[Numpy-discussion] expm

[Timothy Hochberg]

On 7/20/07, Kevin Jacobs <jacobs at bioinformed.com> <bioinformed at gmail.com>
wrote:
>
> On 7/20/07, Kevin Jacobs <jacobs at bioinformed.com> <bioinformed at gmail.com>
> wrote:
> >
> > On 7/20/07, Charles R Harris < charlesr.harris at gmail.com> wrote:
> > >
> > > I expect using sqrt(x) will be faster than x**.5.
> > >
> >
> > I did test this at one point and was also surprised that sqrt(x) seemed
> > slower than **.5.  However I found out otherwise while preparing a timeit
> > script to demonstrate this observation.  Unfortunately, I didn't save the
> > precise script I used to explore this issue the first time.  On my system
> > for arrays with more than 2 elements, sqrt is indeed faster.  For smaller
> > arrays, the different is negligible, but inches out in favor of ** 0.5.
> >
>
>
> This is just not my day.  My observations above are valid for integer
> arrays, but not float arrays:
>
> sqrt(int array)   :  6.98 usec/pass
> (int array)**0.5  : 22.75 usec/pass
> sqrt(float array) :  6.70 usec/pass
> (float array)**0.5:  4.66 usec/pass
>


>From the source, it appears that powers [-1, 0, 0.5, 1, 2] are optimized for
float and complex types, while one power, 2, is optimized for other types. I
can't recall why that is however.


Generated by:
>
> import timeit
>
> n=100000
>
> t=timeit.Timer(stmt='sqrt(arange(3))',setup='from numpy import
> arange,array,sqrt\nx=arange(100)')
> print 'sqrt(int array)   : %5.2f usec/pass' % (1000000*t.timeit
> (number=n)/n)
>
> t=timeit.Timer(stmt='x**0.5',setup='from numpy import
> arange,array\nx=arange(100)')
> print '(int array)** 0.5  : %5.2f usec/pass' % (1000000*t.timeit
> (number=n)/n)
>
> t=timeit.Timer(stmt='sqrt(arange(3))',setup='from numpy import
> arange,array,sqrt\nx=arange(100,dtype=float)')
> print 'sqrt(float array) : %5.2f usec/pass' % (1000000* t.timeit
> (number=n)/n)
>
> t=timeit.Timer(stmt='x**0.5',setup='from numpy import
> arange,array\nx=arange(100,dtype=float)')
> print '(float array)**0.5: %5.2f usec/pass' % (1000000*t.timeit(number=n)/n)
>
>
> -Kevin
>
>
>
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>
>


-- 
.  __
.   |-\
.
.  tim.hochberg at ieee.org
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