Sum the digits of a number

Answer #1:

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 #2:

If you want to keep summing the digits until you get a single-digit number (one of my favorite characteristics of numbers divisible by 9) you can do:

def digital_root(n):
    x = sum(int(digit) for digit in str(n))
    if x < 10:
        return x
    else:
        return digital_root(x)

Which actually turns out to be pretty fast itself…

%timeit digital_root(12312658419614961365)

10000 loops, best of 3: 22.6 µs per loop

Answer #3:

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 #4:

Doing some Codecademy challenges I resolved this like:

def digit_sum(n):
    digits = []
    nstr = str(n)
    for x in nstr:
        digits.append(int(x))
    return sum(digits)

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

Answer #7:

The best way is to use math.
I knew this from school.(kinda also from codewars)

def digital_sum(num):
    return (num % 9) or num and 9

Just don’t know how this works in code, but I know it’s maths

If a number is divisible by 9 then, it’s digital_sum will be 9,
if that’s not the case then num % 9 will be the digital sum.

Answer #8:

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.