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Counting the frequencies in a list using dictionary in Python

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Given an unsorted list of some elements(which may or may not be integers), Find the frequency of each distinct element in the list using a Python dictionary.

Example:

Input:  [1, 1, 1, 5, 5, 3, 1, 3, 3, 1,
                  4, 4, 4, 2, 2, 2, 2]
Output:  1 : 5
         2 : 4
         3 : 3
         4 : 3
         5 : 2
Explanation: Here 1 occurs 5 times, 2 
              occurs 4 times and so on...

The problem can be solved in many ways. A simple approach would be to iterate over the list and use each distinct element of the list as a key of the dictionary and store the corresponding count of that key as values. Below is the Python code for this approach:

Approach 1: Counting the frequencies in a list using a loop

In this method, we will use a Python loop to count the distinct element from the list and append it to the dictionary.

Python3




def CountFrequency(my_list):
 
    # Creating an empty dictionary
    freq = {}
    for item in my_list:
        if (item in freq):
            freq[item] += 1
        else:
            freq[item] = 1
 
    for key, value in freq.items():
        print("% d : % d" % (key, value))
 
 
# Driver function
if __name__ == "__main__":
    my_list = [1, 1, 1, 5, 5, 3, 1, 3, 3, 1, 4, 4, 4, 2, 2, 2, 2]
 
    CountFrequency(my_list)


Output: 

 1 :  5
 2 :  4
 3 :  3
 4 :  3
 5 :  2

Time Complexity: O(N), where N is the length of the list.

Approach 2: Count the frequencies in a list using  list.count() 

An alternative approach can be to use the list.count() method. 

Python3




import operator
 
 
def CountFrequency(my_list):
 
    # Creating an empty dictionary
    freq = {}
    for items in my_list:
        freq[items] = my_list.count(items)
 
    for key, value in freq.items():
        print("% d : % d" % (key, value))
 
 
# Driver function
if __name__ == "__main__":
    my_list = [1, 1, 1, 5, 5, 3, 1, 3, 3, 1, 4, 4, 4, 2, 2, 2, 2]
    CountFrequency(my_list)


Output: 

 1 :  5
 2 :  4
 3 :  3
 4 :  3
 5 :  2

Time Complexity: O(N2), where N is the length of the list. The time complexity list.count() is O(N) alone, and when used inside the loop it will become O(N2). 

Approach 3: Count the frequencies in a list using dict.get()

The dict.get() method, makes the program much shorter and makes understanding how the get method is useful instead of if…else. 

Python3




def CountFrequency(my_list):
 
    # Creating an empty dictionary
    count = {}
    for i in [1, 1, 1, 5, 5, 3, 1, 3, 3, 1, 4, 4, 4, 2, 2, 2, 2]:
        count[i] = count.get(i, 0) + 1
    return count
 
 
# Driver function
if __name__ == "__main__":
    my_list = [1, 1, 1, 5, 5, 3, 1, 3, 3, 1, 4, 4, 4, 2, 2, 2, 2]
    print(CountFrequency(my_list))


Output: 

{1: 5, 5: 2, 3: 3, 4: 3, 2: 4}

Related Article: Count frequencies of all elements in array in Python using collections module

Count the frequencies in a list using  operator.countOf() method

Approach 4: 

  1. Created a dictionary with keys as unique list values and values as unique element count in list using for loop and operator.countOf() method.
  2. Displayed the keys and values of the dictionary.

Python3




import operator
 
def CountFrequency(my_list):
 
    # Creating an empty dictionary
    freq = {}
    for items in my_list:
        freq[items] = operator.countOf(my_list, items)
 
    for key, value in freq.items():
        print("% d : % d" % (key, value))
 
# Driver function
if __name__ == "__main__":
    my_list = [1, 1, 1, 5, 5, 3, 1, 3, 3, 1, 4, 4, 4, 2, 2, 2, 2]
    CountFrequency(my_list)


Output

 1 :  5
 5 :  2
 3 :  3
 4 :  3
 2 :  4

Time Complexity : O(N*N) 
N -length of list
Auxiliary Space : O(1)



Last Updated : 24 Mar, 2023
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