For loop slows down on row 25 of 3D matrix and does not advance and original speed. What may be the issue?

Asked by mashtine on 24 Jun 2015

Latest activity Commented on by mashtine on 24 Jun 2015

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Hi everyone,

I am using the following code to calculate a given parameter. I have included a progress bar for testing purposes and I noted that it always sticks at 19%. Upon further snooping around, I realize the for loop does not stop but greatly slows down after 19% even though the progress bar says only 2 min 33 secs left.

I am not running out of memory (checking with top command in the terminal) and there seems to be nothing odd about the data/anything different from surrounding data.

testcase1 = single(zeros(length(lon),length(lat),length(time)));
testcase2 = single(zeros(length(lon),length(lat),length(time)));
progressbar
for i = 1:length(lon);
    for j = 1:length(lat);
        for t = 1:length(time);
              if C(i,j,t) >= -500 && C(i,j,t) <= -12
                  testcase1(i,j,t) = A(i,j,t)*(0.35*((-(B(i,j,t)/(K*C(i,j,t))))^(2/3)) + (2 -(10/B(i,j,t))))^(1/2);       
              else              
                  testcase2(i,j,t) = 2*A(i,j,t)*((1 -(10/B(i,j,t)))^(1/2));
              end
          end
      end
      progressbar(i/121) % Update progress bar
  end

Any ideas why this would stick/slow down here?

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1 Answer

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Answer by Guillaume on 24 Jun 2015

Accepted Answer

If that's the above code that you're running, I don't see any reason for it to hang.

However, there are many ways to speed it up. For a start, passing the output class (single) to zeros instead of generating a double matrix and then converting it to single would be faster:

testcase1 = zeros(numel(lon), numel(lat), numel(time), 'single');
testcase2 = zeros(numel(lon), numel(lat), numel(time), 'single');

Secondly, the loop is completely unnecessary (and is a performance killer). Use vectorised operations:

tf = C >= -500 & C <= -12;
testcase1(tf) = A(tf).*(0.35*((-(B(tf)./(K*C(tf)))).^(2/3)) + (2 -(10./B(tf)))).^(1/2);
testcase2(~tf) = 2*A(~tf).*((1 -(10./B(~tf))).^(1/2));

Just three lines!

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mashtine on 24 Jun 2015

The vectorized version quickly increased the memory being used and after killing the script after a controlled time, a lot less was written to the output variables. Not sure why. I did not know that the for loop slowed down versus stopping at the time of my comment so forgive the confusion there! Nonetheless, your last suggestion (thanks a million) does the trick. My memory surges but if produces an output in a fraction of the time!

The superfluous brackets were just an editing issues sorry!

Thanks again.

Guillaume on 24 Jun 2015

The vectorised version will either produces the whole output or nothing at all if it runs out of memory. There's no in between.

mashtine on 24 Jun 2015

Ah ok, makes sense.

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