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

Charles R Harris charlesr.harris at gmail.com
Sat Jul 29 12:20:51 CDT 2006


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.

The precision of the new routine is admittedly suspect.  Convolve can also
be used:

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)]

Which gives

numpy convolve took: 0.260000 sec.


You might actually want mode="valid" which would only return averages over
runs of length n.

One oddity here is that I think convolution should be slower than cumsum and
that raises my suspicions about the latter, it seems much too slow.

Chuck

On 7/29/06, Phil Ruggera <pruggera at gmail.com> wrote:
>
> I rewrote some python code using numpy to do a performance comparison.
> The results were the opposite of what I wanted.  Numpy was slower
> than Python without numpy.  Is there something wrong with my approach?
>
> # 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):
>     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):
>     numpy_nmean(t,50)
> print "numpy took: %f sec."%(time.clock() - start)
>
>
> Results:
> regular python took: 1.215274 sec.
> numpy took: 2.499299 sec.
>
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