Nested List and count()

I want to get the number of times x appears in the nested list.

if the list is:

list = [1, 2, 1, 1, 4]
list.count(1)
>>3

This is OK. But if the list is:

list = [[1, 2, 3],[1, 1, 1]]

How can I get the number of times 1 appears? In this case, 4.

Solution 1:

>>> L = [[1, 2, 3], [1, 1, 1]]
>>> sum(x.count(1) for x in L)
4

Solution 2:

itertools and collections modules got just the stuff you need (flatten the nested lists with itertools.chain and count with collections.Counter

import itertools, collections

data = [[1,2,3],[1,1,1]]
counter = collections.Counter(itertools.chain(*data))
print counter[1]

Use a recursive flatten function instead of itertools.chain to flatten nested lists of arbitrarily level depth

import operator, collections

def flatten(lst):
    return reduce(operator.iadd, (flatten(i) if isinstance(i, collections.Sequence) else [i] for i in lst))

reduce with operator.iadd has been used instead of sum so that the flattened is built only once and updated in-place

Solution 3:

Here is yet another approach to flatten a nested sequence. Once the sequence is flattened it is an easy check to find count of items.

def flatten(seq, container=None):
    if container is None:
        container = []

    for s in seq:
        try:
            iter(s)  # check if it's iterable
        except TypeError:
            container.append(s)
        else:
            flatten(s, container)

    return container


c = flatten([(1,2),(3,4),(5,[6,7,['a','b']]),['c','d',('e',['f','g','h'])]])
print(c)
print(c.count('g'))

d = flatten([[[1,(1,),((1,(1,))), [1,[1,[1,[1]]]], 1, [1, [1, (1,)]]]]])
print(d)
print(d.count(1))

The above code prints:

[1, 2, 3, 4, 5, 6, 7, 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']
1
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
12

Solution 4:

Try this:

reduce(lambda x,y: x+y,list,[]).count(1)

Basically, you start with an empty list [] and add each element of the list list to it. In this case the elements are lists themselves and you get a flattened list.

PS: Just got downvoted for a similar answer in another question!

PPS: Just got downvoted for this solution as well!