storing output (which are matrices) of for loops

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Sumera Yamin
Sumera Yamin 2018-6-6
评论： Sumera Yamin 2018-6-6
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.

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采纳的回答

Stephen23
Stephen23 2018-6-6
编辑：Stephen23 2018-6-6
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

更多回答（1 个）

Ankita Bansal
Ankita Bansal 2018-6-6
编辑：Ankita Bansal 2018-6-6
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 个评论
Stephen23
Stephen23 2018-6-6
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 2018-6-6
Thank you Ankita for your time and answer. This gives me good idea of my problem.

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