Counting depth or the deepest level a nested list goes to

问题

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

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...

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

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...

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

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

回答1:

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)))

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

回答2:

Here is one way to write the function

depth = lambda L: isinstance(L, list) and max(map(depth, L))+1

I think the idea you are missing is to use max()

回答3:

Let's first rephrase your requirements slightly.

The depth of a list is one more than the maximum depth of its sub-lists.

Now, this can be translated directly to code:

def depth(l):
    if isinstance(l, list):
        return 1 + max(depth(item) for item in l)
    else:
        return 0

回答4:

easy with recursion

def flat(l):
    depths = []
    for item in l:
        if isinstance(item, list):
            depths.append(flat(item))
    if len(depths) > 0:
        return 1 + max(depths)
    return 1

回答5:

Did it in one line of python :)

enjoy

def f(g,count=0): return count if not isinstance(g,list) else max([f(x,count+1) for x in g])

回答6:

Abusive way: Say your list is called mylist
mybrackets = map(lambda x: 1 if x=='[' else -1, [x for x in str(mylist) if x=='[' or x==']'])
maxdepth = max([sum(mybrackets[:i+1]) for i in range(len(mybrackets))])

This converts your list to a list of opening and closing brackets, then finds the largest number of opening brackets that occur before the corresponding closing bracket occurs.

回答7:

A way that does not need any additional modules and has the same speed, no matter what depth:

def depth(nested):
    instring = False
    count = 0
    depthlist = []
    for char in repr(nested):
        if char == '"' or char == "'":
            instring = not instring
        elif not instring and ( char == "[" or char == ")" ):
            count += 1
        elif not instring and ( char == "]" or char == ")" ):
            count -= 1
        depthlist.append(count)
    return(max(depthlist))

Basically, what this does is convert the list to a string using repr(). Then for every character in this string equal to "(" or "[" it increases the variable count. for the closing brackets it decreases count. It then returns the maximum that count has reached.

回答8:

I extended the hammar's answer for every iterable (strings disabled by default):

def depth(arg, exclude=None):
    if exclude is None:
        exclude = (str, )

    if isinstance(arg, tuple(exclude)):
        return 0

    try:
        if next(iter(arg)) is arg:  # avoid infinite loops
            return 1
    except TypeError:
        return 0

    try:
        depths_in = map(lambda x: depth(x, exclude), arg.values())
    except AttributeError:
        try:
            depths_in = map(lambda x: depth(x, exclude), arg)
        except TypeError:
            return 0

    try:
        depth_in = max(depths_in)
    except ValueError:
        depth_in = 0

    return 1 + depth_in

回答9:

A short addition to what has been said so it can handle empty lists too:

def list_depth(list_of_lists):
    if isinstance(list_of_lists, list):
        if(len(list_of_lists) == 0):
            depth = 1
        else:
            depth = 1 + max([list_depth(l) for l in list_of_lists])
    else:
        depth = 0
    return depth

回答10:

@John's solution is excellent, but to address the empty list cases, like [], [[]], you may need to do something like this

depth = lambda L: isinstance(L, list) and (max(map(depth, L)) + 1) if L else 1

回答11:

In Numpy, you can convert the data structure to a numpy array and use its library functions. arr.shape gives length per layer, so we can len() the shape and get the depth of the structure:

import numpy as np

def f( lists )
  arr = np.array( lists )
  return len(arr.shape)

f( [[[1,2],[3,4]],[[3,4],[5,6]]] ) # results in 3
f( [[1,2],[3,4]] ) # results in 2
f( [1,2] )  # results in 1
f( [] )  # results in 1

Numpy docs for shape: https://docs.scipy.org/doc/numpy/reference/generated/numpy.ndarray.shape.html

来源：https://stackoverflow.com/questions/6039103/counting-depth-or-the-deepest-level-a-nested-list-goes-to