Given that A is a sparse matrix, norm(A(i,:)) takes a very long time. Why and can one do better ?

I am computing norm(A(i,:)) within a loop, where A is a sparse matrix. This seems to be a bottleneck, is there anyway to perform this faster ? Thank you.

[Walter Roberson]

Can you recode with the matrix transposed? Accessing sparse by columns is faster than by row.

Answer 1

[John D'Errico]

You really don't want to use loops to do something like this.

>> A = sprand(10000,10000,0.001);

>> timeit(@() norm(A(1,:)))
ans =
            0.000121650132
            
>> timeit(@() sqrt(sum(A.^2,2)))
ans =
            0.000442972132

So, the time to compute the norm of just one of the rows was a significant fraction of the total time to compute the norm of all rows at once.

Answer 2

[Walter Roberson]

If you were going to do this repeatedly, then

%one time cost
At = A.';

%then
norm(At(:,1))

is over 100 times faster than

norm(A(1,:))

For John's suggestion of A = sprand(10000,10000,0.001); then the cost of transposing A is roughly the same as the cost of calculating 11 row norms of A.

... And of course, it might happen to be practical to work with the transposed version instead of the original, so in some cases there would be no computation cost, just the cost of changing the code.

Answer 3

[James Tursa]

Will you eventually need all of the rows? E.g., do this once at the beginning outside the loop

n = sqrt(sum(A.^2,2))

And then just pick off the element you want each iteration.

Alternatively, you could do this calculation one row at a time in a mex routine in order to avoid the data copy associated with forming the A(i,:) row explicitly.