem (and a headache) with an assignment...

我正在学习编程入门课，我必须编写一个函数，给定一个列表，该函数将返回其达到的最大"深度. 例如:[1,2,3]将返回1，[1，[2,3]]将返回2 ...

I'm in an introductory programming class, and I have to write a function that, given a list, will return the "maximum" depth it goes to... For example: [1,2,3] will return 1, [1,[2,3]] will return 2...

我已经编写了这段代码(这是我能得到的最好的T_T)

I've written this piece of code (it's the best I could get T_T)

def flat(l): count=0 for item in l: if isinstance(item,list): count+= flat(item) return count+1

但是，它显然不能像应有的那样工作，因为如果存在不计入最大深度的列表，它仍然会增加计数器...

However, It obviously doens't work like it should, because if there are lists that do not count for the maximum deepness, it still raises the counter...

例如:当我将函数与[1,2，[3,4]，5，[6]，7]一起使用时，它应该返回2，但它返回3 ...

For example: when I use the function with [1,2,[3,4],5,[6],7] it should return 2, but it returns 3...

任何想法或帮助将不胜感激^^非常感谢！！我已经为此困扰了好几周...

Any ideas or help would be greatly appreciated ^^ thanks a lot!! I've been strugling with this for weeks now...

推荐答案

宽度优先，无需递归，它也适用于其他序列类型:

Breadth-first, without recursion, and it also works with other sequence types:

from collections import Sequence from itertools import chain, count def depth(seq): for level in count(): if not seq: return level seq = list(chain.from_iterable(s for s in seq if isinstance(s, Sequence)))

相同的想法，但内存消耗少得多:

The same idea, but with much less memory consumption:

from collections import Sequence from itertools import chain, count def depth(seq): seq = iter(seq) try: for level in count(): seq = chain([next(seq)], seq) seq = chain.from_iterable(s for s in seq if isinstance(s, Sequence)) except StopIteration: return level