For loop with a big matrix of 800 MB

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Trung Ngo
Trung Ngo el 26 de Jun. de 2019
Editada: Trung Ngo el 27 de Jun. de 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 comentarios
Jan
Jan el 27 de Jun. de 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 el 27 de Jun. de 2019
@Jan: Sorry for being unclear. It is 850MB matrix under a computer with 16GB RAM.

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Guillaume
Guillaume el 26 de Jun. de 2019
Editada: Guillaume el 26 de Jun. de 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);

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