Python – Factors Frequency Dictionary

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}

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}