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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!

KSSV
on 18 Jun 2020

Edited: KSSV
on 18 Jun 2020

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.

KSSV
on 18 Jun 2020

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.

madhan ravi
on 18 Jun 2020

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{:})

madhan ravi
on 18 Jun 2020

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

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