Count every instance of an item in a list of lists and sum them

Question

In Python 3.x, I have a list of lists:

[['a','b','c'], ['a', 'c'], ['c', 'd'] ]

I saw this answer, but it only applies to a single list. How would I do this for a list of lists? My desired output would be a sorted list (or a sortable list) of the frequency of a particular item in a list.

Something like:

{a: 2, b: 1, c: 3, d: 1 }

[How to make a flat list out of list of lists?](https://stackoverflow.com/questions/952914/how-to-make-a-flat-list-out-of-list-of-lists) — Sayse, Feb 23 '20 at 20:29
Okay, so make a flat list, then run the Collections method on it? — Jeff, Feb 23 '20 at 20:35
Yes, make a flat [iterable](https://stackoverflow.com/a/953097/1324033) — Sayse, Feb 23 '20 at 20:38
@Salt - Correct, you make an iterator that you can use within counter, no need to make a for loop either. — Sayse, Feb 23 '20 at 20:41

score 3 · Answer 1 · answered Feb 23 '20 at 20:39

3

You could use itertools.chain(*l) as an input to the Counter.

>>> l= [['a','b','c'], ['a', 'c'], ['c', 'd'] ]
>>> Counter(itertools.chain(*l))
Counter({'c': 3, 'a': 2, 'b': 1, 'd': 1})

answered Feb 23 '20 at 20:39

abc

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score 2 · Accepted Answer · answered Feb 23 '20 at 20:35

This may be solved using Counter. Counter creates a dictionary of the counts of the elements in the lists.

L = [['a','b','c'], ['a', 'c'], ['c', 'd'] ]
>>> from collections import Counter
>>> d = Counter()
>>> for sub in L:
    d.update(sub)


>>> d
Counter({'c': 3, 'a': 2, 'b': 1, 'd': 1})

score 1 · Answer 3 · answered Feb 23 '20 at 20:34

You can use Counter from collections to do this very efficiently:


In [161]: from collections import Counter
     ...: 
     ...: count = Counter()
     ...: 
     ...: lists = [['a','b','c'], ['a', 'c'], ['c', 'd']]
     ...: 
     ...: for sublist in lists:
     ...:     count += Counter(sublist)
     ...: 
     ...: print(count)
Counter({'c': 3, 'a': 2, 'b': 1, 'd': 1})

This is "one-line-able" using python's builtin sum:

In [163]: from collections import Counter
     ...: lists = [['a','b','c'], ['a', 'c'], ['c', 'd']]
     ...: 
     ...: count = sum(map(Counter, lists), start=Counter())
     ...: 
     ...: print(count)
Counter({'c': 3, 'a': 2, 'b': 1, 'd': 1})

score 0 · Answer 4 · answered Feb 23 '20 at 20:35

You can flatten the list and use Counter. If your list is arbitrarily nested.

from collections import Counter
def flatten(lst):
    if not isinstance(lst,list):
        return [lst]
    else:
        return [j for i in lst for j in flatten(i)]

print(Counter(flatten([['a','b','c'], ['a', 'c'], ['c', 'd'] ])))
#Counter({'c': 3, 'a': 2, 'b': 1, 'd': 1})

score 0 · Answer 5 · answered Feb 23 '20 at 20:37

0

You can do by itering in each sublist:

dict={} 
for sublist in list:
    for item in sublist:
        if item in dict.keys():
            dict[item] +=1
        else:
            dict[item] =1

answered Feb 23 '20 at 20:37

Renaud

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score 0 · Answer 6 · answered Feb 23 '20 at 22:15

0

you can flatten your list and apply collections.Counter:

from collections import Counter

l = [['a','b','c'], ['a', 'c'], ['c', 'd'] ]
Counter((e for i in l for e in i))

answered Feb 23 '20 at 22:15

kederrac

15,932
5
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Count every instance of an item in a list of lists and sum them

6 Answers6