Nested List and count()

Question

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.

Asked By: Yugo Kamo

||

Source

Answer 1

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!

Answered By: manojlds

Answer 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

Answered By: Imran

Answer 3

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

Answered By: Thomas Edleson

Answer 4

If there is only one level of nesting flattening can be done with this list comprenension:

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

Answered By: dting

Answer 5

For the heck of it: count to any arbitrary nesting depth, handling tuples, lists and arguments:

hits = lambda num, *n: ((1 if e == num else 0)
    for a in n
        for e in (hits(num, *a) if isinstance(a, (tuple, list)) else (a,)))

lst = [[[1,(1,),((1,(1,))), [1,[1,[1,[1]]]], 1, [1, [1, (1,)]]]]]
print sum(hits(1, lst, 1, 1, 1))

15

Answered By: samplebias

Answer 6

def nested_count(lst, x):
    return lst.count(x) + sum(
        nested_count(l,x) for l in lst if isinstance(l,list))

This function returns the number of occurrences, plus the recursive nested count in all contained sub-lists.

>>> data = [[1,2,3],[1,1,[1,1]]]
>>> print nested_count(data, 1)
5

Answered By: shang

Answer 7

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

Answered By: sateesh

Answer 8

The following function will flatten lists of lists of any depth^(a) by adding non-lists to the resultant output list, and recursively processing lists:

def flatten(listOrItem, result = None):
    if result is None: result = []      # Ensure initial result empty.

    if type(listOrItem) != type([]):    # Handle non-list by appending.
        result.append(listOrItem)
    else:
        for item in listOrItem:         # Recursively handle each item in a list.
            flatten(item, result)
    return result                       # Return flattened container.

mylist = flatten([[1,2],[3,'a'],[5,[6,7,[8,9]]],[10,'a',[11,[12,13,14]]]])
print(f'Flat list is {mylist}, count of "a" is {mylist.count("a")}')

print(flatten(7))

Once you have a flattened list, it’s a simple matter to use count on it.
The output of that code is:

Flat list is [1, 2, 3, 'a', 5, 6, 7, 8, 9, 10, 'a', 11, 12, 13, 14], count of "a" is 2
[7]

Note the behaviour if you don’t pass an actual list, it assumes you want a list regardless, one containing just the single item.

If you don’t want to construct a flattened list, you can just use a similar method to get the count of any item in the list of lists, with something like:

def deepCount(listOrItem, searchFor):
    if type(listOrItem) != type([]): # Non-list, one only if equal.
        return 1 if listOrItem == searchFor else 0

    subCount = 0                     # List, recursively collect each count.
    for item in listOrItem:
        subCount += deepCount(item, searchFor)
    return subCount

deepList = [[1,2],[3,'a'],[5,[6,7,[8,9]]],[10,'a',[11,[12,13,14]]]]
print(f'Count of "a" is {deepCount(deepList, "a")}')
print(f'Count of 13  is {deepCount(deepList, 13)}')
print(f'Count of 99  is {deepCount(deepList, 99)}')

As expected, the output of this is:

Count of "a" is 2
Count of 13  is 1
Count of 99  is 0

^(a) Up to the limits imposed by Python itself of course, limits you can increase by just adding this to the top of your code:

import sys
sys.setrecursionlimit(1001) # I believe default is 1000.

I mention that just in case you have some spectacularly deeply nested structures but you shouldn’t really need it. If you’re nesting that deeply then you’re probably doing something wrong 🙂

Answered By: paxdiablo

Nested List and count()

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