Recursively counting occurrences in a nested list of numbers

Question

I'm finally getting around to recursion in Python and trying to count the number of occurrences of a target number in a list. However, I'm running into issues with counting occurrences in a nested list of numbers.

For example

def count(lst, target):

    if lst == []:
        return 0
    if lst[0] == target:
        return 1 + count(lst[1:], target)
    else:
        return 0 + count(lst[1:], target)

Output

>>> count( [1,2,3,[4,5,5],[[5,2,1],4,5],[3]], 1 )

Output: 1

Expected output: 2

Is there an easy way to flatten nested-lists in Python? Or a simple way for me to account for the fact that there is a nested-list in my code?

http://stackoverflow.com/questions/11377208/recursive-generator-for-flattening-nested-lists?rq=1 This might be of help. — Krash, Feb 17 '16 at 05:09

score 5 · Accepted Answer · edited Oct 14 '18 at 19:38

5

def count(lst, target):
    n = 0
    for i in lst:
        if i == target:
            n += 1
        elif type(i) is list:
            n += count(i, target)
    return n

edited Oct 14 '18 at 19:38

Aran-Fey

39,665
11
104
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answered Feb 17 '16 at 05:16

jfsturtz

397
4
13

score 4 · Answer 2 · answered Feb 17 '16 at 05:13

You just need an extra case to deal with lst[0] being a sublist, like:

def count(lst, target):

    if lst == []:
        return 0
    if lst[0] == target:
        return 1 + count(lst[1:], target)
    # If first element is a list, descend into it to count within it,
    #    and continue with counts of remaining elements
    elif type(lst[0]) == list:
        return count(lst[0], target) + count(lst[1:], target)
    else:
        return 0 + count(lst[1:], target)

Recursively counting occurrences in a nested list of numbers

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