On 5/22/12 9:08 PM, Massimo Di Pierro wrote:
This problem is linear so probably Ram IO bound. I do not think I
would benefit much for multiple cores. But I will give it a try. In
the short term this is good enough for me.
Yeah, this what common sense seems to indicate, that RAM IO bound
hello everybody,
first of all thanks to the developed for bumpy which is very useful. I am
building a software that uses numpy+pyopencl for lattice qcd computations. One
problem that I am facing is that I need to perform most operations on arrays in
place and I must avoid creating temporary
On 05/22/2012 04:25 PM, Massimo DiPierro wrote:
hello everybody,
first of all thanks to the developed for bumpy which is very useful. I am
building a software that uses numpy+pyopencl for lattice qcd computations.
One problem that I am facing is that I need to perform most operations on
Thank you. I will look into numexpr.
Anyway, I do not need arbitrary expressions. If there were something like
numpy.add_scaled(a,scale,b)
with support for scale in int, float, complex, this would be sufficient for me.
Massimo
On May 22, 2012, at 9:32 AM, Dag Sverre Seljebotn wrote:
On
For now I will be doing this:
import numpy
import time
a=numpy.zeros(200)
b=numpy.zeros(200)
c=1.0
# naive solution
t0 = time.time()
for i in xrange(len(a)):
a[i] += c*b[i]
print time.time()-t0
# possible solution
n=10
t0 = time.time()
for i in xrange(0,len(a),n):
a[i:i+n]
On Tue, May 22, 2012 at 3:47 PM, Massimo DiPierro
massimo.dipie...@gmail.com wrote:
Thank you. I will look into numexpr.
Anyway, I do not need arbitrary expressions. If there were something like
numpy.add_scaled(a,scale,b)
with support for scale in int, float, complex, this would be
Thank you this does it.
On May 22, 2012, at 9:59 AM, Robert Kern wrote:
On Tue, May 22, 2012 at 3:47 PM, Massimo DiPierro
massimo.dipie...@gmail.com wrote:
Thank you. I will look into numexpr.
Anyway, I do not need arbitrary expressions. If there were something like
On 05/22/2012 04:54 PM, Massimo DiPierro wrote:
For now I will be doing this:
import numpy
import time
a=numpy.zeros(200)
b=numpy.zeros(200)
c=1.0
# naive solution
t0 = time.time()
for i in xrange(len(a)):
a[i] += c*b[i]
print time.time()-t0
# possible solution
On 5/22/12 8:47 PM, Dag Sverre Seljebotn wrote:
On 05/22/2012 04:54 PM, Massimo DiPierro wrote:
For now I will be doing this:
import numpy
import time
a=numpy.zeros(200)
b=numpy.zeros(200)
c=1.0
# naive solution
t0 = time.time()
for i in xrange(len(a)):
a[i] += c*b[i]
Thank you Dag,
I will look into it. Is there any documentation about ufunc?
Is this the file core/src/umath/ufunc_object.c
Massimo
On May 22, 2012, at 1:47 PM, Dag Sverre Seljebotn wrote:
On 05/22/2012 04:54 PM, Massimo DiPierro wrote:
For now I will be doing this:
import numpy
import
This problem is linear so probably Ram IO bound. I do not think I
would benefit much for multiple cores. But I will give it a try. In
the short term this is good enough for me.
On May 22, 2012, at 1:57 PM, Francesc Alted wrote:
On 5/22/12 8:47 PM, Dag Sverre Seljebotn wrote:
On 05/22/2012
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