For loop Optimization?

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denis bertin
denis bertin 2017-9-19
评论： denis bertin 2017-10-13
Hi everybody, i want to optmize(make faster loop) this function:
for k=1:2001
integrando=exp(i*2*pi*var_spaziale(k)*var_spettrale).*funzionefor k=1:N
integrando=exp(i*2*pi*var_spaziale(k)*var_spettrale).*funzione;
f(k)=du*sum(integrando);
end
How is it possible to do please?
Many thanks.
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denis bertin
denis bertin 2017-10-1
Hi everyone,
I have these attached file name: datineltempo_ciclico.m that read file name malta1t.mat,
in th line 95, there is the peace of code like this:
NPF = 942;
N=2001;
v=381x1;(complex numbers)
f=381x1;
t=1x2001;
for(k = 1:NPF)
v = datif2(:,k);
g = zeros(2001,1);
du=f(2)-f(1);
for k1=1:N
integrando=exp(i*2*pi*t(k1)*f).*v;
g(k1)=du*sum(integrando);
end
v1 = g;
v1 = 2 * real(v1);
datit(:,k) = v1;
%k
end
.....
The problem is that is take more time to load the 942*2001 matrix;
Please somebody can i help me optimize(make faster) this loop?
Many thanks.
I need the maximum time reduce execution...
Please Help Me....
denis bertin
denis bertin 2017-10-3
编辑：denis bertin 2017-10-3
Hi,
The modification is to put for example 'tic' and 'toc' in the init and end of code to debug the execution time for all the program.
Now it take too many time to execute. just launch a function call: datineltempo_ciclico.m , then the will open a pop up to select the malta1t.mat file, and enter 0.02. and the execution will start. But is too long time take.
Thank you. If you have another problem to execute it,let me know about please.

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采纳的回答

Jan
Jan 2017-10-1
编辑：Jan 2017-10-1
The exp() function is expensive. So start with avoiding repeated calls with the same argument:
C = exp((2i * pi * t) .* f); % Auto-expanding, >= R2016b
for k = 1:NPF
v = datif2(:,k);
g = zeros(2001,1);
du = f(2)-f(1);
for k1 = 1:N
integrando = C(:, k1) .* v;
g(k1) = du * sum(integrando);
end
datit(:, k) = 2 * real(g);
end
Is datit pre-allocated before the loop? If not, add this.
For Matlab < 2016b:
C = exp(2i * pi * bsxfun(@times, t, f));
Try if this is faster:
C = exp((2i * pi) * t.' .* f.'); % Auto-expaning, >= R2016b
for k = 1:NPF
du = f(2)-f(1);
g = du * C * datif2(:,k);
datit(:, k) = 2 * real(g);
end
Unfortunately I cannot try this by my own, because your posted code misses the function "interplin". What is it?
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denis bertin
denis bertin 2017-10-13
编辑：denis bertin 2017-10-13
Hi JAN, Sorry, i check the two Output:
C = exp((2i * pi * t) .* f); % Auto-expanding, >= R2016b
for k = 1:NPF
v = datif2(:,k);
g = zeros(2001,1);
du = f(2)-f(1);
for k1 = 1:N
integrando = C(:, k1) .* v;
g(k1) = du * sum(integrando);
end
datit(:, k) = 2 * real(g);
end
with:
C = exp((2i * pi) * t.' .* f.'); % Auto-expaning, >= R2016b
for k = 1:NPF
du = f(2)-f(1);
g = du * C * datif2(:,k);
datit(:, k) = 2 * real(g);
end
but the two datit matrix are differents.
How is it possible please?
denis bertin
denis bertin 2017-10-13
May be is missing sum(...) function on the second code?

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更多回答（1 个）

denis bertin
denis bertin 2017-10-3
编辑：denis bertin 2017-10-3
Hi JAN SIMON, sorry for this missing information. dammi il passo di misura desiderato? enter: 0.02 . it's all.
The modification is to put for example 'tic' and 'toc' in the init and end of code to debug the execution time for all the program.
Now it take too many time to execute. just launch a function call: datineltempo_ciclico.m , then the will open a pop up to select the malta1t.mat file, and enter 0.02. and the execution will start. But is too long time take.
Thank you. If you have another problem to execute it, i can just rename file italian->to English, this is not a problem.

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