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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
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    if doing this millions of times might be better to use numpy over vanilla python Commented Oct 16, 2020 at 11:40
  • 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? Commented Oct 16, 2020 at 11:40
  • Have you actually timed a million such operations? Commented Oct 16, 2020 at 11:48
  • 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. Commented Oct 16, 2020 at 12:00
  • I advise you to give a look to stackoverflow.com/questions/5558492/… if you plan to use Numpy, Numba, Cython or PyPy. Commented Oct 16, 2020 at 13:23

2 Answers 2

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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
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Not faster, but more concise: return x if x < 10 else x // 10
2

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.

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6 Comments

the same (for 100000 only) on repl.it/languages/python3 gives 1.6996094920032192 0.9015585849992931 0.7443572329939343 0.5641695599915693
The timing include the generation of the random number and the import which are much slower. Thus, the timings are not very accurate.
You're right. With a fixed integer (x = 57), I get these results on an iMac i7 4ghz : 5.5298409462 5.52825784683 0.631176948547 0.556658983231 while the first was more than 19ms using a random number
@JérômeRichard Yes, the timings are not accurate but the relative timings between methods are, as both import et number generation statements appear in all the timeit calls. Or am I missing something ?
@BastienBroussard Not really. On my machine import random; n = random.randint(10, 99); x = n // 10 % 10 gives 5.23 while import random; n = random.randint(10, 99); x = n // 10 % 10; y = n // 10 % 10 gives 5.54. We expect the second result to be twice the first. The variation of the random generation might be significant compared to the computation of x. As a result, you cannot be sure n // 10 % 10 is faster than n//10 if n>9 else n statistically speaking because of the very slight gap between the two last timings. I advise to use the setup argument of timeit.
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