On Mon, Sep 29, 2008 at 10:50 PM, Robert Kern <robert.kern@gmail.com> wrote:
On Mon, Sep 29, 2008 at 23:02, Charles R Harris <charlesr.harris@gmail.com> wrote:
On Mon, Sep 29, 2008 at 9:02 PM, David Cournapeau <david@ar.media.kyoto-u.ac.jp> wrote:
Charles R Harris wrote:
So the proposition is, sign, max, min return nan when any of the arguments is nan.
Note that internally, signbit (the C function) returns an integer.
That is the signature of the ufunc. It could be changed... I believe the actual signbit of nan is undefined but I suppose we could return -1 in
the
nan case. That would be a fairly typical error signal for integers.
numpy.signbit() should work like C99 signbit() (where possible), IMO. It can only return (integer) 0 or 1, and it does differentiate between NAN and -NAN. I don't think we should invent new semantics if we can avoid it. I think we can change what the platform provides, but only in the direction of C99, IMO. I see signbit() as more along the lines of functions like isnan() than log().
Sounds reasonable.
There is no C99 cognate for numpy.sign(), and it is a float->float function, so I think we could make it return NAN. C99's copysign(x,y) is almost a cognate (e.g. numpy.sign(y) == copysign(1.0,y) except for y==+/-0.0), but since it does fall down on y==0, I don't think it's determinative for y==NAN.
Sign doesn't distinguish +/-0 . The sign bit of 0 is explicitly cleared in the current (and former) code by adding +0 to the result.
[~]$ man copysign COPYSIGN(3) BSD Library Functions Manual COPYSIGN(3)
NAME copysign -- changes the sign of x to that of y
SYNOPSIS #include <math.h>
double copysign(double x, double y); ... [~]$ gcc --version i686-apple-darwin9-gcc-4.0.1 (GCC) 4.0.1 (Apple Inc. build 5465) Copyright (C) 2005 Free Software Foundation, Inc. This is free software; see the source for copying conditions. There is NO warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
[~]$ cat foo.c #include <stdio.h> #include <math.h>
int main(int argc, char **argv) { printf("signbit(NAN) = %d\n", signbit(NAN)); printf("signbit(-NAN) = %d\n", signbit(-NAN)); printf("copysign(1.0, NAN) = %g\n", copysign(1.0, NAN)); printf("copysign(1.0, -NAN) = %g\n", copysign(1.0, -NAN)); return 0; } [~]$ gcc -std=c99 -o foo foo.c -lm [~]$ ./foo signbit(NAN) = 0 signbit(-NAN) = 1 copysign(1.0, NAN) = 1 copysign(1.0, -NAN) = -1
Hmm, signbit(NAN) = 0 signbit(-NAN) = -2147483648 copysign(1.0, NAN) = 1 copysign(1.0, -NAN) = -1 signbit(0.0) = 0 signbit(-0.0) = -2147483648 copysign(1.0, 0.0) = 1 copysign(1.0, -0.0) = -1 Looking at the standard, signbit is only required to return a non-zero value for negatives. I think we need to be more explicit for numpy. How about 1? Chuck