What is the fastest way to do repeated element wise matrix multiplication?

9 Ansichten (letzte 30 Tage)
Given a matrix `A`, I need to multiply with another constant vector `B`, N times (N > 1 million). The size of `A` is `9000x1` and `B` is `9000x1000`.
The code is currently evaluated in the following way (random values taken for example):
B = rand(9000,1000); % B is fixed, does not depend on i
N = 1000000;
for i=1:N
rand('seed',i);
A = rand(9000,1); % A is 9000x1 matrix which varies with i
% prod = A.*B; % prod is 9000 x 1000 matrix
% sum_temp = sum(product); % sum_temp is 1 x 1000 matrix
% Edit
sum_temp = A.' * B;
% do multiple pperations with sum_temp
% result(i) = some_constant;
end
I used Profiler to see which line is taking the most time and it is the 2nd line (prod = A.*R;). The problem is that N is very large and the code is taking over several days to complete.
I am about to try the parallel computing toolbox (GPU computing), but are there any suggestions on what I can do in the basic version?
How can I reduce the run-time of such codes in MATLAB?

Akzeptierte Antwort

Stephen23
Stephen23 am 19 Feb. 2024
Verschoben: Stephen23 am 19 Feb. 2024
Reduce the number of operations inside the loop by replacing TIMES and SUM with MTIMES (of course adjusting the matrix/vector orientations to suit).
  5 Kommentare
Stephen23
Stephen23 am 19 Feb. 2024
Bearbeitet: Stephen23 am 19 Feb. 2024
@Aravind Varma Dantuluri: did that make enough difference to the speed?
Another possibilty would be to look at some kind of parallel processing:
Aravind Varma Dantuluri
Aravind Varma Dantuluri am 19 Feb. 2024
@Stephen23, using MTIMES definitely helped. Now I am checking out pagemtimes and then will check parallel computing capabilities. Thanks for your help!

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Kategorien

Mehr zu Linear Algebra finden Sie in Help Center und File Exchange

Produkte


Version

R2023b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by