通过Matlab进行图像旋转而不使用imrotate

Question

我试图用Matlab旋转图像而不使用imrotate函数.我实际上是通过使用变换矩阵来制作的.但它还不够好.问题是,旋转的图像是"滑动的".让我用图片告诉你.

这是我想要旋转的图像:

但是当我旋转它时,例如45度,它变为:

我在问为什么会发生这种情况.这是我的代码,是否有任何数学或编程错误？

image=torso;

%image padding
[Rows, Cols] = size(image); 
Diagonal = sqrt(Rows^2 + Cols^2); 
RowPad = ceil(Diagonal - Rows) + 2;
ColPad = ceil(Diagonal - Cols) + 2;
imagepad = zeros(Rows+RowPad, Cols+ColPad);
imagepad(ceil(RowPad/2):(ceil(RowPad/2)+Rows-1),ceil(ColPad/2):(ceil(ColPad/2)+Cols-1)) = image;

degree=45;

%midpoints
midx=ceil((size(imagepad,1)+1)/2);
midy=ceil((size(imagepad,2)+1)/2);

imagerot=zeros(size(imagepad));

%rotation
for i=1:size(imagepad,1)
    for j=1:size(imagepad,2)

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

         if (x>=1 && y>=1)
              imagerot(x,y)=imagepad(i,j); % k degrees rotated image         
         end

    end
end

 figure,imagesc(imagerot);
 colormap(gray(256));

Answer 1

您在图像中出现漏洞的原因是因为您正在计算imagerot每个像素的位置imagepad.你需要以相反的方式进行计算.也就是说,对于imagerot插值中的每个像素imagepad.要做到这一点,你只需要应用逆变换,在旋转矩阵的情况下,它只是矩阵的转置(只需改变每个的符号sin并转换另一个方式).

循环在像素中imagerot:

imagerot=zeros(size(imagepad)); % midx and midy same for both

for i=1:size(imagerot,1)
    for j=1:size(imagerot,2)

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

         if (x>=1 && y>=1 && x<=size(imagepad,2) && y<=size(imagepad,1))
              imagerot(i,j)=imagepad(x,y); % k degrees rotated image         
         end

    end
end

还要注意你的midx和midy需要分别用size(imagepad,2)和计算size(imagepad,1),因为第一个维度是指行数(高度)和第二个宽度.

注意:当您决定采用除最近邻居之外的插值方案时,同样的方法适用,如Rody的线性插值示例.

编辑:我假设您使用循环进行演示,但实际上不需要循环.这是最近邻插值(您正在使用的)的示例,保持相同大小的图像,但您可以修改它以生成包含整个源图像的更大图像:

imagepad = imread('peppers.png');
[nrows ncols nslices] = size(imagepad);
midx=ceil((ncols+1)/2);
midy=ceil((nrows+1)/2);

Mr = [cos(pi/4) sin(pi/4); -sin(pi/4) cos(pi/4)]; % e.g. 45 degree rotation

% rotate about center
[X Y] = meshgrid(1:ncols,1:nrows);
XYt = [X(:)-midx Y(:)-midy]*Mr;
XYt = bsxfun(@plus,XYt,[midx midy]);

xout = round(XYt(:,1)); yout = round(XYt(:,2)); % nearest neighbor!
outbound = yout<1 | yout>nrows | xout<1 | xout>ncols;
zout=repmat(cat(3,1,2,3),nrows,ncols,1); zout=zout(:);
xout(xout<1) = 1; xout(xout>ncols) = ncols;
yout(yout<1) = 1; yout(yout>nrows) = nrows;
xout = repmat(xout,[3 1]); yout = repmat(yout,[3 1]);
imagerot = imagepad(sub2ind(size(imagepad),yout,xout,zout(:))); % lookup
imagerot = reshape(imagerot,size(imagepad));
imagerot(repmat(outbound,[1 1 3])) = 0; % set background value to [0 0 0] (black)

为了将上述修改为线性插值,计算每个坐标的4个相邻像素,XYt并使用小数分量乘积作为权重来执行加权和.我将把它作为一种练习,因为它只会使我的答案进一步超出你的问题的范围.:)

Answer 2

您使用的方法(通过采样旋转)是最快和最简单的,但也是最不准确的.

按区域映射旋转,如下所示(这是一个很好的参考),在保留颜色方面要好得多.

但是:请注意,这仅适用于灰度/ RGB图像,但不适用于彩色贴图图像,例如您似乎正在使用的图像.

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);

输出:

Answer 3

根据用户给出的角度旋转彩色图像,无需在matlab中进行任何图像裁剪.

该程序的输出类似于内置命令"imrotate"的输出.该程序根据用户给出的角度输入动态创建背景.通过使用旋转矩阵和原点移位,我们得到初始图像和最终图像的坐标之间的关系.我们现在映射每个像素的强度值.

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);