Efficiency help

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Andy
Andy el 13 de En. de 2012
hi, i need some help in improving this section of code. it takes over 20 seconds to run. Because i have over 150 samples to run, a 10 second improvement would mean close to 1 hour of time saved. Thanks!
for p = 1: length(unique_cell(1,:)) for f = 1:length(ions{1,p})
max_inloop(f)= (ions{2,p}(1,f))/(ions{1,p}(1,f));
max_inloop2(f)= (ions{1,p}(1,f))/(ions{2,p}(1,f));
end
max_ion_list{p} = max_inloop;
max_ion_list2{p} = max_inloop2;
end
  1 comentario
Alex
Alex el 13 de En. de 2012
One issue you are having with speed is based off of memory, each time through the loop, MATLAb is having to re-allocate memory for a larger cell matrix. Simply defining the size of the cell before you start the loop will save a LOT of time.

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Walter Roberson
Walter Roberson el 13 de En. de 2012
for p = 1 : length(unique_cell(1,:))
max_inloop = ions{2,p}(1,:) ./ ions{1,p}(1,:);
max_ion_list{p} = max_inloop;
max_ion_list2{p} = 1 ./ max_inloop;
end
  1 comentario
Andy
Andy el 13 de En. de 2012
thanks! works perfectly

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Más respuestas (1)

Alex
Alex el 13 de En. de 2012
This should work. I can't test it out though. Also, note the comment I made on your main post. Always pre-allocate memory if you can. Currently, within your loop, each time, Matlab has to take time to make the matrix larger.
max_ion_list = cell(length(unique_cell(1,:)), length(ions{1,p}));
max_ion_list2 = cell(length(unique_cell(1,:)), length(ions{1,p}));
f = 1:length(ions{1,p});
for p = 1: length(uniue_cell(1,:))
max_ion_list{p} = (ions{2,p}(1,f))/(ions{1,p}(1,f));
max_ion_list2{p} = (ions{1,p}(1,f))/(ions{2,p}(1,f));
end
  5 comentarios
Walter Roberson
Walter Roberson el 20 de En. de 2012
In R2011b, MATLAB apparently does better on memory allocation; I do not have access to that version to experiment with.
max_ion_list{p} = []; will not help with pre-allocation, as all of max_ion_list{p} is overwritten.
Anyhow, my earlier comment to you had nothing to do with preallocation, and only had to do with the difference between the "/" operator that you coded and the "./" operator that is needed.
Alex
Alex el 23 de En. de 2012
Ahh, I understand now.

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