Python | Sort list elements by frequency

Given a list containing repeated and non-repeated elements, the task is to sort the given list on basis of the frequency of elements. Let’s discuss few methods for the same.

Method #1: Using collections.counter()

Python3

# Python code to demonstrate
# sort list by frequency
# of elements

from collections import Counter

ini_list = [1, 2, 3, 4, 4, 5, 5, 5, 5, 7,
            1, 1, 2, 4, 7, 8, 9, 6, 6, 6]

# printing initial ini_list
print ("initial list", str(ini_list))

# sorting on basis of frequency of elements
result = [item for items, c in Counter(ini_list).most_common()
                                      for item in [items] * c]

# printing final result
print("final list", str(result))

Output:

initial list [1, 2, 3, 4, 4, 5, 5, 5, 5, 7, 1, 1, 2, 4, 7, 8, 9, 6, 6, 6]
final list [5, 5, 5, 5, 1, 1, 1, 4, 4, 4, 6, 6, 6, 2, 2, 7, 7, 3, 8, 9]

Time Complexity: O(n), where n is the length of the list ini_list 
Auxiliary Space: O(n) additional space of size n is created where n is the number of elements in the ini list 

Method #2: Using iterables

Python3

# Python code to demonstrate
# sort list by frequency
# of elements

from collections import Counter
from itertools import repeat, chain

ini_list = [1, 2, 3, 4, 4, 5, 5, 5, 5, 7,
            1, 1, 2, 4, 7, 8, 9, 6, 6, 6]

# printing initial ini_list
print ("initial list", str(ini_list))

# sorting on basis of frequency of elements
result = list(chain.from_iterable(repeat(i, c)
         for i, c in Counter(ini_list).most_common()))

# printing final result
print("final list", str(result))

Output:

initial list [1, 2, 3, 4, 4, 5, 5, 5, 5, 7, 1, 1, 2, 4, 7, 8, 9, 6, 6, 6]
final list [5, 5, 5, 5, 1, 1, 1, 4, 4, 4, 6, 6, 6, 2, 2, 7, 7, 3, 8, 9]

Time Complexity: O(n), where n is the length of the list ini_list 
Auxiliary Space: O(n) additional space of size n is created where n is the number of elements in the ini list 

Method #3: Using sorted

Python3

# Python code to demonstrate
# sort list by frequency
# of elements

ini_list = [1, 1, 2, 2, 2, 3, 3, 3, 3, 3,
               5, 5, 5, 4, 4, 4, 4, 4, 4]

# printing initial ini_list
print ("initial list", str(ini_list))

# sorting on basis of frequency of elements
result = sorted(ini_list, key = ini_list.count,
                                reverse = True)

# printing final result
print("final list", str(result))

Output:

initial list [1, 1, 2, 2, 2, 3, 3, 3, 3, 3, 5, 5, 5, 4, 4, 4, 4, 4, 4]
final list [4, 4, 4, 4, 4, 4, 3, 3, 3, 3, 3, 2, 2, 2, 5, 5, 5, 1, 1]

Time Complexity: O(n*logn)
Auxiliary Space: O(n), where n is length of list

Method #4: Using dictionary comprehension

Here is an example of how the to use dictionary comprehension for creating the frequency dictionary:

Python3

ini_list = [1, 2, 3, 4, 4, 5, 5, 5, 5, 7, 1, 1, 2, 4, 7, 8, 9, 6, 6, 6]
# create a dictionary to store the frequency of each element using dictionary comprehension
frequency_dict = {element: ini_list.count(element) for element in ini_list}

# sort the dictionary by values in descending order
sorted_dict = {k: v for k, v in sorted(frequency_dict.items(), key=lambda item: item[1], reverse=True)}

# create a new list with the elements in the sorted order
result = []
for element, frequency in sorted_dict.items():
    result.extend([element] * frequency)

# print the final result
print(result)
#This code is contributed by Edula Vinay Kumar Reddy

Output

[5, 5, 5, 5, 1, 1, 1, 4, 4, 4, 6, 6, 6, 2, 2, 7, 7, 3, 8, 9]

Using dictionary comprehension allows us to create the frequency dictionary in a single line of code, which can make the code more concise and easier to read. 

Time complexity: O(nlogn)

Auxiliary Space: O(n)