Azzera filtri
Azzera filtri

For loop with a big matrix of 800 MB

5 visualizzazioni (ultimi 30 giorni)
Trung Ngo
Trung Ngo il 26 Giu 2019
Modificato: Trung Ngo il 27 Giu 2019
Hi all,
I am searching through a big matrix file ( nearly 850 MB) and tried to loop through that matrix but it take a massive amount of time. I wonder if are there any method to run a simple for loop in this "big data" matrix. My code is at belown
clearvars;
load('landorocean_250_W.mat')
tA_world = tall(A_world);
tx_world = tall(x_world);
ty_world = tall(y_world);
[m_1,n_1]=size(A_world);
for i=1:m_1
for j=1:n_1
if gather(tA_world(i,j) == -32768)
island_250m_1(i,j)=0;
else
island_250m_1(i,j)=1;
end
end
end
Thank you for your time,
Sincerely,
  4 Commenti
Jan
Jan il 27 Giu 2019
@Trung Ngo: In the opriginal question you wrote ">800mb". Of course this might mean 800 TeraByte also. Then only a tall array can catch this. Please post the maximum size and how much free RAM you have. It matters of ">800MB" means 850MB on a computer with 8GB of RAM, or if it means 800GB on a machine with 2 GB RAM.
Trung Ngo
Trung Ngo il 27 Giu 2019
@Jan: Sorry for being unclear. It is 850MB matrix under a computer with 16GB RAM.

Accedi per commentare.

Risposta accettata

Guillaume
Guillaume il 26 Giu 2019
Modificato: Guillaume il 26 Giu 2019
Using gather as you have done on each individual element of the tall array completely nullifies the advantage of tall arrays. You may as well not have used tall arrays, it would have been faster.
The loop was not needed in the first place:
island_250m_1 = tA_world == -32768; %creates a tall array the same size as tA_world
%now you can call gather if you want to convert that tall array into a concrete array
island_250m_1 = gather(island_250m_1);

Più risposte (0)

Categorie

Scopri di più su Programming in Help Center e File Exchange

Prodotti


Release

R2019a

Community Treasure Hunt

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

Start Hunting!

Translated by