Sum the digits of a number

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

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

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