This is fun. Two more variations with 1000 reps instead of 100 for better timing:
def numpy_nmean_conv_nl_tweak1(list,n):
b = numpy.ones(n,dtype=float)
a = numpy.convolve(list,b,mode="full")
a[:n] /= numpy.arange(1, n + 1)
a[n:] /= n
return a[:len(list)]
def numpy_nmean_conv_nl_tweak2(list,n):
b = numpy.ones(n,dtype=float)
a = numpy.convolve(list,b,mode="full")
a[:n] /= numpy.arange(1, n + 1)
a[n:] *= 1.0/n
return a[:len(list)]
Which gives
numpy convolve took: 2.630000 sec.
numpy convolve noloop took: 0.320000 sec.
numpy convolve noloop tweak1 took: 0.250000 sec.
numpy convolve noloop tweak2 took: 0.240000 sec.
Chuck
On 8/2/06, Phil Ruggera <
[EMAIL PROTECTED]> wrote:
A variation of the proposed convolve routine is very fast:
regular python took: 1.150214 sec.
numpy mean slice took: 2.427513 sec.
numpy convolve took: 0.546854 sec.
numpy convolve noloop took: 0.058611 sec.
Code:
# mean of n values within an array
import numpy, time
def nmean(list,n):
a = []
for i in range(1,len(list)+1):
start = i-n
divisor = n
if start < 0:
start = 0
divisor = i
a.append(sum(list[start:i])/divisor)
return a
t = [1.0*i for i in range(1400)]
start = time.clock()
for x in range(100):
reg = nmean(t,50)
print "regular python took: %f sec."%( time.clock() - start)
def numpy_nmean(list,n):
a = numpy.empty(len(list),dtype=float)
for i in range(1,len(list)+1):
start = i-n
if start < 0:
start = 0
a[i-1] = list[start:i].mean(0)
return a
t = numpy.arange(0,1400,dtype=float)
start = time.clock()
for x in range(100):
npm = numpy_nmean(t,50)
print "numpy mean slice took: %f sec."%(time.clock() - start)
def numpy_nmean_conv(list,n):
b = numpy.ones(n,dtype=float)
a = numpy.convolve(list,b,mode="full")
for i in range(0,len(list)):
if i < n :
a[i] /= i + 1
else :
a[i] /= n
return a[:len(list)]
t = numpy.arange(0,1400,dtype=float)
start = time.clock()
for x in range(100):
npc = numpy_nmean_conv(t,50)
print "numpy convolve took: %f sec."%( time.clock() - start)
def numpy_nmean_conv_nl(list,n):
b = numpy.ones(n,dtype=float)
a = numpy.convolve(list,b,mode="full")
for i in range(n):
a[i] /= i + 1
a[n:] /= n
return a[:len(list)]
t = numpy.arange(0,1400,dtype=float)
start = time.clock()
for x in range(100):
npn = numpy_nmean_conv_nl(t,50)
print "numpy convolve noloop took: %f sec."%(time.clock () - start)
numpy.testing.assert_equal(reg,npm)
numpy.testing.assert_equal(reg,npc)
numpy.testing.assert_equal(reg,npn)
On 7/29/06, David Grant <[EMAIL PROTECTED] > wrote:
>
>
>
> On 7/29/06, Charles R Harris <[EMAIL PROTECTED]> wrote:
> >
> > 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
>
> 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.
>
> --
> David Grant
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