Counting integer in a nested list

Question

Say I have a list [1, [2, 1], 1, [3, [1, 3]], [4, [1], 5], [1], 1, [[1]]] And I want to count the number of 1's in the list. How do i do that with .count ?? Is there anyway to remove [] like enumerate(seq) which removes () then count the number of 1's in the list?

A) recursion B) flatten – Karoly Horvath Mar 14 '14 at 11:36 — Karoly Horvath, Mar 14 '14 at 11:36
C) use regex :) `len(re.findall(r'\b1\b', str(xs)))` – behzad.nouri Mar 14 '14 at 12:03 — behzad.nouri, Mar 14 '14 at 12:03

santosh-patil · Answer 1 · 2014-03-14T12:08:58.227

4

This might not be the coolest way, but it works:

l=[1, [2, 1], 1, [3, [1, 3]], [4, [1], 5], [1], 1, [[1]]]
>>> from compiler.ast import flatten
>>> flatten(l).count(1)
8

Here, as name suggests, flatten() converts the nested list into simple single level list. And counting the number of 1s from resulting list achieves the task.

edited Mar 14 '14 at 12:08

answered Mar 14 '14 at 11:50

santosh-patil

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It wont work in python 3 though as the compiler module has been removed. – Reite Mar 14 '14 at 12:20

Reite · Accepted Answer · 2014-03-14T11:58:26.883

You need a function to traverse an arbitrarily nested list. Then the counting is trivial. The traversing can be accomplished with a recursive generator:

def traverse(val):
    if isinstance(val, list):
        for v in val:
            for x in traverse(v):
                yield x 
    else:
        yield val


>>> list(traverse([1, [2, 1], 1, [3, [1, 3]], [4, [1], 5], [1], 1, [[1]]]))
[1, 2, 1, 1, 3, 1, 3, 4, 1, 5, 1, 1, 1]

This definition would be even nicer with the new yield from syntax in python 3.3, with it we could replace one of the loops:

def traverse(val):
    if isinstance(val, list):
        for v in val:
            yield from traverse(v)
    else:
        yield val

Counting integer in a nested list

2 Answers2