Efficiency help

2 次查看(过去 30 天)
Andy
Andy 2012-1-13
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 个评论
Alex
Alex 2012-1-13
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.

请先登录,再进行评论。

采纳的回答

Walter Roberson
Walter Roberson 2012-1-13
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

更多回答(1 个)

Alex
Alex 2012-1-13
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 个评论
Walter Roberson
Walter Roberson 2012-1-20
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 2012-1-23
Ahh, I understand now.

请先登录,再进行评论。

类别

Help CenterFile Exchange 中查找有关 Logical 的更多信息

标签

Community Treasure Hunt

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

Start Hunting!

Translated by