Recursive function to count all items in a nested list?

Question

I'm trying to create a method that will count all the items in a nested list. So count([[3, 2] , [2]]) == 3. However, it's a Class attribute so I can't just simply do:

def count(L, target):
    s = 0
    for i in L: 
        if isinstance(i, list): 
            s += count(i, target)
        else: 
            if i == target: 
                s += 1

    return s

Rather, I tried to do this, but I get a max recursion depth error. I'm not sure why. Before you look at the code, there's a few things to keep in mind: (1) I expect the base list given to only contain lists so it will have the format: [ [], ]. Also (2) the sub lists will not contain anything except items: [ [item, item], [item] ] :

def count(self, stack=None):
    n = 0
    if stack:
        n += len(stack)
    else:
        for i in self._items:
            if isinstance(i, list):
                n += self.count(i)

    return n

user2357112 · Accepted Answer · 2014-01-26 22:55:24Z

2

    if stack:

Empty lists are considered false in a boolean context. You want if stack is not None.

Why use recursion, though? You don't need it.

def count(self):
    return sum(len(item) for item in self._items)

answered Jan 26, 2014 at 22:55

user2357112

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1 Comment

user2690972 Over a year ago

Lmao ... this is what happens when I learn a new concept and try to apply it to everything. What an easy solution. I appreciate you knocking my head back into gear :P

senshin · Accepted Answer · 2014-01-26 22:55:49Z

0

If your lists are only nested one level deep, this is easy.

class MyClass:
    def __init__(self, items):
        self.items = items

    def count(self):
        return sum(len(x) for x in self.items)

a = MyClass([[3,2],[2]])
b = MyClass([[1,2,3],[4,5,6],[7],[]])
print(a.count()) # 3
print(b.count()) # 7

answered Jan 26, 2014 at 22:55

senshin

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