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));
At the end of the article you can find the matlab code. First I tried to rotate an Image by using built in functions in Matlab. Matlab built_in function rot90 (A,k) can be used to rotate images in 90 degrees. Here is an example using rot90: Assign K=1 for 90 degree, 2 for 180, 3 for 270 and 4 for 360. The output image will be rotated 90 degrees.
check this site for rotate image without imrotate. Matrix rotation is the best solution. And imrotate.m also use the matrix rotation. You can always apply a 2d rotation matrix to get the rotated coordinates of your image. Or in simple cases (90, 180 degrees) just flip the image matrix.
By default, imrotate uses nearest neighbor interpolation, setting the values of pixels in J that are outside the rotated image to 0. J = imrotate (I,angle,method) rotates image I using the interpolation method specified by method.
And imrotate.m also use the matrix rotation. You can always apply a 2d rotation matrix to get the rotated coordinates of your image. Or in simple cases (90, 180 degrees) just flip the image matrix. But why would you try to avoid imrotate? Yes, Matrix rotation seems to viable solution. A = [... ( (C (2)-source (2))* (C (1)-source (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. :)
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:
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