How do I count the occurrences of a list item?

If you only want a single item's count, use the count method:

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

Important: this is very slow if you are counting multiple different items

Each count call goes over the entire list of n elements. Calling count in a loop n times means n * n total checks, which can be catastrophic for performance.

If you want to count multiple items, use Counter, which only does n total checks.

Use Counter if you are using Python 2.7 or 3.x and you want the number of occurrences for each element:

>>> from collections import Counter
>>> z = ['blue', 'red', 'blue', 'yellow', 'blue', 'red']
>>> Counter(z)
Counter({'blue': 3, 'red': 2, 'yellow': 1})

Counting the occurrences of one item in a list

For counting the occurrences of just one list item you can use count()

>>> l = ["a","b","b"]
>>> l.count("a")
1
>>> l.count("b")
2

Counting the occurrences of all items in a list is also known as "tallying" a list, or creating a tally counter.

Counting all items with count()

To count the occurrences of items in l one can simply use a list comprehension and the count() method

[[x,l.count(x)] for x in set(l)]

(or similarly with a dictionary dict((x,l.count(x)) for x in set(l)))

Example:

>>> l = ["a","b","b"]
>>> [[x,l.count(x)] for x in set(l)]
[['a', 1], ['b', 2]]
>>> dict((x,l.count(x)) for x in set(l))
{'a': 1, 'b': 2}

Counting all items with Counter()

Alternatively, there's the faster Counter class from the collections library

Counter(l)

Example:

>>> l = ["a","b","b"]
>>> from collections import Counter
>>> Counter(l)
Counter({'b': 2, 'a': 1})

How much faster is Counter?

I checked how much faster Counter is for tallying lists. I tried both methods out with a few values of n and it appears that Counter is faster by a constant factor of approximately 2.

Here is the script I used:

from __future__ import print_function
import timeit

t1=timeit.Timer('Counter(l)', \
                'import random;import string;from collections import Counter;n=1000;l=[random.choice(string.ascii_letters) for x in range(n)]'
                )

t2=timeit.Timer('[[x,l.count(x)] for x in set(l)]',
                'import random;import string;n=1000;l=[random.choice(string.ascii_letters) for x in range(n)]'
                )

print("Counter(): ", t1.repeat(repeat=3,number=10000))
print("count():   ", t2.repeat(repeat=3,number=10000)

And the output:

Counter():  [0.46062711701961234, 0.4022796869976446, 0.3974247490405105]
count():    [7.779430688009597, 7.962715800967999, 8.420845870045014]

Another way to get the number of occurrences of each item, in a dictionary:

dict((i, a.count(i)) for i in a)