# Archives for 4 Sep,2009

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## Python: sum of digits in a string

I have a function I wrote for a project euler that calculates the sum of the digits in a number. This is my first attempt which simply converts each letter to an integer and sums them.

Method 1:

def digitsum(x):
total=0
for letter in str(x):
total+=int(letter)

I thought that this could be improved using ord, which converts a letter into its decimal ascii number. Numbers ‘0’, ‘1’, ‘2’ … ‘9’ correspond to the ascii values of 48 – 57 and then took the moduli of this with 48 to give the integer value. I later realised that this was completely nonsensical and should have just subtracted 48, but I decided to include it for the purposes of the speed test.

Method 2:

def digitsum2(x):
total=0
for letter in str(x):
total+=ord(letter)%48

Method 3:

def digitsum3(x):
total=0
for letter in str(x):
total+=ord(letter)-48

Speed Test:
The test uses a long number and one million repetitions for each method.

from time import time

# .. functions go here

# Nice long number to sum
x=981234153134415646571899783156122451653

tic = time()
for i in range(0,1000000):
digitsum(x)
print time() - tic, 'Seconds elapsed'

tic = time()
for i in range(0,1000000):
digitsum2(x)
print time() - tic, 'Seconds elapsed'

tic = time()
for i in range(0,1000000):
digitsum3(x)
print time() - tic, 'Seconds elapsed'

Results:

#Method 1
29.3496568203 Seconds elapsed
#Method 2
12.185685873 Seconds elapsed
#Method 3
9.59367895126 Seconds elapsed

So we can see that the first method is much slower, avoiding the integer conversion by using ord speeds it up the function by ~60% and that using subtraction rather than modulus (a division based operation) saves a further ~20% on top of this.