I believe that Tony is suggesting manually applying the sparse operator rather than explicitly constructing it and then applying it. This is a common (and significant) performance optimization when a sparse operator is only used once or twice, as the construction of the sparse matrix is often *more* expensive than applying it once.
The answer to your question (since you are applying the sparse matrix from the right to a row vector) would be a loop of the form: forall nonzero triplets (i,j,value), y[j] += x[i] * value. On Tuesday, September 15, 2015 at 8:52:46 AM UTC-7, Frank Kampas wrote: > > Could you post a link to the part of the documentation that describes how > to do that? > > On Tuesday, September 15, 2015 at 3:53:11 AM UTC-4, Tony Kelman wrote: >> >> Instead of constructing a sparse matrix in the inner loop it would be >> more efficient to write an in place stencil kernel function to perform the >> equivalent operation. > >
