How to fix "Error using == Arrays have incompatible sizes for this operation." in this code?

411 vues (au cours des 30 derniers jours)
Hi, I am writing the following code:
for k = 1 : length(theFiles) % files are alreaady defined in original code
p=1;
t=1;
for q=1:43 % there are 43 inputs for arr
c=corr2(B0,mask);
arr(p)=abs(c);
if(arr(p)<1)
if(arr(p)>0.85)
cor(t)=arr(p);
t=t+1;
end
end
p=p+1;
end %end of q loop
mx=max(cor); %conditioned arr
ixa=find(arr==mx,1) %find the location for which mx matches arr value the first instance
But when run, the result shows following:
Error using ==
Arrays have incompatible sizes for this operation.
Error in Untitled12 (line 264)
ixa=find(arr==mx,1);
It is noteworthy that, the error shows only for certain image inputs. For other inputs, the code works fine.
How can I fix it?
  1 commentaire
kanika bhalla
kanika bhalla le 7 Juil 2021
@Tawsif Mostafiz Can you add those images on which your code gives error so that we can try and check what is a problem?

Connectez-vous pour commenter.

Réponse acceptée

KSSV
KSSV le 7 Juil 2021
Error is clear, you are trying to equate arr and mx. Check do they have same dimensions? They have different dimensions that's why error.
A = rand(2) ; B = rand(2) ;
A == B % no error as the dimensions are same
ans = 2×2 logical array
0 0 0 0
C = rand(3) ;
A == C % error as the dimensions are different
Arrays have incompatible sizes for this operation.
  6 commentaires
Walter Roberson
Walter Roberson le 9 Fév 2023
If you change the calls to
h11=integral(h, 0, 10, 'arrayvalued', true); %inf
h22=integral(v, 0, 10, 'arrayvalued', true);%inf
then the integrals can proceed -- but they take a long time. Long enough that I do not know yet whether the steps after that would be compatible with the values returned by those statements.
Walter Roberson
Walter Roberson le 9 Fév 2023
psi1=psi1+(Ai.*(r.^(-i+1)+h11)+(r.^(1./2).*besselk(i-1./2,r.*alpha)+h22).*Bi).*gegenbauerC(i,-1./2, cos(t));
Variable psi1 is not defined at that point in the code. For debugging purposes we can initialize it to 0 before that line.
[DH1,hh2]=contour(x,y,psi,10,'-k');
Variable psi is not defined at that point in the code. For debugging purposes we can use psi1 instead.
psi1=psi1+(Ai.*(r.^(-i+1)+h11)+(r.^(1./2).*besselk(i-1./2,r.*alpha)+h22).*Bi).*gegenbauerC(i,-1./2, cos(t));
That comes out as all zero, because Ai and Bi are zero.

Connectez-vous pour commenter.

Plus de réponses (0)

Produits


Version

R2021a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by