[Numpy-discussion] Bug in logaddexp2.reduce

Charles R Harris charlesr.harris@gmail....
Thu Apr 1 00:11:23 CDT 2010


On Wed, Mar 31, 2010 at 11:03 PM, Anne Archibald
<peridot.faceted@gmail.com>wrote:

> On 31 March 2010 16:21, Charles R Harris <charlesr.harris@gmail.com>
> wrote:
> >
> >
> > On Wed, Mar 31, 2010 at 11:38 AM, T J <tjhnson@gmail.com> wrote:
> >>
> >> On Wed, Mar 31, 2010 at 10:30 AM, T J <tjhnson@gmail.com> wrote:
> >> > Hi,
> >> >
> >> > I'm getting some strange behavior with logaddexp2.reduce:
> >> >
> >> > from itertools import permutations
> >> > import numpy as np
> >> > x = np.array([-53.584962500721154, -1.5849625007211563,
> >> > -0.5849625007211563])
> >> > for p in permutations([0,1,2]):
> >> >    print p, np.logaddexp2.reduce(x[list(p)])
> >> >
> >> > Essentially, the result depends on the order of the array...and we get
> >> > nans in the "bad" orders.  Likely, this also affects logaddexp.
> >> >
> >>
> >> Sorry, forgot version information:
> >>
> >> $ python -c "import numpy;print numpy.__version__"
> >> 1.5.0.dev8106
> >> __
> >
> > Looks like roundoff error.
>
> No. The whole point of logaddexp and logaddexp2 is that they function
> correctly - in particular, avoid unnecessary underflow and overflow -
> in this sort of situation. This is a genuine problem.
>
>
Well, it did function correctly for me ;) Just some roundoff error.


> I see it using the stock numpy in Ubuntu karmic:
> In [3]: np.logaddexp2(-0.5849625007211563, -53.584962500721154)
> Out[3]: nan
>
> In [4]: np.logaddexp2(-53.584962500721154, -0.5849625007211563)
> Out[4]: nan
>
> In [5]: np.log2(2**(-53.584962500721154) + 2**(-0.5849625007211563))
> Out[5]: -0.58496250072115608
>
> In [6]: numpy.__version__
> Out[6]: '1.3.0'
>
> In [8]: !uname -a
> Linux octopus 2.6.31-20-generic #58-Ubuntu SMP Fri Mar 12 05:23:09 UTC
> 2010 i686 GNU/Linux
>
> (the machine is a 32-bit Pentium M laptop.)
>
>
I think this is related to 32 bits somehow. The code for the logaddexp and
logaddexp2 functions is almost identical. Maybe the compiler?

Chuck
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