python 3 problem: how to convert an extension method into a class Method

Peter Otten __peter__ at web.de
Wed Feb 27 12:14:00 CET 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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