问不使用不旋转的Matlab实现图像旋转

EN

image = imread('peppers.png');

figure(1), clf, hold on
subplot(1,2,1)
imshow(image);

degree = 45;

switch mod(degree, 360)
    % Special cases
    case 0
        imagerot = image;
    case 90
        imagerot = rot90(image);
    case 180
        imagerot = image(end:-1:1, end:-1:1);
    case 270
        imagerot = rot90(image(end:-1:1, end:-1:1));

    % General rotations
    otherwise

        % Convert to radians and create transformation matrix
        a = degree*pi/180;
        R = [+cos(a) +sin(a); -sin(a) +cos(a)];

        % Figure out the size of the transformed image
        [m,n,p] = size(image);
        dest = round( [1 1; 1 n; m 1; m n]*R );
        dest = bsxfun(@minus, dest, min(dest)) + 1;
        imagerot = zeros([max(dest) p],class(image));

        % Map all pixels of the transformed image to the original image
        for ii = 1:size(imagerot,1)
            for jj = 1:size(imagerot,2)
                source = ([ii jj]-dest(1,:))*R.';
                if all(source >= 1) && all(source <= [m n])

                    % Get all 4 surrounding pixels
                    C = ceil(source);
                    F = floor(source);

                    % Compute the relative areas
                    A = [...
                        ((C(2)-source(2))*(C(1)-source(1))),...
                        ((source(2)-F(2))*(source(1)-F(1)));
                        ((C(2)-source(2))*(source(1)-F(1))),...
                        ((source(2)-F(2))*(C(1)-source(1)))];

                    % Extract colors and re-scale them relative to area
                    cols = bsxfun(@times, A, double(image(F(1):C(1),F(2):C(2),:)));

                    % Assign                     
                    imagerot(ii,jj,:) = sum(sum(cols),2);

                end
            end
        end        
end

subplot(1,2,2)
imshow(imagerot);

img=imread('img.jpg'); 

[rowsi,colsi,z]= size(img); 

angle=45;

rads=2*pi*angle/360;  

%calculating array dimesions such that  rotated image gets fit in it exactly.
% we are using absolute so that we get  positve value in any case ie.,any quadrant.

rowsf=ceil(rowsi*abs(cos(rads))+colsi*abs(sin(rads)));                      
colsf=ceil(rowsi*abs(sin(rads))+colsi*abs(cos(rads)));                     

% define an array withcalculated dimensionsand fill the array  with zeros ie.,black
C=uint8(zeros([rowsf colsf 3 ]));

%calculating center of original and final image
xo=ceil(rowsi/2);                                                            
yo=ceil(colsi/2);

midx=ceil((size(C,1))/2);
midy=ceil((size(C,2))/2);

% in this loop we calculate corresponding coordinates of pixel of A 
% for each pixel of C, and its intensity will be  assigned after checking
% weather it lie in the bound of A (original image)
for i=1:size(C,1)
    for j=1:size(C,2)                                                       

         x= (i-midx)*cos(rads)+(j-midy)*sin(rads);                                       
         y= -(i-midx)*sin(rads)+(j-midy)*cos(rads);                             
         x=round(x)+xo;
         y=round(y)+yo;

         if (x>=1 && y>=1 && x<=size(img,1) &&  y<=size(img,2) ) 
              C(i,j,:)=img(x,y,:);  
         end

    end
end

imshow(C);

img = imread('Koala.jpg');

theta = pi/10;
rmat = [
cos(theta) sin(theta) 0
-sin(theta) cos(theta) 0
0           0          1];

mx = size(img,2);
my = size(img,1);
corners = [
    0  0  1
    mx 0  1
    0  my 1
    mx my 1];
new_c = corners*rmat;

T = maketform('affine', rmat);   %# represents translation
img2 = imtransform(img, T, ...
    'XData',[min(new_c(:,1)) max(new_c(:,1))],...
    'YData',[min(new_c(:,2)) max(new_c(:,2))]);
subplot(121), imshow(img);
subplot(122), imshow(img2);