storing output (which are matrices) of for loops

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Sumera Yamin
Sumera Yamin el 6 de Jun. de 2018
Comentada: Sumera Yamin el 6 de Jun. de 2018
Hello, I am trying to store my output from for loop to be used later in code. The output is a 2x4 matrix for each cycle of loop. here is my code.
B= [0.0250; 0.0500; 0.0750; 0.1000; 0.1250; 0.1500; 0.1750; 0.2000; 0.2250; 0.2500; 0.2750; 0.3000; 0.3250; 0.3500; 0.3750; 0.4000];
rig=0.01
for j=1:length(B)
k= B(j)/(2*rig)
TMP= [cos(k*l)^2, (sin(k*l)*cos(k*l))/k, sin(k*l)*cos(k*l), (sin(k*l)^2)/k;
-k*sin(k*l)*cos(k*l), cos(k*l)^2, -k*sin(k*l)^2, sin(k*l)*cos(k*l);
-sin(k*l)*cos(k*l), -(sin(k*l)^2)/k, cos(k*l)^2, (sin(k*l)*cos(k*l))/k;
k*sin(k*l)^2, -sin(k*l)*cos(k*l), -k*sin(k*l)*cos(k*l), cos(k*l)^2]
TM= TMP*TMP
C1=[TM(1,:);TM(3,:)]
end
I want to store C1 and k for each cycle of loop to be used later in other parts of code. please let me know what would be the best way to do so. I tried to do it by initializing an array (like we do for scalar outputs but this did not workout. Thank you very much for your help.

Respuesta aceptada

Stephen23
Stephen23 el 6 de Jun. de 2018
Editada: Stephen23 el 6 de Jun. de 2018
Array preallocation works for me:
l = 1;
B = [0.0250; 0.0500; 0.0750; 0.1000; 0.1250; 0.1500; 0.1750; 0.2000; 0.2250; 0.2500; 0.2750; 0.3000; 0.3250; 0.3500; 0.3750; 0.4000];
B = 0.025:0.025:0.4;
kV = nan(1,1,numel(B));
CM = nan(2,4,numel(B));
rig = 0.01;
for j = 1:numel(B)
k = B(j)/(2*rig);
TMP = [cos(k*l)^2, (sin(k*l)*cos(k*l))/k, sin(k*l)*cos(k*l), (sin(k*l)^2)/k;
-k*sin(k*l)*cos(k*l), cos(k*l)^2, -k*sin(k*l)^2, sin(k*l)*cos(k*l);
-sin(k*l)*cos(k*l), -(sin(k*l)^2)/k, cos(k*l)^2, (sin(k*l)*cos(k*l))/k;
k*sin(k*l)^2, -sin(k*l)*cos(k*l), -k*sin(k*l)*cos(k*l), cos(k*l)^2];
TMP = TMP*TMP;
CM(:,:,j) = [TMP(1,:);TMP(3,:)];
kV(j) = k;
end

Más respuestas (1)

Ankita Bansal
Ankita Bansal el 6 de Jun. de 2018
Editada: Ankita Bansal el 6 de Jun. de 2018
Hi Sumera, you can do so by changing k to k(j) and C1 to C1(:,:,j).
B= [0.0250; 0.0500; 0.0750; 0.1000; 0.1250; 0.1500; 0.1750; 0.2000; 0.2250; 0.2500; 0.2750; 0.3000; 0.3250; 0.3500; 0.3750; 0.4000];
rig=0.01
for j=1:length(B)
k_1(j)= B(j)/(2*rig)
k=k_1(j);
TMP= [cos(k*l)^2, (sin(k*l)*cos(k*l))/k, sin(k*l)*cos(k*l), (sin(k*l)^2)/k;
-k*sin(k*l)*cos(k*l), cos(k*l)^2, -k*sin(k*l)^2, sin(k*l)*cos(k*l);
-sin(k*l)*cos(k*l), -(sin(k*l)^2)/k, cos(k*l)^2, (sin(k*l)*cos(k*l))/k;
k*sin(k*l)^2, -sin(k*l)*cos(k*l), -k*sin(k*l)*cos(k*l), cos(k*l)^2]
TM= TMP*TMP
C1(:,:,j)=[TM(1,:);TM(3,:)]
end
Here i have stored values in k_1.
Hope this helps.
  2 comentarios
Stephen23
Stephen23 el 6 de Jun. de 2018
Note that this does not preallcoate the output arrays, so will require the arrays to be resized and moved on each iteration.
Sumera Yamin
Sumera Yamin el 6 de Jun. de 2018
Thank you Ankita for your time and answer. This gives me good idea of my problem.

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