[docs] Python 2.7 Profile module documentation

[Quentin de Metz]

Hello again,

I have not received any sign of life since I suggested this modification.
What is your current status?

Is this the right place to ask?

Thanks,

Quentin




On Fri, Apr 19, 2013 at 3:59 PM, Quentin de Metz <quentin at box.com> wrote:

> Hello to the Python documentation team,
>
> I work in a software company in the Silicon Valley and we are using Python
> (2.7) for one of our upcoming releases.
>
> I'm working on performance testing for this software and have been
> fiddling quite a bit with the profiling modules available. I believe there
> is a lack of information in the official documentation and would like to
> report it.
>
> I've been trying to profile our software to see where CPU bottlenecks are.
> Despite reports of being able to use cProfile to measure CPU time, all of
> the documentation I found on the internet seems to portray cProfile
> measuring "wall time" - that is, time elapsed during a function, regardless
> of actual CPU usage. (Have a look at the UNIX "time" command if you don't
> know what I'm talking about.)
>
> For example, take the following snippet:
>
> import cProfile
> import time
>
> def sleep():
>     time.sleep(2)
>
> def count():
>     for i in xrange(100000000):
>         pass
>
> def test():
>     sleep()
>     count()
>
> *profiler = cProfile.Profile()*
> profiler.run('test()')
> profiler.print_stats
>
> Let's take a look at the functions defined here.
> Obviously the sleep function uses no CPU time, however the count function
> should generate some CPU usage.
>
> The output I get from this script is the following:
>
>    ncalls  tottime  percall  cumtime  percall filename:lineno(function)
>         1    2.045    2.045    2.045    2.045 <stdin>:1(count)
>         1    0.000    0.000    2.001    2.001 <stdin>:1(sleep)
> *        1    0.000    0.000    4.046    4.046 <stdin>:1(test)*
>         1    0.000    0.000    4.046    4.046 <string>:1(<module>)
>         1    0.000    0.000    0.000    0.000 {method 'disable' of
> '_lsprof.Profiler' objects}
>         1    2.001    2.001    2.001    2.001 {time.sleep}
>
> And reading the *line* corresponding to the "test" function does not show
> the results I expect (I expect approx 2.045).
> Once again I have scoured the Internet and found no clues as to how to
> leverage cProfile to measure only CPU time.
>
> I wrote to Mike Mueller from Python Academy who was able to provide me
> with some useful information on how to obtain what I was looking for.
>
> *On a UNIX system,* writing
> profiler = cProfile.Profile(time.clock)
> achieves the desired effet:
>
>    ncalls  tottime  percall  cumtime  percall filename:lineno(function)
>         1    1.988    1.988    1.988    1.988 <stdin>:1(count)
>         1    0.000    0.000    0.000    0.000 <stdin>:1(sleep)
> *        1    0.000    0.000    1.989    1.989 <stdin>:1(test)*
>         1    0.000    0.000    1.989    1.989 <string>:1(<module>)
>         1    0.000    0.000    0.000    0.000 {method 'disable' of
> '_lsprof.Profiler' objects}
>         1    0.000    0.000    0.000    0.000 {time.sleep}
>
>
> *On Windows*, one must first define:
> def win_clock():     return os.times()[0]
> and write
> profiler = cProfile.Profile(win_clock)
>
> which also achieves the desired effect:
> ncalls  tottime  percall  cumtime  percall filename:lineno(function)
>     1    1.404    1.404    1.404    1.404 <stdin>:1(count)
>     1    0.000    0.000    0.000    0.000 <stdin>:1(sleep)
> *    1    0.000    0.000    1.404    1.404 <stdin>:1(test)*
>     1    0.000    0.000    1.404    1.404 <string>:1(<module>)
>     1    0.000    0.000    0.000    0.000 {method 'disable' of
> '_lsprof.Profiler' objects}
>     1    0.000    0.000    0.000    0.000 {time.sleep}
>
>
> I sincerely believe some of this information belongs in the official
> Python documentation. Without Mike Mueller I think I would never have found
> it. I hope my description has been thorough and that you will have
> sufficient information to complete the existing documentation which has
> never disappointed me.
>
> Thank you very much for your consideration,
> Hoping to hear from you soon,
>
> Quentin de Metz
> Box, Inc
>
>
>
>
>
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