Sum the digits of a number

Question

If I want to find the sum of the digits of a number, i.e.:

• Input: 932
• Output: 14, which is (9 + 3 + 2)

Answer 1

def digitsum(n):
    result = 0
    for i in range(len(str(n))):
        result = result + int(str(n)[i:i+1])
    return(result)

"result" is initialized with 0.

Inside the for loop, the number(n) is converted into a string to be split with loop index(i) and get each digit. ---> str(n)[i:i+1]

This sliced digit is converted back to an integer ----> int(str(n)[i:i+1])

And hence added to result.

Answer 2

This might help

def digit_sum(n):
    num_str = str(n)
    sum = 0
    for i in range(0, len(num_str)):
        sum += int(num_str[i])
    return sum

Answer 3

Both lines you posted are fine, but you can do it purely in integers, and it will be the most efficient:

def sum_digits(n):
    s = 0
    while n:
        s += n % 10
        n //= 10
    return s

or with divmod:

def sum_digits2(n):
    s = 0
    while n:
        n, remainder = divmod(n, 10)
        s += remainder
    return s

---

Even faster is the version without augmented assignments:

def sum_digits3(n):
   r = 0
   while n:
       r, n = r + n % 10, n // 10
   return r

---

> %timeit sum_digits(n)
1000000 loops, best of 3: 574 ns per loop

> %timeit sum_digits2(n)
1000000 loops, best of 3: 716 ns per loop

> %timeit sum_digits3(n)
1000000 loops, best of 3: 479 ns per loop

> %timeit sum(map(int, str(n)))
1000000 loops, best of 3: 1.42 us per loop

> %timeit sum([int(digit) for digit in str(n)])
100000 loops, best of 3: 1.52 us per loop

> %timeit sum(int(digit) for digit in str(n))
100000 loops, best of 3: 2.04 us per loop

Answer 4

num = 123
dig = 0
sum = 0
while(num > 0):
  dig = int(num%10)
  sum = sum+dig
  num = num/10

print(sum) // make sure to add space above this line

Answer 5

Found this on one of the problem solving challenge websites. Not mine, but it works.

num = 0            # replace 0 with whatever number you want to sum up
print(sum([int(k) for k in str(num)]))

Answer 6

Here is a solution without any loop or recursion but works for non-negative integers only (Python3):

def sum_digits(n):
    if n > 0:
        s = (n-1) // 9    
        return n-9*s
    return 0