Azzera filtri
Azzera filtri

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Creation of matrix from loop

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Renjith V Ravi
Renjith V Ravi il 17 Set 2016
Chiuso: MATLAB Answer Bot il 20 Ago 2021
The following is a code for creation of matrix using for loop
for r = 1:256;
for c=1:256;
m(r,c) = (r+c)^2;
end;
end;
disp(m)
In m(r,c) = (r+c)^2; Suppose, if there are two or more equations like (r+c)^2 and (2r-c)^2 and (r+3c) ,I want to get m(r1,c1)=(r+c)^2 and m(r2,c2)=(2r-c)^2 and m(r3,c2)=(r+3c) and again m(r4,c4)=(r+c)^2 and m(r5,c5)=(2r-c)^2 and ...... etc upto m(r256,c256).

Risposte (1)

Star Strider
Star Strider il 17 Set 2016
I am not certain what you want to do, and you do not say what ‘(r2,c2)’ and the rest are, or what you want to do with them.
You can eliminate the loop with meshgrid:
r = 1:256;
c = 1:256;
[R,C] = meshgrid(r,c);
m = (R + C).^2;
figure(1)
meshc(R, C, m)
grid on
xlabel('\bf\itr')
ylabel('\bf\itc')
zlabel('\bf\itm')
  4 Commenti
Renjith V Ravi
Renjith V Ravi il 17 Set 2016
Yes, But m = (R+C).^2 . I just want to merge r and c to form a matrix of size 256x256
Star Strider
Star Strider il 17 Set 2016
My code does exactly that, without the loop.
Run my code and compare the figure it produces (in figure(1)) with the figure your code produces (in figure(2)):
for r = 1:256;
for c=1:256;
m(r,c) = (r+c)^2;
end;
end;
figure(2)
meshc(m)
grid on
The result is the same, the only difference being that my code is more efficient and easier to work with.

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