Count number of occurrences for each item in a list

Question

I want to count how many times each item is displaying in a list. Here is what I have:

for i in range(len(alist)):
    print(alist[i], alist.count(i))

The issue with this right now is that if the list has for example 7 of the same occurrences, it is printing

a 0
a 0
a 0 
a 0
a 0
a 0
a 0

rather than what I want which is

a 7

Answer 1

You could use a collections.Counter for that:

from collections import Counter

cnt = Counter(['a', 'a', 'b', 'a'])
print(cnt)  # Counter({'a': 3, 'b': 1})

Because a Counter is a dict underneath, you can then do:

for char, count in cnt.items(): 
    print(char, count)

# a 3
# b 1

Answer 2

You've declared the value of i as an integer, so you need to count the list entry of i, not i itself.

print(alist[i], alist.count(alist[i]))

Alternatively, I'd suggest:

your_list = ['A', 'B', 'A', 'C', 'A']
for item in set(list):
    print(f'{item} occurs {your_list.count(item)} number of times.')

This makes your code more readable:) set(list) will return a set, ie all unique values within the list. Or you can use numpy.unique() to the same effect.