Optimized dot product in Python

Question

The dot product of two n-dimensional vectors u=[u1,u2,...un] and v=[v1,v2,...,vn] is is given by u1*v1 + u2*v2 + ... + un*vn.

Answer 1

Just for fun I wrote a "d4" which uses numpy:

from numpy import dot
def d4(v1, v2): 
    check(v1, v2)
    return dot(v1, v2)

My results (Python 2.5.1, XP Pro sp3, 2GHz Core2 Duo T7200):

d0 elapsed:  12.1977242918
d1 elapsed:  13.885232341
d2 elapsed:  13.7929552499
d3 elapsed:  11.0952246724

d4 elapsed: 56.3278584289 # go numpy!

And, for even more fun, I turned on psyco:

d0 elapsed:  0.965477735299
d1 elapsed:  12.5354792299
d2 elapsed:  12.9748163524
d3 elapsed:  9.78255448667

d4 elapsed: 54.4599059378

Based on that, I declare d0 the winner :)

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Update

@kaiser.se: I probably should have mentioned that I did convert everything to numpy arrays first:

from numpy import array
v3 = [array(vec) for vec in v1]
v4 = [array(vec) for vec in v2]

# then
t4 = timeit.Timer("d4(v3,v4)","from dot_product import d4,v3,v4")

And I included check(v1, v2) since it's included in the other tests. Leaving it off would give numpy an unfair advantage (though it looks like it could use one). The array conversion shaved off about a second (much less than I thought it would).

All of my tests were run with N=50.

@nikow: I'm using numpy 1.0.4, which is undoubtedly old, it's certainly possible that they've improved performance over the last year and a half since I've installed it.

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Update #2

@kaiser.se Wow, you are totally right. I must have been thinking that these were lists of lists or something (I really have no idea what I was thinking ... +1 for pair programming).

How does this look:

v3 = array(v1)
v4 = array(v2)

New results:

d4 elapsed:  3.22535741274

With Psyco:

d4 elapsed:  2.09182619579

d0 still wins with Psyco, but numpy is probably better overall, especially with larger data sets.

Yesterday I was a bit bothered my slow numpy result, since presumably numpy is used for a lot of computation and has had a lot of optimization. Obviously though, not bothered enough to check my result :)