[pythran] Re: dimension-independent functions?
- From: serge Guelton <sguelton@xxxxxxxxxxxxx>
- To: pythran@xxxxxxxxxxxxx
- Date: Fri, 5 Sep 2014 14:18:20 +0200
On Fri, Sep 05, 2014 at 07:01:27AM -0400, Neal Becker wrote:
> What is the best way to write a dimension-independent function? Â For example,
> suppose I want to apply the function "sqr" 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):
> Â Â size = x.shape[0]
> Â Â y = np.empty_like (x)
> Â Â for i in range (size):
> Â Â Â Â y[i] = x[i] * x[i]
> Â Â 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.
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