How to use a callable as the setup with timeit.Timer?

Answer 1

In 2.6, you can "just do it", per the docs:

Changed in version 2.6: The stmt and setup parameters can now also take objects that are callable without arguments. This will embed calls to them in a timer function that will then be executed by timeit(). Note that the timing overhead is a little larger in this case because of the extra function calls.

Are you stuck using an old version of Python? In that case, I would suggest taking Python 2.6's timer.py sources and "backporting" them to the version you're stuck with -- should not be difficult if that version is 2.5 (gets harder the farther you need to travel back in time, of course). I would generally not recommend this for "production" code, but it's quite fine for code that supports testing, debugging, performance measurements, and the like.

Answer 2

I've asked a variant of this question here and also got an answer that didn't solve my problem. I believe we are both failing to articulate the problem in terms Pythonic.

I don't know if this is a hack, workaround, or how it is supposed to be done, but it does work:

>>> import timeit
>>> b = range(1, 1001)
>>> t = timeit.Timer('sorted(b)', setup='from __main__ import b')
>>> t.repeat(3, 100)
[0.0024309158325195312, 0.0024671554565429688, 0.0024020671844482422]
# was it really running 'sorted(b)'? let's compare'
>>> t = timeit.Timer('pass', setup='from __main__ import b')
>>> t.repeat(3, 100) 
[3.0994415283203125e-06, 2.1457672119140625e-06, 1.9073486328125e-06] 
# teeny times for 'pass', 'sorted(b)' took more time

I have read timeit.py, it works by constructing a statement and then calls eval() on it using new namespaces which have (seemingly) no connection to your main namespace. Note that since sorted is a built-in, the eval'ed expression inside timeit.repeat() has access to the name; if it was your def you'd have to from __main__ import b, myfunc.

I hope there is more proper way than this to achieve the end.