The most efficient way to get the first number of an integer in Python?

0
votes

Description of problem

I need to get the first number from a given integer. This operation will be done millions of times, therefore I need to make sure that I use the most efficient way of doing this.

If the length of the integer affects the answer, then in my case I know that the integer will always be a 2 digit number.

What I tried

I have tried the methods mentioned below. Method 1 and 2 seems slow since I have to convert back and forth. Method 3 uses //, ** and % which I could assume are also heavy on the system. Is there a better way of performing this seemingly "simple" task?

# Method 1:
first_digit = int(str(x)[0])

# Method 2:
first_digit = int(str(x)[1:])

# Method 3:
first_digit = x // 10 % 10
2
pythonpython-3.xperformanceinteger
if doing this millions of times might be better to use numpy over vanilla python – Umar.H
Did you try and do any timing of these different options? Are you sure that this will be a bottleneck in whatever you're going to use it, and that you're not trying to prematurely optimize? – Thierry Lathuille
Have you actually timed a million such operations? – Abhinav Mathur
I tried to time it now with the functions separated and it runs for a long time. But when I get the actually numbers out it says 0.0 which is weird. However, I tried using cProfile and then it told me that the operator int(str(x)[1:]) was amongst the 5 largest time consumers of my code, therefore my question. – eligolf
I advise you to give a look to stackoverflow.com/questions/5558492/… if you plan to use Numpy, Numba, Cython or PyPy. – Jérôme Richard

2 Answers

2
votes

If the number has never more than 2 digits the % 10 is useless. But could it have a single digit as well ? In that case the result would be zero which is wrong. So, assumed that the number is never more than 2 digits, the formula could be :

if x > 9: 
    return x // 10;
return x
2
votes

I used timeit module to time your methods as well as dspr's on 10 million repeats :

from timeit import timeit

n = 10000000
print(timeit(stmt='import random; n = random.randint(10, 99); x = int(str(n)[0])', number=n))
print(timeit(stmt='import random; n = random.randint(10, 99); x = int(str(n)[1:])', number=n))
print(timeit(stmt='import random; n = random.randint(10, 99); x = n // 10 % 10', number=n))
print(timeit(stmt='import random; n = random.randint(10, 99); x = n//10 if n>9 else n', number=n))

which gave me the following results :

10.7325472
11.0877854
8.493264900000003
8.550117300000004

It seems that x // 10 % 10 method is a little bit faster than the others.