I'm working on implementing the stochastic gradient descent algorithm for recommender systems (see Funk) using sparse matrices with Scipy.

This is how a first basic implementation looks like:

N = self.model.shape[0] #no of users
M = self.model.shape[1] #no of items
self.p = np.random.rand(N, K)
self.q = np.random.rand(M, K)
rows,cols = self.model.nonzero()        
for step in xrange(steps):
    for u, i in zip(rows,cols):
        e=self.model-np.dot(self.p,self.q.T) #calculate error for gradient
        p_temp = learning_rate * ( e[u,i] * self.q[i,:] - regularization * self.p[u,:])
        self.q[i,:]+= learning_rate * ( e[u,i] * self.p[u,:] - regularization * self.q[i,:])
        self.p[u,:] += p_temp

I have two questions and I am hoping some seasoned python coders / recommender systems experts here could provide me with some insight into this:

1) How often should the error e be adjusted? It is not clear from Funk's and Koren's papers if it should be once for every rating or if it should be once per every factor. Meaning, should i take it out of the for loop, leave it there, or put it even deeper (over the k iterations) ?

2) Unfortunately, my code is pretty slow, taking about 20 minutes on ONE iteration over a 100k ratings dataset. This means that computing the factorization over the whole dataset takes about 10 hours. I was thinking that this is probably due to the sparse matrix for loop. I've tried expressing the q and p changes using fancy indexing but since I'm still pretty new at scipy and numpy, I couldn't figure a better way to do it. Do you have any pointers on how i could avoid iterating over the rows and columns of the sparse matrix explicitly?

  • \$\begingroup\$ Which matrix is the scipy sparse matrix? \$\endgroup\$ Nov 15, 2018 at 19:52

2 Answers 2


There may be other optimizations available, but this one alone should go a long way...

When you do e = self.model - np.dot(self.p,self.q.T), the resulting e is a dense matrix the same size as your model. You later only use e[u, i] in your loop, so you are throwing away what probably are millions of computed values. If you simply replace that line with:

e = self.model[u, i] - np.dot(self.p[u, :], self.q[i, :])

and then replace e[u, i] with e, you will spare yourself a huge amount both of computations and memory accesses, which should boost performance by a lot.

  • \$\begingroup\$ Thank you. That's a great improvement. I also did some other improvements. I transposed q from the very beginning and called the column directly, instead of calling and transposing the row. I also integrated the learning rate into the error , because it was being called three times. All in all, I manage to get one step to take only about 10 seconds now! Do you think it's worth optimising further with Cython? \$\endgroup\$
    – Ana Todor
    Nov 20, 2013 at 16:21
  • \$\begingroup\$ Yes, it almost cetainly is... The problem is that you are updating your p and q vectors as you progress over the non-zero entries of model. This makes it very hard to get rid of the for loops, because p and q are different for every entry of model. If you could delay updating of p and q until the full model has been processed, then you would get different results, but you could speed things up a lot without Cython or C. With your algorithm, I cannot think of any way around that... \$\endgroup\$
    – Jaime
    Nov 20, 2013 at 16:33
  • \$\begingroup\$ Thank you for the insight. Time to learn myself some Cython then! I'll be tackling the Netflix dataset soon, so I need all the speed I can get. I'm going to mark this as answered then. \$\endgroup\$
    – Ana Todor
    Nov 20, 2013 at 16:45

Besides the recommendation that Jaime offered, it would be very easy to get performance improvements by using Numba.

There is a tutorial on Numba and matrix factorization here. When Numba works, it requires much less effort than rewriting your code in Cython. In a nutshell, you need to import numba:

from numba.decorators import jit

And then add a decorator to your matrix factorization function, by simply adding this:


Before the method definition.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.