On Wed, Jan 7, 2009 at 23:44, Ryan May <rma...@gmail.com> wrote: > Nicolas ROUX wrote: > > Hi, > > > > I need help ;-) > > I have here a testcase which works much faster in Matlab than Numpy. > > > > The following code takes less than 0.9sec in Matlab, but 21sec in Python. > > Numpy is 24 times slower than Matlab ! > > The big trouble I have is a large team of people within my company is > ready to replace Matlab by Numpy/Scipy/Matplotlib, > > but I have to demonstrate that this kind of Python Code is executed with > the same performance than Matlab, without writing C extension. > > This is becoming a critical point for us. > > > > This is a testcase that people would like to see working without any code > restructuring. > > The reasons are: > > - this way of writing is fairly natural. > > - the original code which showed me the matlab/Numpy performance > differences is much more complex, > > and can't benefit from broadcasting or other numpy tips (I can later give > this code) > > > > ...So I really need to use the code below, without restructuring. > > > > Numpy/Python code: > > ##################################################################### > > import numpy > > import time > > > > print "Start test \n" > > > > dim = 3000 > > > > a = numpy.zeros((dim,dim,3)) > > > > start = time.clock() > > > > for i in range(dim): > > for j in range(dim): > > a[i,j,0] = a[i,j,1] > > a[i,j,2] = a[i,j,0] > > a[i,j,1] = a[i,j,2] > > > > end = time.clock() - start > > > > print "Test done, %f sec" % end > > ##################################################################### > <SNIP> > > Any idea on it ? > > Did I missed something ? > > I think you may have reduced the complexity a bit too much. The python > code > above sets all of the elements equal to a[i,j,1]. Is there any reason you > can't > use slicing to avoid the loops? > > Yes, I think so. I think the testcase is a matter of python loop vs matlab loop rather than python vs matlab.
-- Cheers, Grissiom
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