These 3 codes when run give error
1 visualización (últimos 30 días)
Mostrar comentarios más antiguos
Sadiq Akbar
el 1 de Mayo de 2024
Comentada: Sadiq Akbar
el 4 de Mayo de 2024
I have 3 codes as given in attachment. When I run the "main.m", it gives the following error:
Index in position 1 exceeds array bounds. Index must not exceed 1.
Error in ArGeo (line 17)
cosd(b(2, :)) .* cosd(b(1, :)); % x-component
Error in main>@(b)ArGeo(b,u,Noise)+penaltyTerm(b,u) (line 25)
[time,gBest,gBestScore]=WHO(20,100,lb,ub,dim,@(b) ArGeo(b,u,Noise)+penaltyTerm(b, u));
Error in WHO (line 47)
group(i).cost=fobj(group(i).pos);
Error in main (line 25)
[time,gBest,gBestScore]=WHO(20,100,lb,ub,dim,@(b) ArGeo(b,u,Noise)+penaltyTerm(b, u));
>>
0 comentarios
Respuesta aceptada
Voss
el 1 de Mayo de 2024
In WHO.m lines 45-48, you have
for i=1:Nfoal
group(i).pos=lb+rand(1,dim).*(ub-lb);
group(i).cost=fobj(group(i).pos);
end
where dim is 2, and lb and ub are each 1xdim vectors. Thus, group(i).pos is created as a 1xdim vector on line 46, and the error happens after fobj (which is @(b) ArGeo(b,u,Noise)+penaltyTerm(b, u)) is called on line 47, because inside ArGeo.m, input b, which is group(i).pos in WHO.m, is expected to have at least 2 rows:
kb = pi * [
cosd(b(2, :)) .* cosd(b(1, :)); % x-component
sind(b(2, :)) .* cosd(b(1, :)); % y-component
sind(b(1, :)) % z-component
];
That's the reason for the error.
As far as how to fix it, I suspect that lines 26-29 of WHO.m are not doing what's intended:
if size(ub,1)==1
ub=ones(1,dim).*ub;
lb=ones(1,dim).*lb;
end
Since ub and lb are each 1xdim vectors, you are multiplying each element of ub and lb by 1, which leaves them unchanged. Instead, I guess you want to replicate them so that they have exactly 2 rows (not dim rows), which could be done like:
if size(ub,1)==1
ub=ones(2,1).*ub;
lb=ones(2,1).*lb;
end
Then they are of size 2xdim, which seems like what's intended.
9 comentarios
Más respuestas (0)
Ver también
Categorías
Más información sobre Matrix Indexing en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!