Python - Factors Frequency Dictionary

Last Updated : 04 May, 2023

Given a list with elements, construct a dictionary with frequency of factors.

Input : test_list = [2, 4, 6, 8] Output : {1: 4, 2: 4, 3: 1, 4: 2, 5: 0, 6: 1, 7: 0, 8: 1} Explanation : All factors count mapped, e.g 2 is divisible by all 4 values, hence mapped with 4. Input : test_list = [1, 2] Output : {1: 2, 2 : 1} Explanation : Similar as above, 1 is factor of all.

Method #1 : Using loop

This is brute way in which this task can be performed. In this, the elements are iterated and required number is checked for being a factor, if yes, its frequency is increased in dictionary corresponding to its key.

Python3

# Python3 code to demonstrate working of  # Factors Frequency Dictionary # Using loop  # initializing list test_list = [2, 4, 6, 8, 3, 9, 12, 15, 16, 18]  # printing original list print("The original list : " + str(test_list))  res = dict()  # iterating till max element  for idx in range(1, max(test_list)):     res[idx] = 0     for key in test_list:                  # checking for factor          if key % idx == 0:             res[idx] += 1          # printing result  print("The constructed dictionary : " + str(res))

Output

The original list : [2, 4, 6, 8, 3, 9, 12, 15, 16, 18] The constructed dictionary : {1: 10, 2: 7, 3: 6, 4: 4, 5: 1, 6: 3, 7: 0, 8: 2, 9: 2, 10: 0, 11: 0, 12: 1, 13: 0, 14: 0, 15: 1, 16: 1, 17: 0}

Time Complexity: O(n*n), where n is the elements of dictionary
Auxiliary Space: O(n), where n is the size of dictionary

Method #2 : Using sum() + loop

This is almost similar approach to above problem. The difference being sum() is used for summation rather than a manual loop for solving problem.

Python3

# Python3 code to demonstrate working of  # Factors Frequency Dictionary # Using sum() + loop  # initializing list test_list = [2, 4, 6, 8, 3, 9, 12, 15, 16, 18]  # printing original list print("The original list : " + str(test_list))  res = dict() for idx in range(1, max(test_list)):          # using sum() instead of loop for sum computation     res[idx] = sum(key % idx == 0 for key in test_list)  # printing result  print("The constructed dictionary : " + str(res))

Output

The original list : [2, 4, 6, 8, 3, 9, 12, 15, 16, 18] The constructed dictionary : {1: 10, 2: 7, 3: 6, 4: 4, 5: 1, 6: 3, 7: 0, 8: 2, 9: 2, 10: 0, 11: 0, 12: 1, 13: 0, 14: 0, 15: 1, 16: 1, 17: 0}

Time Complexity: O(n*n) where n is the number of elements in the list “test_list”.
Auxiliary Space: O(n) where n is the number of elements in the list “test_list”.

Method #3: Using collections.Counter() and itertools.chain()

Import the collections module.
Initialize a dictionary res with all the keys as integers from 1 to the maximum value in the test_list.
Convert the test_list into a list of factors using a nested list comprehension.
Flatten the list of factors into a single list using the itertools.chain() method.
Count the frequency of each factor using the collections.Counter() method and store the result in res.
Print the resulting dictionary res.

Python3

import collections import itertools  # initializing list test_list = [2, 4, 6, 8, 3, 9, 12, 15, 16, 18]  # printing original list print("The original list : " + str(test_list))  # using collections.Counter() and itertools.chain() to construct frequency dictionary res = {i: collections.Counter(itertools.chain(*[[j for j in range(1, i+1) if key % j == 0] for key in test_list]))[i] for i in range(1, max(test_list)+1)}  # printing result print("The constructed dictionary : " + str(res))

Output

The original list : [2, 4, 6, 8, 3, 9, 12, 15, 16, 18] The constructed dictionary : {1: 10, 2: 7, 3: 6, 4: 4, 5: 1, 6: 3, 7: 0, 8: 2, 9: 2, 10: 0, 11: 0, 12: 1, 13: 0, 14: 0, 15: 1, 16: 1, 17: 0, 18: 1}

The time complexity O(n^2), where n is the length of the test_list.
The auxiliary space O(n^2), since we are creating a list of factors for each element in test_list, and then flattening it into a single list.

Method #4 : Using list(),set() and count() methods

Approach

Find the factors of each number of test_list using nested for loops and append them to an empty list x
Remove the duplicates from x using list(),set() and store it in y, create an empty dictionary res
Initiate a for loop over list y and initialise the dictionary with elements of y as keys and count of these elements in x as values
Display res

Python3

# Python3 code to demonstrate working of # Factors Frequency Dictionary # Using loop  # initializing list test_list = [2, 4, 6, 8, 3, 9, 12, 15, 16, 18]  # printing original list print("The original list : " + str(test_list))   # iterating till max element x=[] for i in range(1, max(test_list)):     for key in test_list:         if key % i == 0:             x.append(i) y=list(set(x)) res=dict() for i in y:     res[i]=x.count(i)      # printing result print("The constructed dictionary : " + str(res))

Output

The original list : [2, 4, 6, 8, 3, 9, 12, 15, 16, 18] The constructed dictionary : {1: 10, 2: 7, 3: 6, 4: 4, 5: 1, 6: 3, 8: 2, 9: 2, 12: 1, 15: 1, 16: 1}

Time Complexity : O(M*N) M - length of range 1 to max(test_list) N - length of test_list

Auxiliary Space : O(N) N - length of res dictionary