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问题:
I am trying to rotate an image with Matlab without using imrotate function. I actually made it by using transformation matrix.But it is not good enough.The problem is, the rotated image is "sliding".Let me tell you with pictures.
This is my image which I want to rotate:
But when I rotate it ,for example 45 degrees, it becomes this:
I am asking why this is happening.Here is my code,is there any mathematical or programming mistakes about it?
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));
回答1:
The reason you have holes in your image is because you are computing the location in imagerot of each pixel in imagepad. You need to do the computation the other way around. That is, for each pixel in imagerot interpolate in imagepad. To do this, you just need to apply the inverse transform, which in the case of a rotation matrix is just the transpose of the matrix (just change the sign on each sin and translate the other way).
Loop over pixels in 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
Also note that your midx and midy need to be calculated with size(imagepad,2) and size(imagepad,1) respectively, since the first dimension refers to the number of rows (height) and the second to width.
NOTE: The same approach applies when you decide to adopt an interpolation scheme other than nearest neighbor, as in Rody's example with linear interpolation.
EDIT: I'm assuming you are using a loop for demonstrative purposes, but in practice there is no need for loops. Here's an example of nearest neighbor interpolation (what you are using), keeping the same size image, but you can modify this to produce a larger image that includes the whole source image:
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)
To modify the above to linear interpolation, compute the 4 neighboring pixels to each coordinate in XYt and perform a weighted sum using the fractional components product as the weights. I'll leave that as an exercise, since it would only serve to bloat my answer further beyond the scope of your question. :)
回答2:
The method you are using (rotate by sampling) is the fastest and simplest, but also the least accurate.
Rotation by area mapping, as given below (this is a good reference), is much better at preserving color.
But: note that this will only work on greyscale/RGB images, but NOT on colormapped images like the one you seem to be using.
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);
Output:
回答3:
Rotates colored image according to angle given by user without any cropping of image in matlab.
Output of this program is similar to output of inbuilt command "imrotate" .This program dynamically creates background according to angle input given by user.By using rotation matrix and origin shifting, we get relation between coordinates of initial and final image.Using relation between coordinates of initial and final image, we now map the intensity values for each pixel.
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);
回答4:
Check this out.
this is fastest way that you can do.
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);
