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August 2, 2020 04:47 pm GMT

Ways to count occurrences of the list items in Python

Introduction
Count
operator
Counter
Pandas
loop and Dict

Introduction

Hey guys,
hope you all are doing well during this crisis. I have been coding in python for a while and came across a situation where I had to count some occurrences for a particular element in the list and I came across a few cool techniques to do this. So, let's get started.

Count

well, this is the most interesting and easy method to use, let's look at an example

>>> l = ['a', 'a', 'a', 'a', 'a', 'b', 'b', 'b', 'b', 'c', 'c', 'c', 'd', 'd', 'e']>>> l.count('a')5>>> l.count('e')1

we can also use this method to get a count of all distinct values at once.

>>> l = ['a', 'a', 'a', 'a', 'a', 'b', 'b', 'b', 'b', 'c', 'c', 'c', 'd', 'd', 'e']>>> [[x,l.count(x)] for x in set(l)][['d', 2], ['b', 4], ['c', 3], ['e', 1], ['a', 5]]>>> dict((x,l.count(x)) for x in set(l)){'d': 2, 'b': 4, 'c': 3, 'e': 1, 'a': 5}

operator

In this method, you have to import the operator module. Let's have a quick example

>>> l = ['a', 'a', 'a', 'a', 'a', 'b', 'b', 'b', 'b', 'c', 'c', 'c', 'd', 'd', 'e']>>> import operator as op>>> op.countOf(l,'a')5

let's go to the next method

Counter

Here you have to import Counter from the collections module.

>>> l = ['a', 'a', 'a', 'a', 'a', 'b', 'b', 'b', 'b', 'c', 'c', 'c', 'd', 'd', 'e']>>> from collections import Counter>>> Counter(l)Counter({'a': 5, 'b': 4, 'c': 3, 'd': 2, 'e': 1})

This can also be used to get a count for a single element

>>> from collections import Counter>>> c = Counter(l)>>> c['a']5

Pandas

pandas is a very popular library and it can also be used for this purpose

>>> import pandas as pd>>> l = ['a', 'a', 'a', 'a', 'a', 'b', 'b', 'b', 'b', 'c', 'c', 'c', 'd', 'd', 'e']>>> my_count = pd.Series(l).value_counts()>>> my_counta    5b    4c    3d    2e    1dtype: int64

in case if you are looking for a count of a particular element

# you should run steps in the previous snippet before running this i.e. importing pandas, etc>>> my_count['a']5

loop & Dict

This is the fastest way among all the above, let's go to an example

>>> def countElement(a):...     g = {}...     for i in a:...         if i in g:...             g[i] +=1...         else:...             g[i] =1...     return g...>>> countElement(l){'a': 5, 'b': 4, 'c': 3, 'd': 2, 'e': 1}