[pythran] Re: dimension-independent functions?

  • From: Neal Becker <ndbecker2@xxxxxxxxx>
  • To: pythran@xxxxxxxxxxxxx
  • Date: Fri, 5 Sep 2014 08:28:36 -0400

Thanks.

Your version seems much more elegant

def sqrv(x):
        return x * x

I didn't know pythran would produce an element-wise iteration over the
whole array from this code.


On Fri, Sep 5, 2014 at 8:18 AM, serge Guelton <sguelton@xxxxxxxxxxxxx>
wrote:

> On Fri, Sep 05, 2014 at 07:01:27AM -0400, Neal Becker wrote:
> > What is the best way to write a dimension-independent function?  For
> example,
> > suppose I want to apply the function &quot;sqr&quot; element-wise to an
> array
> > of any number
> > of dimensions N.
> >
> > I see one way is to explicitly instantiate the function for each desired
> number
> > of dimensions:
>
> >
> > import numpy as np
> > #pythran export sqr(float64[])
> > #pythran export sqr(float64[][])
> > def sqr (x):
> >     size = x.shape[0]
> >     y = np.empty_like (x)
> >     for i in range (size):
> >         y[i] = x[i] * x[i]
> >     return y
>
> That's the way it should be :-/
>
> I tried this:
>
>     import numpy as np
>     #pythran export sqr(float64[])
>     #pythran export sqr(float64[][])
>     #pythran export sqrv(float64[])
>     #pythran export sqrv(float64[][])
>     #pythran export sqr(float32[])
>     #pythran export sqr(float32[][])
>     #pythran export sqrv(float32[])
>     #pythran export sqrv(float32[][])
>     def sqr(x):
>         size = x.shape[0]
>         y = np.empty_like (x)
>         for i in range(size):
>             y[i] = x[i] * x[i]
>         return y
>     def sqrv(x):
>         return x * x
>
> compiled with:
>
> pythran -DUSE_BOOST_SIMD -march=native sqr.py
>
> and benchmarked with:
>
> python -m timeit -s 'import numpy as np ; r =
> np.asarray(np.random.rand(1000,1000), dtype=np.float64); import sqr'
> 'sqr.sqrv(r)'
>
> I end up with pythran performing roughly as fast as numpy for the
> vectorized version, faster when using explicit loops, and faster on
> float32 thanks to vectorisation.
>
>


-- 
*Those who don't understand recursion are doomed to repeat it*

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