numpy.where behavior

Paul Dubois pfdubois at gmail.com
Mon Nov 13 22:13:39 CST 2006


Unfortunately, where does not have the behavior of not evaluating the second
argument where the first one is true. That would be nice (if the speed were
ok) but it isn't possible unless where is built into the language, since
where doesn't even get called until the arguments have all been calculated.
It was intended as having a different use than avoiding zero-divide. The ma
package can calculate 1/a without problem, resulting in masked results where
a is 0.0.

I put where into numeric after it had proved invaluable in Basis, even
though it has this limitation; it takes care of doing both merge and
compress.

On 13 Nov 2006 20:02:31 -0800, Tim Hochberg <tim.hochberg at ieee.org> wrote:
>
> vallis.35530053 at bloglines.com wrote:
> > Using numpy 1.0, why does
> >
> >
> >
> >
> >>>> a = numpy.array([0.0,1.0,2.0],'d')
> >>>>
> >
> >
> >>>> numpy.where(a
> >>>>
> > == 0.0,1,1/a)
> >
> >
> >
> > give the correct result, but with the warning "Warning: divide
> > by zero encountered in divide"?
> >
> >
> >
> > ? I thought that the point of where was
> > that the second expression is never used for the elements where the
> condition
> > evaluates true.
> >
> >
> >
> > If this is the desired behavior, is there a way to suppress
> > the warning?
> >
> Robert Kern has already pointed you to seterr. If you are using Python
> 2.5, you also have the option using the with statement, which is more
> convenient if you want to temporarily change the error state. You'll
> need a "from __future__ import with_statement" at the top of your file.
> Then you can temporarily disable errors as shown:
>
>      >>> a = zeros([3])
>      >>> b = 1/a  # This will warn
>     Warning: divide by zero encountered in divide
>      >>> with errstate(divide='ignore'): # But this will not
>     ...     c = 1/a
>     ...
>      >>> d = 1/a # And this will warn again since the error state is
>     restored when we exit the block
>     Warning: divide by zero encountered in divide
>
>
> Another little tidbit: this is not as general as where, and could
> probably be considered a little too clever to be clear, but:
>
>         b = 1 / (a + (a==0.0))
>
> is faster than using where in this particular case and sidesteps the
> divide by zero issue altogether.
>
> -tim
>
>
>
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