loadind data loop fastly?

2 views (last 30 days)
denis bertin
denis bertin on 10 Oct 2017
Commented: OCDER on 11 Oct 2017
Hi everybody, Please i have this code that load data too long time, i want to read it fastly(reduce time execution), somebody can i help me?
%Data is the 381x247x3 matrix; if true
LastGood = floor((122.5 - 50) / Header.Fstep) + 1 - 1; % value =29
FirstGood = ceil((140 - 50) / Header.Fstep) + 1 + 1; % value =38
disp(LastGood); % value =29
disp(FirstGood); % value =38
disp(size(Data, 2)); % value =247
for(k = 1:size(Data, 2))
InterpolazioneReale1 = fit(cat(2,[1:LastGood],[FirstGood:381])',cat(1,real(Data(1:LastGood,k,1)),real(Data(FirstGood:end,k,1))),'splineinterp');
InterpolazioneReale2 = fit(cat(2,[1:LastGood],[FirstGood:381])',cat(1,real(Data(1:LastGood,k,2)),real(Data(FirstGood:end,k,2))),'splineinterp');
InterpolazioneReale3 = fit(cat(2,[1:LastGood],[FirstGood:381])',cat(1,real(Data(1:LastGood,k,3)),real(Data(FirstGood:end,k,3))),'splineinterp');
InterpolazioneImmaginaria1 = fit(cat(2,[1:LastGood],[FirstGood:381])',cat(1,imag(Data(1:LastGood,k,1)),imag(Data(FirstGood:end,k,1))),'splineinterp');
InterpolazioneImmaginaria2 = fit(cat(2,[1:LastGood],[FirstGood:381])',cat(1,imag(Data(1:LastGood,k,2)),imag(Data(FirstGood:end,k,2))),'splineinterp');
InterpolazioneImmaginaria3 = fit(cat(2,[1:LastGood],[FirstGood:381])',cat(1,imag(Data(1:LastGood,k,3)),imag(Data(FirstGood:end,k,3))),'splineinterp');
Data(LastGood + 1:FirstGood - 1,k,1) = InterpolazioneReale1(LastGood + 1:FirstGood - 1) + i * InterpolazioneImmaginaria1(LastGood + 1:FirstGood - 1);
Data(LastGood + 1:FirstGood - 1,k,2) = InterpolazioneReale2(LastGood + 1:FirstGood - 1) + i * InterpolazioneImmaginaria2(LastGood + 1:FirstGood - 1);
Data(LastGood + 1:FirstGood - 1,k,3) = InterpolazioneReale3(LastGood + 1:FirstGood - 1) + i * InterpolazioneImmaginaria3(LastGood + 1:FirstGood - 1);
end
end
  6 Comments
Jan
Jan on 10 Oct 2017
Edited: Jan on 10 Oct 2017
Note:
cat(2,[1:LastGood],[FirstGood:381])'
can be simplified to:
[1:LastGood, FirstGood:381]'
Then store this in a variable instead of creating it 6 times.
Performing 6 spline fits for the same locations seems rather inefficient. Could this be simplified?
denis bertin
denis bertin on 10 Oct 2017
Edited: denis bertin on 10 Oct 2017
Hi JAN SIMON, Yes, could be simplified, but i don't know how! I changed cat(2,[1:LastGood],[FirstGood:381])' to one variable [1:LastGood, FirstGood:381]', there is just a few change performance.
Thank you.

Sign in to comment.

Accepted Answer

OCDER
OCDER on 10 Oct 2017
Edited: OCDER on 10 Oct 2017
Not sure if this is faster or not, but your code could be simplified a lot. If you have parallel computing toolbox, you could divide slow jobs across N cores.
LastGood = floor((122.5 - 50) / Header.Fstep) + 1 - 1; % value =29
FirstGood = ceil((140 - 50) / Header.Fstep) + 1 + 1; % value =38
disp(LastGood); % value = 29
disp(FirstGood); % value = 38
disp(size(Data, 2));% value = 247
X = [1:LastGood, FirstGood:381]';
parfor z = 1:3
Tdata = Data(:, :, z);
for k = 1:size(Tdata, 2) %Or do parfor here instead
InterpReal = fit(X, real(Tdata(X, k)), 'splineinterp');
InterpImag = fit(X, imag(Tdata(X, k)), 'splineinterp');
Range = LastGood + 1:FirstGood - 1;
Tdata(Range, k) = InterpReal(Range) + 1j * InterpImag(Range); %NOTE: 1j is now sqrt(-1)
end
Data(:, :, z) = Tdata;
end
  2 Comments
denis bertin
denis bertin on 11 Oct 2017
Many Thank's DONALD LEE.
OCDER
OCDER on 11 Oct 2017
You're welcome!

Sign in to comment.

More Answers (0)

Community Treasure Hunt

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

Start Hunting!