CODEWITHSUNDEEP
matlabloopsstored-procedurestemporary
zongxian
zongxian

Reputation: 123

Matlab take more time when storage a temporary variable

When I run this example codes, I found it take more time on the 8 line. I don't why, and is there any tips let it faster??

The x and y value is not important, I just curious about the red line why take so much time, in the real codes, they are two complex expressions, why storage take so much time?

result = zeros(2, 1000);
x=0;y=0;
tic
for i =1:200
    for j =1:200
        for k =1:1000
            x=x+x/2+2*y; y=y+x*y/2;
            result(:, k) = [x;y];
        end
        sum(result); % or other operations, just for example
    end
end
toc

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Upvotes: 0

Views: 68

Answers (1)

bg2b
bg2b

Reputation: 1979

Explicit looping is generally going to be slow compared to vector operations. In this case you can do:

x=0;y=0;
tic
for i =1:200
    for j =1:200
      result=[x+(1:1000)*i; y+(1:1000)*j];
      x=x+1000*i;
      y=y+1000*j;
      sum(result); % or other operations, just for example
    end
end
toc

Edit: In your revised example, do explicit updating instead of vector construction and slice assignment:

result = zeros(2, 1000);
x=0;y=0;
tic
for i =1:200
    for j =1:200
        for k =1:1000
            x=x+x/2+2*y; y=y+x*y/2;
            result(1, k) = x;
            result(2, k) = y;
        end
        sum(result); % or other operations, just for example
    end
end
toc

Upvotes: 2

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