Sum the digits of a number

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

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)]))

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

---

Slightly faster is using a single assignment statement:

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

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