How to use python timeit when passing variables to functions?

Make it a callable:

if __name__=='__main__':    from timeit import Timer    t = Timer(lambda: superMegaIntenseFunction(10))    print(t.timeit(number=1))

Should work

Timer(superMegaIntenseFunction(10)) means "call superMegaIntenseFunction(10), then pass the result to Timer". That's clearly not what you want. Timer expects either a callable (just as it sounds: something that can be called, such as a function), or a string (so that it can interpret the contents of the string as Python code). Timer works by calling the callable-thing repeatedly and seeing how much time is taken.

Timer(superMegaIntenseFunction) would pass the type check, because superMegaIntenseFunction is callable. However, Timer wouldn't know what values to pass to superMegaIntenseFunction.

The simple way around this, of course, is to use a string with the code. We need to pass a 'setup' argument to the code, because the string is "interpreted as code" in a fresh context - it doesn't have access to the same globals, so you need to run another bit of code to make the definition available - see @oxtopus's answer.

With lambda (as in @Pablo's answer), we can bind the parameter 10 to a call to superMegaIntenseFunction. All that we're doing is creating another function, that takes no arguments, and calls superMegaIntenseFunction with 10. It's just as if you'd used def to create another function like that, except that the new function doesn't get a name (because it doesn't need one).

You should be passing a string. i.e.

t = Timer('superMegaIntenseFunction(10)','from __main__ import superMegaIntenseFunction')