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 listtest_list = [2, 4, 6, 8, 3, 9, 12, 15, 16, 18]# printing original listprint("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)) |
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 listtest_list = [2, 4, 6, 8, 3, 9, 12, 15, 16, 18]# printing original listprint("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)) |
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 collectionsimport itertools# initializing listtest_list = [2, 4, 6, 8, 3, 9, 12, 15, 16, 18]# printing original listprint("The original list : " + str(test_list))# using collections.Counter() and itertools.chain() to construct frequency dictionaryres = {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 resultprint("The constructed dictionary : " + str(res)) |
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 listtest_list = [2, 4, 6, 8, 3, 9, 12, 15, 16, 18]# printing original listprint("The original list : " + str(test_list))# iterating till max elementx=[]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 resultprint("The constructed dictionary : " + str(res)) |
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
