[Gretl-users] Re: One large matrix or many small matrices for storing simulation results?

On Fri, 23 Apr 2021, Alecos Papadopoulos wrote: > Thanks for both good advises, fortunately I have been following them > all along, namely, declare the matrices before hand, and if slice, > then slice by column. > > However, none replies to my question. So I run an experiment based on > Jack's example > > <hansl> > #case 1 > X = zeros(N, 3) > set stopwatch > loop i = 1..N >     X[i,1] = sqrt(i) >     X[i,2] = sqrt(i) >     X[i,3] = sqrt(i) > endloop > t1 = $stopwatch > > #case 2 > X1 = zeros(N, 1) > X2 = zeros(N, 1) > X3 = zeros(N, 1) > set stopwatch > loop i = 1..N >     X1[i,1] = sqrt(i) >     X2[i,1] = sqrt(i) >     X3[i,1] = sqrt(i) > endloop > t2 = $stopwatch > </hansl> > > and I got > > <output> > t1 = 0.641354 > t2 = 0.653886 > </output> > > So, perhaps contrary to first a priori impressions, it appears that > slicing down the column dimension, if anything, makes gretl slow down > a bit. When timings are as close as in this case, you should try executing the two variants in both orders. With very intensive computation the registers will heat up and run faster by the time you get to the second variant. Cutting your big matrix into a set of column vectors can simplify the indexation, but you're (partially) missing out on that by providing a redundant second index: skip that, and you may find the column-vectors case runs slightly faster. I replicated your example with N = 100000; gave just the row index in the column-vectors case, as in X1[i] = sqrt(i) and named the times t_mat (using a single matrix) and t_vec (using per-column vectors). What I'm seeing is: t_vec 0.423s t_mat 0.441s t_vec 0.420s There's not much in it, but the set-of-vectors approach seems marginally faster. I very much doubt you'd see a speed-up if your "slicing" didn't yield vectors (that is, if it just reduced an m x n matrix to a set of m x k matrices with k > 1), since then there would be no reduction in indexation overhead. However, it we take the computational task to be literally what's in your example (which we probably don't want to do!), there's a MUCH faster way of doing it in whole-matrix mode: precompute a vectorized solution: matrix y = sqrt(seq(1,N))' loop j = 1..3     X[,j] = y endloop The timing for that with N = 1000000 is 0.0149 seconds. Allin _______________________________________________ Gretl-users mailing list -- gretl-users(a)gretlml.univpm.it To unsubscribe send an email to gretl-users-leave(a)gretlml.univpm.it Website: https://gretlml.univpm.it/postorius/lists/gretl-users.gretlml.univpm.it/