If I want to find the sum of the digits of a number, i.e.:
932
14
, which is (9 + 3 + 2)
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