On Mon, Jul 26, 2010 at 4:18 PM, Skipper Seabold <jsseab...@gmail.com> wrote: > On Mon, Jul 26, 2010 at 7:38 PM, Charles R Harris > <charlesr.har...@gmail.com> wrote: >> >> >> On Mon, Jul 26, 2010 at 5:05 PM, Skipper Seabold <jsseab...@gmail.com> >> wrote: >>> >>> On Mon, Jul 26, 2010 at 5:48 PM, Alan G Isaac <ais...@american.edu> wrote: >>> > On 7/26/2010 12:45 PM, Skipper Seabold wrote: >>> >> Right now np.linalg.det does not handle scalars or 1d (scalar) arrays. >>> > >>> > I don't have a real opinion on changing this, but I am curious >>> > to know the use case, as the current behavior seems >>> >>> Use case is just so that I can have less atleast_2d's in my code, >>> since checks are done in linalg.det anyway. >>> >>> > a) correct and b) to provide an error check. >>> > >>> >>> Isn't the determinant defined for a scalar b such that det(b) == >>> det([b]) == det([[b]])? >>> >> >> Well, no ;) Matrices have determinants, scalars don't. Where are you >> running into a problem? Is something returning a scalar where a square array >> would be more appropriate? > > No, linalg.det always returns a scalar, and I, of course, could be > more careful and always ensure that whatever the user supplies it > becomes a 2d array, but I don't like putting atleast_2d everywhere if > I don't need to. I thought that the determinant of a scalar was by > definition a scalar (e.g, google "determinant of a scalar is"), hence > > np.linalg.det(np.array([[2]])) > #2.0 > > which should either fail or if not, then I think np.linalg.det should > handle scalars and scalars as 1d arrays. > > So instead of me having to do > > b = np.array([2]) > b = np.atleast_2d(b) > np.linalg.det(b) > #2.0 > > I could just do > b = np.array([2]) > np.linalg.det(b) > #2.0 > > Regardless, doing asarray, checking if something is 2d, and then > checking if its square seems redundant and could be replaced by an > atleast_2d in linalg.slogdet which 1) takes a view as an array, 2) > ensures that the we have a 2d array, and 3) handles the scalar case. > Then we check if it's square. It doesn't really change much except > keeping me from having to put atleast_2d's in my code. > > Skipper > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@scipy.org > http://mail.scipy.org/mailman/listinfo/numpy-discussion >
imo, the determinant of a scalar should be defined as itself, based on the definition of the determinant. I don't have a vested interest in linalg's behavior in this respect, though. --Josh _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
