[Numpy-discussion] Iterative Matrix Multiplication
Mon Mar 1 19:39:09 CST 2010
Excellent--this setup works perfectly! In the areas I was concentrating on,
the the speed increased an order of magnitude.
However, the overall speed seems to have dropped. I believe this may be
because the heavy indexing that follows on the result is slower in numpy.
Is this a correct analysis?
What would be amazing is if I could port the other half of the algorithm to
numpy too . . .
The algorithm is for light volume extrusion of geometry. Pseudocode of the
rest of the algorithm:
1 v_array #array of beautifully transformed vertices from last step
2 n_array #array of beautifully transformed normals from last step
3 f_list #list of vec4s, where every vec4 is three vertex indices and a
4 #normal index. These indices together describe a triangle--
5 #three vertex indices into "v_array", and one normal from
6 edges = 
7 for v1index,v2index,v3index,nindex in f_list:
8 v1,v2,v3 = [v_array[i] for i in [vi1index,v2index,v3index]]
9 if angle_between(n_array[nindex],v1-a_constant_point)<90deg:
10 for edge in
11 #add "edge" to "edges"
12 #remove pairs of identical edges (also note that things like
13 #[831,326] are the same as [326,831])
14 edges2 = 
15 for edge in edges:
17 return edges2
If that needs clarification, let me know.
The goal with this is obviously to remove as many looping operations as
possible. I think the slowdown may be in lines 8, 9, and 16, as these are
the lines that index into v_array or n_array.
In line 9, the normal "n_array[nindex]" is tested against any vector from a
vertex (in this case "v1") to a constant point (here, the light's position)
see if it is less than 90deg--that is, that the triangle's front is towards
the light. I thought this might be a particularly good candidate for a
The edge pair removal is something that I have worked fairly extensively
on. By testing and removing pairs as they are added (line 11), a bit more
performance can be gained. I have put it here in 12-13 because I'm not sure
this can be done in numpy. My current algorithm works using Python's
standard sets, and using the "edge" as a key. If an edge is already in the
set of edges (in the same or reversed form), then the edge is removed from
I may be able to do something with lines 14-17 myself--but I'm not sure.
If my code above can't be simplified using numpy, is there a way to
efficiently convert numpy arrays back to standard Python lists?
As mentioned before, I'm certainly no pro with numpy--but I am learning!
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