How to find the number of nested lists in a list?

Question

The function takes a list and returns an int depending on how many lists are in the list not including the list itself. (For the sake of simplicity we can assume everything is either an integer or a list.)

For example:

x=[1,2,[[[]]],[[]],3,4,[1,2,3,4,[[]] ] ]

count_list(x) # would return 8

I think using recursion would help but I am not sure how to implement it, this is what I have so far.

def count_list(a,count=None, i=None):

    if count==None and i==None:
        count=0
        i=0
    if i>len(a)
        return(count)
    if a[i]==list
       i+=1
       count+=1
       return(count_list(a[i][i],count))
    else:
        i+=1
        return(count_list(a[i]))

Answer 1

You can use a recursion function as following:

In [14]: def count(l):
    ...:     return sum(1 + count(i) for i in l if isinstance(i,list))
    ...: 

In [15]: 

In [15]: x=[1,2,[[[]]],[[]],3,4,[1,2,3,4,[[]] ] ]

In [16]: count(x)
Out[16]: 8

Answer 2

This seems to do the job:

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

Answer 3

Here is a non-recursive solution:

1. First, put every items of the list into a stack
2. Keep popping an item off the stack until it is exhausted
3. If the item is a list: a) count it, b) push every items in in to the stack

The code:

def count_list(lst):
    """ Given a master list, count the number of sub-lists """
    stack = lst[:]
    count = 0
    while stack:
        item = stack.pop()
        if isinstance(item, list):
            # If the item is a list, count it, and push back into the
            # stack so we can process it later
            count += 1
            stack.extend(item)
    return count