Issues with using timeit in ipython

Question

I was quickly trying to time 2 functions in ipython, m1() and m2() doing the same task with 2 different implementation.

In [23]: %timeit for x in range(100): m1(a)
10000 loops, best of 3: 57.6 us per loop

In [24]: %timeit for x in range(100): m2(a)
10000 loops, best of 3: 108 us per loop

Result: the first implementation is almost 2x faster. So far, so good.

Out of curiousity, I changed the range of the for loop above, and now I am at a loss making sense of the output.

In [25]: %timeit for x in range(1): m2(a)
1000000 loops, best of 3: 1.29 us per loop

In [26]: %timeit for x in range(10): m2(a)
100000 loops, best of 3: 10.8 us per loop

In [27]: %timeit for x in range(1000): m2(a)
1000 loops, best of 3: 1.06 ms per loop

What exactly is the for loop doing here? And why do the value of the number of loops decrease on increasing the range value?

PS: I was using this as the reference. Also, please edit the title to something better if it doesn't exactly convey my question.

Why are you using the for loop here? It is totally enough to do %timeit m1(a)! Otherwise you time the for loop as well. — Jakob
– Jakob, Commented Oct 20, 2013 at 6:38
Yup, I didn't realise initially what the for loop was doing :| — Anshul Goyal
– Anshul Goyal, Commented Oct 20, 2013 at 6:39

SheetJS · Accepted Answer · 2013-10-20 04:04:13Z

2

timeit is counting the execution time for the entire block.

So what you are seeing is:

running m2(a) 1 time takes 1.29 us
running m2(a) 10 times takes 10.8 us
running m2(a) 1000 times takes 1.06 ms

Which makes some sense, since 1.06ms = 1060 us, roughly 1000x the baseline (and 10.8 us is roughly 10x the baseline)

As for the number of loops, timeit aims to run within a reasonable time:

$ python -mtimeit -h
...
If -n is not given, a suitable number of loops is calculated by trying
successive powers of 10 until the total time is at least 0.2 seconds.

answered Oct 20, 2013 at 4:04

SheetJS

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3 Comments

Anshul Goyal Over a year ago

That explains the output, but doesn't answer either of my questions - What exactly is the for loop doing here? and why do the value of the number of loops decrease on increasing the range value?

SheetJS Over a year ago

@ansh0l updated my response -- it's explained in the help for timeit

Anshul Goyal Over a year ago

Thanks, that explains it. I wrote an answer for detailed explanation for my future self :)

Anshul Goyal · Accepted Answer · 2013-10-20 06:20:23Z

So I finally figured it out what is happening, thanks to @Nirk's answer.

In [26]: %timeit for x in range(100): m2(a)
10000 loops, best of 3: 108 us per loop

Here,

%timeit => ipython magic call

for x in range(100): m2(a) => the statement being executed. Based on the range value, time for execution increases/decreases for each run

10000 loops => Minimum number of loops that timeit will implicilty run, based on timeit modules constraint of minimum 0.2 s of total time

best of 3: 108 us per loop => average time taken by the best 3 loop run by timeit.

Assuming time for each run is same as average of best 3, time for each run = 108 us

minimum loops needed = 10^x, where x is minimum positive integer satisfying 1.08 * (10^-4) * (10^x) > 2 * (10^-1)

i.e, the minimum x for which (10^x) > 1.85 * (10^3) => x = 4

Hence minimum loops needed = 10^x = 10000 loops.

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