how to rotate image using function notation

7 vues (au cours des 30 derniers jours)
mark palmer
mark palmer le 7 Fév 2023
Commenté : Chris le 7 Fév 2023
How do I write Matlab code using functions to make a rotation on an image? THis is all I've got so far, and I'm probably way off...
I = 'cameraman.jpg';
fa = 45;
f1 = @(xy) xy(:,1) .* cosd(fa) + xy(:,2) .* sind(fa);
f2 = @(xy) -xy(:,1) .* sind(fa) + xy(:,2) .* cosd(fa);
ifcn1 = @(xy) f1(xy);
tform1 = geometricTransform2d(ifcn1);
Rin = imref2d(size(I),[-1 1],[-1 1]);
Rout = imref2d(size(I),[-1 1],[-1 1]);
rho1a = imwarp(I,Rin,tform1,'OutputView',Rout,FillValues=0);

Réponse acceptée

Chris
Chris le 7 Fév 2023
If you're using imwarp, you might as well use imrotate, no? You'd be using the function imrotate() to rotate an image, if that's really your goal.
fa = 45;
I = imread('cameraman.tif');
imrot = imrotate(I,fa);
imshow(imrot)
If you want an anonymous function, like you're using...
rotateImage = @(im, ang) imrotate(im, ang);
imrot = rotateImage(I, fa);
imshow(imrot)
You could also make a standalone function, in a file of the same name or at the bottom of a script, that accepts inputs and outputs the rotated image, and does whatever else you want with the input.
imrot = rotImFun(I, fa);
imshow(imrot);
function out = rotImFun(im, ang)
out = imrotate(im, ang);
end
  3 commentaires
mark palmer
mark palmer le 7 Fév 2023
Modifié(e) : mark palmer le 7 Fév 2023
IS there an alternate way to do this with trig functions within an anonymous function? I know it isn't the easiest way, I'm just trying to learn.
Chris
Chris le 7 Fév 2023
I'm unaware of a method that doesn't require a complex series of operations, which a single anonymous function isn't really suited for. That doesn't mean it doesn't exist--it just means I don't have an answer.

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur Images dans Help Center et File Exchange

Produits


Version

R2022a

Community Treasure Hunt

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

Start Hunting!

Translated by