### 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?

##
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.