Is %timeit applied to a function or reference via parameters in command line?

Question

I'm following a guide that suggests running %timeit following where following is a defined function.

I've tried using import timeit but for some reason %timeit does not run. I get a syntax error so clearly I'm not using it correctly. I've made a brief search that yielded pages on timeit library but this confuses me even more in terms of usage.

Answer 1

I suspect that you are confusing %timeit with timeit:

• %timeit is an IPython "magic" command that will only work within an IPython shell session. Example useage would be:

  In [1]: %timeit myfunc()
  

• timeitis a standard Python module - you can import timeit in your script, and use timeit.timeit("expression") etc. See the documentation for full details.

---

Here is an example showing one way you could use timeit.timeit from within an IPython session:

In [2]: def foo(): pass

In [3]: import timeit

In [4]: timeit.timeit("foo()", setup="from __main__ import foo", number=10000)
Out[4]: 0.004509925842285156

In this case, our function foo is defined within the global namespace of the IPython session, so we import it from __main__. If it was defined in some external module then it would be necessary to modify the import statement to reflect this, e.g.:

In [5]: timeit.timeit("pow(10, 3)", setup="from math import pow", number=10000)
Out[5]: 0.00642085075378418

Here I'm importing the pow function from the math module.

Answer 2

To try to reproduce %timeit from IPython's magic with timeit try this:

timeit.Timer(my_function).repeat(3, 1000)

%timeit takes the best of 3 count with n executions, where n is chosen internally (so 1000 in repeat() might not be a good choice)

reference:

-n N: execute the given statement N times in a loop. If this value is not given, a fitting value is chosen.

-r R: repeat the loop iteration R times and take the best result. Default: 3