create a matrix with a for loop
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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!
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Réponse acceptée
KSSV
le 18 Juin 2020
Modifié(e) : KSSV
le 18 Juin 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.
5 commentaires
KSSV
le 18 Juin 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.
Plus de réponses (1)
madhan ravi
le 18 Juin 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{:})
3 commentaires
madhan ravi
le 18 Juin 2020
Modifié(e) : madhan ravi
le 18 Juin 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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