How to count the frequency of the elements in an unordered list?

Question

In Python 2.7 (or newer), you can use collections.Counter:

import collections

a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
counter=collections.Counter(a)

print(counter)
# Counter({1: 4, 2: 4, 3: 2, 5: 2, 4: 1})

print(counter.values())
# [4, 4, 2, 1, 2]

print(counter.keys())
# [1, 2, 3, 4, 5]

print(counter.most_common(3))
# [(1, 4), (2, 4), (3, 2)]

print(dict(counter))
# {1: 4, 2: 4, 3: 2, 5: 2, 4: 1}

If you are using Python 2.6 or older, you can download it here.

Note: You should sort the list before using groupby.

You can use groupby from itertools package if the list is an ordered list.

a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
from itertools import groupby
[len(list(group)) for key, group in groupby(a)]

Output:

[4, 4, 2, 1, 2]

update: Note that sorting takes O(n log(n)) time.

Python 2.7+ introduces Dictionary Comprehension. Building the dictionary from the list will get you the count as well as get rid of duplicates.

>>> a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
>>> d = {x:a.count(x) for x in a}
>>> d
{1: 4, 2: 4, 3: 2, 4: 1, 5: 2}
>>> a, b = d.keys(), d.values()
>>> a
[1, 2, 3, 4, 5]
>>> b
[4, 4, 2, 1, 2]

To count the number of appearances:

from collections import defaultdict

appearances = defaultdict(int)

for curr in a:
    appearances[curr] += 1

To remove duplicates:

a = set(a) 

In Python 2.7+, you could use collections.Counter to count items

>>> a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
>>>
>>> from collections import Counter
>>> c=Counter(a)
>>>
>>> c.values()
[4, 4, 2, 1, 2]
>>>
>>> c.keys()
[1, 2, 3, 4, 5]