python 3 problem: how to convert an extension method into a class Method
Peter Otten
__peter__ at web.de
Wed Feb 27 06:14:00 EST 2013
Robin Becker wrote:
> On 26/02/2013 18:38, Peter Otten wrote:
>> Robin Becker wrote:
> ...........3:
>>
>> $ python -m timeit -s 'from new import instancemethod
>>> from math import sqrt
>>> class A(int): pass
>>> A.m = instancemethod(sqrt, None, A)
>>> a = A(42)
>>> ' 'a.m()'
>> 1000000 loops, best of 3: 0.5 usec per loop
>> $ python -m timeit -s 'from math import sqrt
>>> class A(int):
>>> def m(self):
>>> return sqrt(self)
>>> a = A(42)
>>> ' 'a.m()'
>> 1000000 loops, best of 3: 0.473 usec per loop
>>
>>
> this analysis might be relevant if I wanted to use sqrt. However, in my
> case the method takes
>
>
>
> py C
> utf8 bytes 50 20 usec
> unicode 39 15
>
> here py refers to a native python method and C to the extension method
> after adding to the class. Both are called via an instance of the class.
I think you misunderstood. You compare the time it takes to run the function
coded in C and its Python equivalent -- that difference is indeed
significant.
But what I was trying to measure was the difference between two ways to wrap
the C function:
Given a function cfunc implemented in C and the two ways of turning it into
a method
(1)
class A(object):
pass
def method(self, ...):
return cfunc(self, ...)
A.method = method
(2)
class A(object):
pass
A.method = new.instancemethod(cfunc, None, A)
I interpreted my timeit results as an indication that both ways have roughly
the same overhead (method (1) being 0.027 usec faster).
I don't have your code available, that's why I picked math.sqrt as an
example for cfunc.
I expect that you will get a similar result with your actual cfunc and
therefore can (and should IMO) use method (1) in both Python 2 and 3
-- but of course not being able to measure it myself it may turn out I'm
wrong.
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