[SciPy-user] how to make a sinc(x) function - "divide by zero" !
Scott Ransom
sransom at nrao.edu
Mon Dec 12 13:05:08 CST 2005
Short answer is no. At least not for arrays:
In [25]:x = asarray([-0.1, -0.05, -0.01, -0.005, -0.001, 0.0, \
0.001, 0.005, 0.01, 0.05, 0.1])
In [26]:sinc(x)
Out[26]:
array([ 0.98363164, 0.99589274, 0.99983551, 0.99995888,
0.99999836, 1. , 0.99999836, 0.99995888,
0.99983551, 0.99589274, 0.98363164])
Scalars, however, seem to be a different story. I guess that I
must have only used this function with arrays...
In [27]:sinc(0.0)
---------------------------------------------------------------------------
exceptions.ZeroDivisionError Traceback
(most recent call last)
/home/ransom/<console>
/home/ransom/<console> in sinc(xs)
ZeroDivisionError: float division
Scott
On Mon, Dec 12, 2005 at 10:30:08AM -0800, Sebastian Haase wrote:
> Scott,
> Thanks for your reply. (As Perry points out - my code actually works on
> numarray)
> But in your code: Don't you get a "division by zero" in Numeric here ?
> "where" is evaluation BOTH* results FIRST (over the ENTIRE array) and only
> AFTERWARDS "selects" based on the on the first argument
> (* actually all "three" arrays)
>
> Thanks,
> Sebastian Haase
>
>
> On Monday 12 December 2005 10:04, Scott Ransom wrote:
> > Here is a simple version that I use. It uses the Taylor
> > expansion when the argument is near zero. Note that this
> > is using Numeric and umath right now. This should be
> > trivial to change to new scipy_core (which I am not using yet):
> >
> > def sinc(xs):
> > """
> > sinc(xs):
> > Return the sinc function [i.e. sin(pi * xs)/(pi * xs)]
> > for the values xs.
> > """
> > pxs = umath.pi*xs
> > return Numeric.where(umath.fabs(pxs)<1e-3,
> > 1.0-pxs*pxs/6.0, umath.sin(pxs)/pxs)
> >
> > Scott
> >
> > On Monday 12 December 2005 12:39 pm, Arnd Baecker wrote:
> > > On Mon, 12 Dec 2005, Sebastian Haase wrote:
> > > > Hi,
> > > > I was trying to implement a "sinc" [sin(x)/x] in numarray. But the
> > > > "where"-semantics makes it choke on the x=0: 0/0 case.
> > > > (This should of course work for x being an array - so "if" is no
> > > > option) The best I could come up with was:
> > > >
> > > > def sinc(r):
> > > > na.Error.pushMode(all="ignore")
> > > > a = na.where(r, na.divide(na.sin(r),r), 1)
> > > > na.Error.popMode()
> > > > return a
> > > >
> > > > but I still seem to get a warning...
> > > >
> > > > >>> F.sinc(0)
> > > >
> > > > Warning: Encountered invalid numeric result(s) in divide
> > > > 1.0
> > > >
> > > > What is a better way of doing this ?
> > >
> > > Yuo could try to use `vectorize` (warning: untested code)
> > >
> > > def sinc(x):
> > > if x==0.0: # presumably better to check for small
> > > x return 0.0 # here ...
> > > else:
> > > return sin(x)/x
> > >
> > > sinc_vectorized=scipy.vectorize(sinc)
> > >
> > > HTH,
> > >
> > > Arnd
> > >
> > > _______________________________________________
> > > SciPy-user mailing list
> > > SciPy-user at scipy.net
> > > http://www.scipy.net/mailman/listinfo/scipy-user
>
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Scott M. Ransom Address: NRAO
Phone: (434) 296-0320 520 Edgemont Rd.
email: sransom at nrao.edu Charlottesville, VA 22903 USA
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