[Numpy-discussion] What is the logical value of nan?

David Cournapeau david@ar.media.kyoto-u.ac...
Wed Mar 11 00:26:34 CDT 2009


Charles R Harris wrote:
>
>
> On Tue, Mar 10, 2009 at 4:19 PM, Timothy Hochberg
> <tim.hochberg@ieee.org <mailto:tim.hochberg@ieee.org>> wrote:
>
>
>
>     On Tue, Mar 10, 2009 at 2:49 PM, Charles R Harris
>     <charlesr.harris@gmail.com <mailto:charlesr.harris@gmail.com>> wrote:
>
>
>
>         On Tue, Mar 10, 2009 at 3:16 PM, Stéfan van der Walt
>         <stefan@sun.ac.za <mailto:stefan@sun.ac.za>> wrote:
>
>             2009/3/10 Pauli Virtanen <pav@iki.fi <mailto:pav@iki.fi>>:
>             > Nonzero Python object, hence True. Moreover, it's also
>             True in Python:
>
>             Also in C:
>
>             #include <math.h>
>             #include <stdio.h>
>
>             int main() {
>                double nan = sqrt(-1);
>                printf("%f\n", nan);
>                printf("%i\n", bool(nan));
>                return 0;
>             }
>
>             $ ./nan
>             nan
>             1
>
>
>         So resolved, it is True.
>
>
>     I appear to be late to the party, but IMO it should raise an
>     exception in those cases where it's feasible to do so.
>
>
> That also seems reasonable to me. There is also the unresolved issue
> of whether casting nan to an integer should raise an exception,
> currently it is just converted to 0.

I think it is reasonable as well - but I am worried about the
integration with seterr (not just for this case, but in general in our
way toward better handling of this kind of things). I note that matlab
convert nan to 0 as well - presumably they did not handle it besides
what C guarantees (that is not much in that case I believe):

a = nan;
int32(a); % gives 0

in C:

#define _ISOC99_SOURCE       
                                                                                         

#include
<stdio.h>                                                                       

#include
<math.h>                                                                        

                                                                                         

int
main(void)                                                                           

{                                                                                        

                                                                                         

        printf("nan is %f\n",
NAN);                                                      
        printf("nan is %d\n",
(int)NAN);                                                 
        return
0;                                                                        
}

prints nan and 0 respectively - it may well be implementation dependent,
but it seems that (int)nan simply gives back the nan binary representation.

cheers,

David


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