create a matrix with a for loop

Hello all, i am trying to calculate a bunch of matrices using a for loop. I have written the following code

T = 60;
for t = 60:60:12940
    phi(:,t) = [(n1(1,t+T)-n2(1,t+T)) / (n1(1,t)-n2(1,t)), (n3(1,t+T)-n4(1,t+T)) / (n3(2,t)-n4(2,t)); (n1(2,t+T)-n2(2,t+T)) / (n1(1,t)-n2(1,t)), (n3(2,t+T)-n4(2,t+T)) / (n3(2,t)-n4(2,t))];
end 

i am getting an error code that says the left side of the equation is a 2-by-1 and the right side is a 2-by-2. both sides of the equation are supposed to be a 2-by-2!

How would i go about writing the left side of the eqution so it would work with the for loop and result in a 2-by-2?

Thanks for any help!

Answer 1

[KSSV]

T = 60;
phi = cell([],1) ; 
count = 0 ; 
for t = 60:60:12940
    count = count+1 ; 
    phi{count} = [(n1(1,t+T)-n2(1,t+T)) / (n1(1,t)-n2(1,t)), (n3(1,t+T)-n4(1,t+T)) / (n3(2,t)-n4(2,t)); (n1(2,t+T)-n2(2,t+T)) / (n1(1,t)-n2(1,t)), (n3(2,t+T)-n4(2,t+T)) / (n3(2,t)-n4(2,t))];
end 

If you think that always the RHS is a 2x2 matrix, ot can be saved into matrix by initlaizing it into a 3D matrix.

[Adam Levschuk]

hi KSSV, thank you for your reply. Could you please explain what you did there... i would have never had thought to do anything like it.

I am now getting an error message saying "Index in position 2 exceeds array bounds (must not exceed 12000)."

here is my exact code....

T = 60 % this is also 1.2 seconds 
psi = cell([],1);
count = 0;
for t = 60:60:12940
    count = count+1;
    phi{count} = [(n1(1,t+T)-n2(1,t+T)) / (n1(1,t)-n2(1,t)), (n3(1,t+T)-n4(1,t+T)) / (n3(2,t)-n4(2,t)); (n1(2,t+T)-n2(2,t+T)) / (n1(1,t)-n2(1,t)), (n3(2,t+T)-n4(2,t+T)) / (n3(2,t)-n4(2,t))];
end 

im not sure what is wrong now.

Thank you for the help!

[KSSV]

What is size of n1? You are accessing more number of elements than present..To the index, you are adding 60 i.e (t+T) so you are accessing more number of elements than present.

Check the dimensions.

Initilization check it..it should be phi not psi.

[Adam Levschuk]

n1 is a 2x12000 array. same with n2, n3, and n4. What do you think?

Thank you so much for your help,

Adam

[KSSV]

Then you see...you are accessing 12940....it is clear....

You can access elements only upto index 12000....you need to change the loop indexing.

[Adam Levschuk]

such a silly mistake -- thanks you so much for the help KSSV.

Very much appreciated!!!

Answer 2

[madhan ravi]

T = 60;
phi = cell(numel(t),1);
t = 60:60:12940;
for k = 1:numel(t)
    phi{k} = [(n1(1,t(k)+T)-n2(1,t(k)+T)) / ...
        (n1(1,t(k))-n2(1,t(k))), (n3(1,t(k)+T)-n4(1,t(k)+T)) / ...
        (n3(2,t(k))-n4(2,t(k))); (n1(2,t(k)+T)-n2(2,t(k)+T)) / ...
        (n1(1,t(k))-n2(1,t(k))), (n3(2,t(k)+T)-n4(2,t(k)+T)) / (n3(2,t(k))-n4(2,t(k)))];
end
celldisp(phi)
phi = cat(3,phi{:})

[Adam Levschuk]

Hello madhan, i am getting an error code taht says,

"Index in position 2 exceeds array bounds (must not exceed 12000)."

ive never undertsood this error code when making for loops. Any idea of how this would be solved??

thank you so much for your help!

Adam

[madhan ravi]

I don’t have time to get into details here but to illustrate with an example:

x = rand(2,2)% has two columns
x(1,3) % but you’re trying to access the third column, which doesn’t make sense

[Adam Levschuk]

thanks for your help Madhan, hopefully ill crack this soon!

Adam