[Numpy-discussion] Mean of n values within an array

Phil Ruggera pruggera at gmail.com
Thu Aug 3 00:41:13 CDT 2006


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 <davidgrant at gmail.com> wrote:
>
>
>
> On 7/29/06, Charles R Harris <charlesr.harris at gmail.com> 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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