Re: [Numpy-discussion] Overloading sqrt(5.5)*myvector
On Dec 26, 2007 3:49 AM, Gary Ruben [EMAIL PROTECTED] wrote: Sorry this isn't an answer, just noise, but for those here who don't know, Bruce is the chief maintainer of the vpython project. I have found vpython aka the visual module to be a highly attractive and useful module for teaching physics. It would be great if someone with Boost experience would try to help him out. I wouldn't want him to get falsely disillusioned with this list as I for one have been looking forward to a fully numpy-compatible version of vpython. I think the problem is that few of us are that familiar with boost/python. I have used it myself, but only for interfacing C++ classes or accessing Numpy arrays through their buffer interface. I always avoided the Numeric machinery because it looked clumsy and inefficient to me, I didn't want an embedded version of Numeric (Numarray, Numpy), I wanted speed. Anyway, I think numpy.float64 hides a normal C double, so the slowdown probably comes from the boost/python machinery. I don't think boost/python was ever updated to use Numpy in particular and the Numpy data type may be throwing it through an interpretive loop. The speed of double vs float arithmetic on Intel hardware is pretty much a wash, so should not show much difference otherwise. You can get a float32 result from sqrt In [2]: type(sqrt(float32(5.5))) Out[2]: type 'numpy.float32' But I suspect it will make little difference, probably what is needed is the corresponding C/Python data type. I think that information is available somewhere, but am not sure of the details. Someone else can probably help you there (Travis?) Chuck snip ___ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] Overloading sqrt(5.5)*myvector
Gary Ruben wrote: Sorry this isn't an answer, just noise, but for those here who don't know, Bruce is the chief maintainer of the vpython project. I have found vpython aka the visual module to be a highly attractive and useful module for teaching physics. It would be great if someone with Boost experience would try to help him out. I wouldn't want him to get falsely disillusioned with this list as I for one have been looking forward to a fully numpy-compatible version of vpython. Please keep in mind that for many of us, this is the holiday season and we are on vacation. While I'm happy to check the list and give answers that are at the front of my head, deeper answers that require exploration or experimentation are beyond my available time. I'm sure others are in a similar situation. -- Robert Kern I have come to believe that the whole world is an enigma, a harmless enigma that is made terrible by our own mad attempt to interpret it as though it had an underlying truth. -- Umberto Eco ___ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
[Numpy-discussion] Overloading sqrt(5.5)*myvector
Sorry to repeat myself and be insistent, but could someone please at
least comment on whether I'm doing anything obviously wrong, even if you
don't immediately have a solution to my serious problem? There was no
response to my question (see copy below) which I sent to both the numpy
and Boost mailing lists.
To the numpy experts: Is there something wrong, or something I
could/should change in how I'm trying to overload multiplication of a
numpy square root (or other numpy function) times my own vector
object? I'm seeing a huge performance hit in going from Numeric to numpy
because Numeric sqrt returned float whereas numpy sqrt returns
numpy.float64, so that the result is not one of my vector objects. I
don't have a problem with myvector*sqrt(5.5).
Desperately,
Bruce Sherwood
---
I'm not sure whether this is a Numpy problem or a Boost problem, so I'm
posting to both communities. (I'm subscribed to both lists, but an
attempt to post yesterday to this Boost list seems never have gotten to
the archives, so I'm trying again. My apologies if this shows up twice
here.)
In old Numeric, type(sqrt(5.5)) was float, but in numpy, type(sqrt(5.5))
is numpy.float64. This leads to a big performance hit in calculations in
a beta version of VPython, using the VPython 3D vector class, compared
with the old version that used Numeric (VPython is a 3D graphics module
for Python; see vpython.org).
Operator overloading of the VPython vector class works fine for
vector*sqrt(5.5) but not for sqrt(5.5)*vector. The following free
function catches 5.5*vector but fails to catch sqrt(5.5)*vector, whose
type ends up as numpy.ndarray instead of the desired vector, with
concomitant slow conversions in later vector calculations:
inline vector
operator*( const double s, const vector v)
{ return vector( s*v.x, s*v.y, s*v.z); }
I've thrashed around on this, including trying to add this:
inline vector
operator*( const npy_float64 s, const vector v)
{ return vector( s*v.x, s*v.y, s*v.z); }
But the compiler correctly complains that this is in conflict with the
version of double*vector, since in fact npy_float64 is actually double.
It's interesting and presumably meaningful to the knowledgeable (not me)
that vector*sqrt(5.5) yields a vector, even though the overloading
speaks of double, not a specifically numpy name:
inline vector
operator*( const double s) const throw()
{ return vector( s*x, s*y, s*z); }
VPython uses Boost, and the glue concerning vectors includes the following:
py::class_vector(vector, py::init py::optionaldouble, double,
double ())
.def( self * double())
.def( double() * self)
As far as I can understand from the Boost Python documentation, this is
the proper way to specify the left-hand and right-hand overloadings. But
do I have to add something like .def( npy_float64() * self)? Help would
be much appreciated.
Bruce Sherwood
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