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 Kommentare

per isakson
per isakson am 10 Okt. 2017
Bearbeitet: per isakson am 10 Okt. 2017
Make a minimal working example, MWE. This MWE should focus one your question.
We cannot run your m-files without the m-files called, data files and furthermore we don't know what speed you expect.
KSSV
KSSV am 10 Okt. 2017
It depends on what data your text file has....post your text file....to get help.
per isakson
per isakson am 10 Okt. 2017
Bearbeitet: per isakson am 10 Okt. 2017
"loadind data loop fastly" How do you currently load/read the data files? Your way might be as fast as it gets.
Rik
Rik am 10 Okt. 2017
Sometimes you can even wee it in task manager: if your disk load is 100%, further optimizing your code will not yield any benefit.
Jan
Jan am 10 Okt. 2017
Bearbeitet: Jan am 10 Okt. 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 am 10 Okt. 2017
Bearbeitet: denis bertin am 10 Okt. 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.

Melden Sie sich an, um zu kommentieren.

 Akzeptierte Antwort

OCDER
OCDER am 10 Okt. 2017
Bearbeitet: OCDER am 10 Okt. 2017

0 Stimmen

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 Kommentare

denis bertin
denis bertin am 11 Okt. 2017
Many Thank's DONALD LEE.
OCDER
OCDER am 11 Okt. 2017
You're welcome!

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Kategorien

Gefragt:

am 10 Okt. 2017

Kommentiert:

am 11 Okt. 2017

Community Treasure Hunt

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

Start Hunting!

Translated by