Le 17/12/2012 14:42, Ronnie Ghose a écrit :
Ehh these are sort of trivial things now but why not use a
multidimensional array for the coords [n,2] and then use the numpy
operators to find the difference?
Well I guess that would reduce some overloading, but the reason I keep
my coordinates separated is for clarity. The coordinate system is a bit
more complicated than just (x,y), it's more cynlindrical like, with two
curvilinear coordinates for the surface, plus a radius corresponding to
the distance of the surface w/r to the central axis... I think that
explicitly writing down every coordinate separately avoids mistakes,
though the code looks a bit redondant then.
G.
Just trying to reduce the things you do and offload more to the numpy
fortran library. Uhm hmm you can use 4 cores then ~ just split it into
4 chunks to subtract? I find personally numpy does for loop operations
tons faster ~ so a vectorized function rather than a for loop...
On 17 December 2012 08:38, Guillaume Gay
<[email protected] <mailto:[email protected]>>
wrote:
Le 17/12/2012 13:14, Ronnie Ghose a écrit :
also i assume you're using a numpy / C structure for all of this
right? Also depending on the number of cores you have you could
try parallelizing all of your loops with multiprocessing / if
applicable ~ GPU processing
Well I rely on the `propertymap.a` and `propertymap.fa`
attributes, which are subclasses of numpy ndarray. As for
multiprocessing, of course, but I only have 4 cores, so... As for
GPU processing, I don't know whether there is a simple way to do
this in python.
I guess the real efficient way would be to dive into the
underlying C++ API, but I am not very C++ litterate and look for
something simpler.
But avoiding the inelegant loop over all edges is already
interesting...
Thanks for the suggestion anyway!
Guillaume
On 17 December 2012 05:59, Guillaume Gay
<[email protected]
<mailto:[email protected]>> wrote:
Le 17/12/2012 11:29, Tiago de Paula Peixoto a écrit :
On 12/17/2012 11:28 AM, Guillaume Gay wrote:
Hi Tiago,
Thanks for the blazing fast answer!
It will sure do the job. I can't find it in the documentation though
(it's accessible online via ipython autocompletion and magic '?', but doesn't
appear on your web site).
Oops... Indeed. I'll add it in the next release.
Cheers,
Tiago
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Ok cool..
One more point:
Is there a way to do the following differently:
edge_x = g.new_edge_property()
edge_x.a = np.array([x[e.source()] for e in g.edges()])
with x being a vector property map.
I plain English, is there a (more efficient) way to copy the
vertex property map of every edge source to an edge property
map?
Guillaume
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