Python program to find the sum of all even and odd digits of an integer list

Last Updated : 01 Nov, 2020

The following article shows how given an integer list, we can produce the sum of all its odd and even digits.

Input : test_list = [345, 893, 1948, 34, 2346]
Output :
Odd digit sum : 36
Even digit sum : 40
Explanation : 3 + 5 + 9 + 3 + 1 + 9 + 3 + 3 = 36, odd summation.
Input : test_list = [345, 893]
Output :
Odd digit sum : 20
Even digit sum : 12
Explanation : 4 + 8 = 12, even summation.

Method 1 : Using loop, str() and int()

In this, we first convert each element to string and then iterate for each of its element, and add to respective summation by conversion to integer.

Python3

# initializing list 
test_list = [345, 893, 1948, 34, 2346] 
  
# printing original list 
print("The original list is : " + str(test_list)) 
  
odd_sum = 0
even_sum = 0
  
for sub in test_list: 
    for ele in str(sub): 
          
        # adding in particular summation according to value  
        if int(ele) % 2 == 0: 
            even_sum += int(ele) 
        else: 
            odd_sum += int(ele) 
  
# printing result  
print("Odd digit sum : " + str(odd_sum)) 
print("Even digit sum : " + str(even_sum)) 

Output:

The original list is : [345, 893, 1948, 34, 2346]

Odd digit sum : 36

Even digit sum : 40

Method 2 : Using loop and sum()

In this, we perform task of getting summation using sum(), and loop is used to perform the task of iterating through each element.

Python3

# initializing list 
test_list = [345, 893, 1948, 34, 2346] 
  
# printing original list 
print("The original list is : " + str(test_list)) 
  
odd_sum = 0
even_sum = 0
  
for sub in test_list: 
      
    # sum() used to get summation of even and odd elements 
    odd_sum += sum([int(ele) for ele in str(sub) if int(ele) % 2 == 1]) 
    even_sum += sum([int(ele) for ele in str(sub) if int(ele) % 2 == 0]) 
  
# printing result  
print("Odd digit sum : " + str(odd_sum)) 
print("Even digit sum : " + str(even_sum)) 