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

Answer 1

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

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

A functional-style solution without loops. Recursively processes the first element of a list, and the tail of a list. Add one for each empty-list encountered (that is, once we finish processing some list, its tail becomes empty, and we add 1 to the result). And subtract 1 for the list itself.

def number_of_lists(x):
    f = lambda x: 0 if not isinstance(x,list) else (f(x[0]) + f(x[1:]) if len(x) else 1)
    return f(x) - 1

Results:

x=[1,2,[[[]]],[[]],3,4,[1,2,3,4,[[]] ] ]
number_of_lists(x)
>> 8

Answer 4

lst = [1,2,[[[]]],[[]],3,4,[1,2,3,4,["[[[[[][[[[[[[[[[[[["] ] ]

strlst = re.sub("'.*?'", '',str(lst))
print(strlst.count("[")-1)

Making the list a string will allow you to use count function on the number of [ or ] giving you an answer

But if a string within the list contains either [ or ] they will be included so removing all strings within the list using regex eliminates this problem

Answer 5

I like this tail recursive solution, although it's not much use in Python...

def count_lists(l, counter):
    if (len(l) == 0):
        return counter
    else:
        e = l.pop(0)
        if (isinstance(e, list)):
            l.extend(e)
            return count_lists(l, 1 + counter)
        else:
            return count_lists(l, counter)

x=[1,2,[[[]]],[[]],3,4,[1,2,3,4,[[]]]]
print(count_lists(x, 0))

Answer 6

You can do it with a recursion function :

def count(l):
    return sum(1+count(i) for i in l if isinstance(i,list))

Demo:

>>> x=[1,2,[[[]]],[[]],3,4,[1,2,3,4,[[]] ] ]
>>> count(x)
8