compute sums by accumulating in a for-loop
1 visualización (últimos 30 días)
Mostrar comentarios más antiguos
Kyle
el 3 de Dic. de 2023
Comentada: Kyle
el 3 de Dic. de 2023
I am trying to accumulate sums (phi(l) for lags " l"in a for loop given the code below. I keep getting "Unable to perform assignment because the left and right sides have a different number of elements." I have tried multiple ways to try and fix it but nothing is working.
x=normrnd(0,1,100,1);
xprime=detrend(x,'constant');
phi=zeros(14,1);
L= [0:13]';
for l=1:14
phi(l)=(sum(xprime(1:100-l)).*xprime(1+l:100)/(100-(l-1)-1));
end
0 comentarios
Respuesta aceptada
Image Analyst
el 3 de Dic. de 2023
Look at this:
x=normrnd(0,1,100,1);
xprime=detrend(x,'constant');
phi=zeros(14,1);
L= [0:13]';
for l=1:14
temp = (sum(xprime(1:100-l)).*xprime(1+l:100)/(100-(l-1)-1))
fprintf('The size of temp = %d.\n', numel(temp))
phi(l)=(sum(xprime(1:100-l)).*xprime(1+l:100)/(100-(l-1)-1));
end
So you're trying to stuff 99 values into a slot meant for only one value, phi(l). Not sure how to fix it because I'm not sure what your intent is. Did you mean for phi to be a 99 by 14 matrix and you want to put temp into the columns of phi?
3 comentarios
Image Analyst
el 3 de Dic. de 2023
Watch your parentheses. Maybe you meant either
phi(l) = sum(xprime(1:(100-l)) .* xprime(l:100) / (100-(l-1)-1));
or
phi(l) = sum(xprime(1:(100-l) .* xprime(l:100)) / (100-(l-1)-1);
And l (ell) is not a good variable name - it looks too much like 1 (one) and I (capital I). Use k instead.
Más respuestas (1)
Torsten
el 3 de Dic. de 2023
Editada: Torsten
el 3 de Dic. de 2023
sum(xprime(1:100-l))
This is a scalar.
sum(xprime(1:100-l)).*xprime(1+l:100)
This is a vector of length 100-(1+l)+1.
(sum(xprime(1:100-l)).*xprime(1+l:100)/(100-(l-1)-1))
This remains a vector of length 100-(1+l)+1.
phi(l)
This is s acalar.
Thus you try to assign a vector to a scalar which is not possible.
Maybe you mean
phi(l)=sum(xprime(1:100-l).*xprime(1+l:100))/(100-(l-1)-1));
0 comentarios
Ver también
Categorías
Más información sobre Loops and Conditional Statements en Help Center y File Exchange.
Productos
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!