# Sum the digits of a number - python

### 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)`

What is the fastest way of doing this?

I instinctively did:

``````sum(int(digit) for digit in str(number))
``````

and I found this online:

``````sum(map(int, str(number)))
``````

Which is best to use for speed, and are there any other methods which are even faster?

1
61
1/6/2017 5:44:48 PM

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
``````

But both lines you posted are fine.

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
``````
95
1/7/2016 5:32:31 PM

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
``````