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

On Fri, 23 Apr 2021, Sven Schreiber wrote: > Am 23.04.2021 um 12:34 schrieb Alecos Papadopoulos: >> >> These estimates will be stored in matrices. Does it make any >> difference in computational speed if I use one matrix of [replications >> /times/ 20] dimension, or 20 matrices of [replications /times/ 1] >> dimension? >> > Not directly an answer to your question (sorry), but perhaps still > relevant: In my experience, when creating or filling a large matrix in a > loop, it's a lot faster in gretl if the matrix of the final dimension is > pre-initialized and then the rows' or columns' values are reassigned, > instead of recursively stacking (concatenating) new rows or columns to a > small initial matrix. This is very good advice: especially when matrices are large, the computational impac of memory allocation can be noticeable; consider for example this script: <hansl> set verbose off N = 10000 # method 1 set stopwatch X = {} loop i = 1 .. N     X = X | sqrt(i) endloop t1 = $stopwatch # method 2 set stopwatch X = zeros(N, 1) loop i = 1 .. N     X[i] = sqrt(i) endloop t2 = $stopwatch print t1 t2 </hansl> On my system, the output is <output>              t1 =  0.17871549              t2 =  0.0025337400 </output> As for your original question: the most efficient format depends in many cases on the nature of your problem (see below). > Also the gretl gurus told me that it's faster to work (with hansl > scripting) on entire columns instead of entire rows, due to the internal > memory layout of a gretl matrix. (I hope I got this right and didn't mix > it up.) That's absolutely correct: to be more specific, the operation of matrix slicing takes much less CPU if it involves columns instead of rows, on both sides of the assignment operator. For example, <hansl> set verbose off K = 1000 h = round(K/10) N = 1000 X = mnormal(K, K) y = zeros(K, K) # do some arbitrary slicing # method 1 set stopwatch loop i = 1 .. N     # generate a random slice and slice by column     sel = mrandgen(i, 1, K, 1, h)     y[,sel] = X[,sel] endloop t1 = $stopwatch # method 2 set stopwatch loop i = 1 .. N     # generate a random slice and slice by row     sel = mrandgen(i, 1, N, 1, h)     y[sel,] = X[sel,] endloop t2 = $stopwatch print t1 t2 </hansl> gives, on my system <output>              t1 =  0.091952580              t2 =  0.82560153 </output> -------------------------------------------------------   Riccardo (Jack) Lucchetti   Dipartimento di Scienze Economiche e Sociali (DiSES)   Università Politecnica delle Marche   (formerly known as Università di Ancona)   r.lucchetti(a)univpm.it   http://www2.econ.univpm.it/servizi/hpp/lucchetti ------------------------------------------------------- _______________________________________________ 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/