Calling a function is faster than not calling it?
Terry Reedy
tjreedy at udel.edu
Sun May 10 12:37:10 EDT 2015
On 5/10/2015 5:58 AM, Steven D'Aprano wrote:
> Why is calling a function faster than bypassing the function object and
> evaluating the code object itself? And not by a little, but by a lot?
>
> Here I have a file, eval_test.py:
>
> # === cut ===
> from timeit import Timer
>
> def func():
> a = 2
> b = 3
> c = 4
> return (a+b)*(a-b)/(a*c + b*c)
>
>
> code = func.__code__
> assert func() == eval(code)
>
> t1 = Timer("eval; func()", setup="from __main__ import func")
> t2 = Timer("eval(code)", setup="from __main__ import code")
eval has 3 parameters. I believe omitting the last two results in
globals() and locals() calls, but at least one of them. If {} is
passed, access to builtins is added.
> # Best of 10 trials.
> print (min(t1.repeat(repeat=10)))
> print (min(t2.repeat(repeat=10)))
Adding
g = globals()
t3 = Timer("eval(code, g)", setup="from __main__ import code, g")
print (min(t3.repeat(repeat=10)))
>>>
0.3992733933018515
0.6967548323372563
0.49210603735894587
2/3s of the extra time disappears, but there is still extra time needed
for processing the two extra arguments. Passing g twice has no effect
> [steve at ando ~]$ python2.7 eval_test.py
> 0.804041147232
> 1.74012994766
> Directly eval'ing the code object is easily more than twice as expensive
> than calling the function, but calling the function has to eval the code
Calling the function executes the code via its .__call__ method.
Perhaps eval wraps the code object in a function object with .__call__.
I don't know the comparative internals.
--
Terry Jan Reedy
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