How do I plot many rows from many matrices made in a loop?

Hello, I need to make an (11,32,8) 3D array from an equation. I figured out how to get around indexing non-integer numbers and using three for loops to fill in the zeros matrix. However, the output gives a (11,8,9) array... I've attached my code for you to look at. Thank you!
if true
% code
end
ns=[1.5:0.01:1.6]';
B=ns./1.335;
l=[65:95];
v=l+0.5;
TEM=zeros(11,31,8);
R=[4.4:0.1:5.1]*10^-6;
hold on
for j=1:length(R)
for i=1:length(B)
for k=1:length(v)
TEM(i,j,k)=(2*pi.*(B(i).*1.335).*R(j)).*((v(k)+(1.8557.*v(k).^(1/3))-(B(i)./sqrt(B(i).^2-1))+(1.0331.*(v(k).^(-1/3)))-(((0.6186.*B(i).^3)./((B(i).^2-1).^(3/2))).*(v(k).^(-2/3)))+0).^-1);
end
end
end
K=TEM
hold off

1 commentaire

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Réponses (2)

Your function looks like you could use meshgrid to generate the matrices that you need.
ns=[1.5:0.01:1.6]';
B=ns./1.335;
l=[65:95];
v=l+0.5;
R=[4.4:0.1:5.1]*10^-6;
[v,B,R]=meshgrid(v,B,R);%[x,y,z] results in length(y)-by-length(x)-by-length(z)
TEM=(2*pi.*(B.*1.335).*R).*((v+(1.8557.*v.^(1/3))-(B./sqrt(B.^2-1))+(1.0331.*(v.^(-1/3)))-(((0.6186.*B.^3)./((B.^2-1).^(3/2))).*(v.^(-2/3)))+0).^-1);

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Commenté :

Rik
le 25 Avr 2018

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