Speed up my "for" loop
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Hi Everyone,
Could you please help to speed up this piece of code? What it does, it basically converts 50 Hz data to a 5 Hz data, averaging each 10 values of each field. Second and last field names are not under this conversion. And the last # of values of each field is less than ten, therefore , I use last two lines inside the loop to take care of it.
so, in total there are 11 fieldnames, each containing 12666529 values
tic
for n = [1, 3:length(names)-1]
for k = 0:count
data.(names{n})(k+1) = mean(data.(names{n})((1+10*k):(10*(k+1))));
end
data.(names{n})(count + 2) = mean(data.(names{n})(total - (total - (count+1)*10 + 1):total));
data.(names{n})(count+3:total) = [];
toc
end
The time I spent right now is 304 secs:
each loop takes around 25 secs
Elapsed time is 25.251421 seconds.
Elapsed time is 50.271446 seconds.
Elapsed time is 75.096751 seconds.
Elapsed time is 100.987307 seconds.
Elapsed time is 126.373710 seconds.
Elapsed time is 151.917404 seconds.
Elapsed time is 177.191153 seconds.
Elapsed time is 202.475424 seconds.
Elapsed time is 227.890811 seconds.
Elapsed time is 253.500255 seconds.
Elapsed time is 278.895494 seconds.
Elapsed time is 304.218428 seconds.
Elapsed time is 304.268410 seconds.
Thanks a lot! Nurlan
1 Comment
Mazin Mustafa
on 17 Jul 2016
Edited: Mazin Mustafa
on 17 Jul 2016
Hi,
I think that if you can use vectors instead of the for loop, this might speed up the calculations. Matrix operations are the fastest in MATLAB.
Accepted Answer
More Answers (1)
Ken Atwell
on 11 Sep 2013
0 votes
3 Comments
Nurlan Mukanov
on 11 Sep 2013
Ken Atwell
on 11 Sep 2013
I have not used is much, but from the doc:
- First argument is the signal
- Second argument is the desired sample rate
- Third argument is the current sample rate
It could actually be slower because it may be doing something more mathematically sophisticated than taking the mean, but it is worth trying. More on resampling here.
Nurlan Mukanov
on 12 Sep 2013
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