Efficiently accumulating a collection of sparse scipy matrices

Question

I've got a collection of O(N) NxN scipy.sparse.csr_matrix, and each sparse matrix has on the order of N elements set. I want to add all these matrices together to get a regular NxN numpy array. (N is on the order of 1000). The arrangement of non-zero elements within the matrices is such that the resulting sum certainly isn't sparse (virtually no zero elements left in fact).

At the moment I'm just doing

reduce(lambda x,y: x+y,[m.toarray() for m in my_sparse_matrices])

which works but is a bit slow: of course the sheer amount of pointless processing of zeros which is going on there is absolutely horrific.

Is there a better way ? There's nothing obvious to me in the docs.

Update: as per user545424's suggestion, I tried the alternative scheme of summing the sparse matrices, and also summing sparse matrices onto a dense matrix. The code below shows all approaches to run in comparable time (Python 2.6.6 on amd64 Debian/Squeeze on a quad-core i7)

import numpy as np
import numpy.random
import scipy
import scipy.sparse
import time

N=768
S=768
D=3

def mkrandomsparse():
    m=np.zeros((S,S),dtype=np.float32)
    r=np.random.random_integers(0,S-1,D*S)
    c=np.random.random_integers(0,S-1,D*S)
    for e in zip(r,c):
        m[e[0],e[1]]=1.0
    return scipy.sparse.csr_matrix(m)

M=[mkrandomsparse() for i in xrange(N)]

def plus_dense():
    return reduce(lambda x,y: x+y,[m.toarray() for m in M])

def plus_sparse():
    return reduce(lambda x,y: x+y,M).toarray()

def sum_dense():
    return sum([m.toarray() for m in M])

def sum_sparse():
    return sum(M[1:],M[0]).toarray()

def sum_combo():  # Sum the sparse matrices 'onto' a dense matrix?
    return sum(M,np.zeros((S,S),dtype=np.float32))

def benchmark(fn):
    t0=time.time()
    fn()
    t1=time.time()
    print "{0:16}:  {1:.3f}s".format(fn.__name__,t1-t0)

for i in xrange(4):
    benchmark(plus_dense)
    benchmark(plus_sparse)
    benchmark(sum_dense)
    benchmark(sum_sparse)
    benchmark(sum_combo)
    print

and logs out

plus_dense      :  1.368s
plus_sparse     :  1.405s
sum_dense       :  1.368s
sum_sparse      :  1.406s
sum_combo       :  1.039s

although you can get one approach or the other to come out ahead by a factor of 2 or so by messing with N,S,D parameters... but nothing like the order of magnitude improvement you'd hope to see from considering the number of zero adds it should be possible to skip.

Answer 1

I think I've found a way to speed it up by a factor of ~10 if your matrices are very sparse.

In [1]: from scipy.sparse import csr_matrix

In [2]: def sum_sparse(m):
   ...:     x = np.zeros(m[0].shape)
   ...:     for a in m:
   ...:         ri = np.repeat(np.arange(a.shape[0]),np.diff(a.indptr))
   ...:         x[ri,a.indices] += a.data
   ...:     return x
   ...: 

In [6]: m = [np.zeros((100,100)) for i in range(1000)]

In [7]: for x in m:
   ...:     x.ravel()[np.random.randint(0,x.size,10)] = 1.0
   ...:     

        m = [csr_matrix(x) for x in m]

In [17]: (sum(m[1:],m[0]).todense() == sum_sparse(m)).all()
Out[17]: True

In [18]: %timeit sum(m[1:],m[0]).todense()
10 loops, best of 3: 145 ms per loop

In [19]: %timeit sum_sparse(m)
100 loops, best of 3: 18.5 ms per loop

Answer 2

@user545424 has already posted what will likely be the fastest solution. Something in the same spirit that is more readable and ~same speed... nonzero() has all kinds of useful applications.

def sum_sparse(m):
        x = np.zeros(m[0].shape,m[0].dtype)
        for a in m:
            # old lines
            #ri = np.repeat(np.arange(a.shape[0]),np.diff(a.indptr))
            #x[ri,a.indices] += a.data
            # new line
            x[a.nonzero()] += a.data
        return x

Answer 3

Can't you just add them together before converting to a dense matrix?

>>> sum(my_sparse_matrices[1:],my_sparse_matrices[0]).todense()

Answer 4

This is not a complete answer (and I too would like to see a more detailed response), but you can gain an easy factor of two or more improvement by not creating intermediate results:

def sum_dense():
    return sum([m.toarray() for m in M])

def sum_dense2():
    return sum((m.toarray() for m in M))

On my machine (YMMV), this results in being the fastest computation. By placing the summation in a () rather than a [], we construct a generator rather than the whole list before the summation.