How to vectorize the following python code?

Question

I am trying to use Numpy and vectorization operations to make a section of code run faster but I don't succeed to find a solution. If somebody has an idea... Thanks.

Here's the working code with loops:

y = np.zeros(len(tab))
for i in range(len(tab)):
    s =  0
    for n in range(len(coef[0])):
        s += coef[0][n] * ((a + b * np.dot(tab[i], vectors[n])) ** d)
    y[i] = s

Where,

tab: numpy.array(N,M)
vectors: numpy.array(P,M)
coef: numpy.array(1,P)
a, b, c: constants (a = 0, if it's more easy)

Can you include a sentence at the beginning of your question describing what the code is supposed to accomplish? It's hard to give a pure programmatic assessment. — mprat
– mprat, Commented May 17, 2015 at 19:29
@mprat - it's a part of SVM algorithm implementation, similar to this implementation link - lines 83-88 — oam31
– oam31, Commented May 17, 2015 at 20:28

Divakar · Accepted Answer · 2015-05-17 20:58:39Z

You can use an approach based on np.einsum and matrix-multiplication with np.dot as listed below -

# Calculate "((a + b * np.dot(tab[i], vectors[n])) ** d)" part
p1 = (a + b*np.einsum('ij,kj->ki',tab,vectors))**d

# Include "+= coef[0][n] *" part to get the final output
y_vectorized = np.dot(coef,p1)

Runtime test

Dataset #1:

This is a quick runtime test comparing the original loopy approach with the proposed approach for some random values -

In [168]: N = 50
     ...: M = 50
     ...: P = 50
     ...: 
     ...: tab = np.random.rand(N,M)
     ...: vectors = np.random.rand(P,M)
     ...: coef = np.random.rand(1,P)
     ...: 
     ...: a = 3.233
     ...: b = 0.4343
     ...: c = 2.0483
     ...: d = 3
     ...: 

In [169]: %timeit original_approach(tab,vectors,coef,a,b,c,d)
100 loops, best of 3: 4.18 ms per loop

In [170]: %timeit proposed_approach(tab,vectors,coef,a,b,c,d)
10000 loops, best of 3: 136 µs per loop

Dataset #2:

With N, M and P as 150 each, the runtimes were -

In [196]: %timeit original_approach(tab,vectors,coef,a,b,c,d)
10 loops, best of 3: 37.9 ms per loop

In [197]: %timeit proposed_approach(tab,vectors,coef,a,b,c,d)
1000 loops, best of 3: 1.91 ms per loop

Artsiom Miksiuk · Accepted Answer · 2015-05-17 19:39:48Z

0

Looks terrible. But is it what you need?

y = array([ sum( [coef[0][n] * ((a + b * np.dot(tab[i], vectors[n])) ** d) 
            for n in range(len(vectors[0]))] ) for i in range(len(tab)) ])

answered May 17, 2015 at 19:39

Artsiom Miksiuk

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3 Comments

oam31 Over a year ago

With the same dataset, my implementation: total runtime = 28min and for your implementation: total runtime = 15min. It's better, but isn't there a better solution, without using these 2 loops?

Matthew Over a year ago

this isn't vectorizing, you still have 2 for loops even though you use numpy to speed it up a bit..

hpaulj Over a year ago

This just rewrites the 2 loop as list comprehensions. Still a Python loop, just a bit more efficient.

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