On 7/29/06, Charles R Harris <[EMAIL PROTECTED]> wrote:
I got rid of the for loop entirely. Usually this is the thing to do, at least this will always give speedups in Matlab and also in my limited experience with Numpy/Numeric:
def numpy_nmean2(list,n):
a = numpy.empty(len(list),dtype=float)
b = numpy.cumsum(list)
c = concatenate((b[n:],b[:n]))
a[:n] = b[:n]/(i+1)
a[n:] = (b[n:] - c[n:])/(i+1)
return a
I got no noticeable speedup from doing this which I thought was pretty amazing. I even profiled all the functions, the original, the one written by Charles, and mine, using hotspot just to make sure nothing funny was going on. I guess plain old Python can be better than you'd expect in certain situtations.
Hmmm,
I rewrote the subroutine a bit.
def numpy_nmean(list,n):
a = numpy.empty(len(list),dtype=float)b = numpy.cumsum(list)
for i in range(0,len(list)):
if i < n :
a[i] = b[i]/(i+1)
else :
a[i] = (b[i] - b[i-n])/(i+1)
return a
and got
regular python took: 0.750000 sec.
numpy took: 0.380000 sec.
I got rid of the for loop entirely. Usually this is the thing to do, at least this will always give speedups in Matlab and also in my limited experience with Numpy/Numeric:
def numpy_nmean2(list,n):
a = numpy.empty(len(list),dtype=float)
b = numpy.cumsum(list)
c = concatenate((b[n:],b[:n]))
a[:n] = b[:n]/(i+1)
a[n:] = (b[n:] - c[n:])/(i+1)
return a
--
David Grant
http://www.davidgrant.ca
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