Matrix multiplication vs equations for image transformations in MATLAB

Question

I am trying to write some code to rotate an image in MATLAB, that is, equivalent to imrotate. I use matrix multiplication to perform an inverse mapping of the new image to the input image. However, it takes a lot longer than explicitly writing out the equivalent equation. Is there a better way to perform this multiplication?

I would prefer to use matrix multiplication, because I could use the same code for other transformations by replacing the transformation matrix, RT.

im1 = imread('file.jpg');
[h, w, p] = size(im1);
theta = -pi/6;
hh = round( h*cos(theta) + w*abs(sin(theta)));
ww = round( w*cos(theta) + h*abs(sin(theta)));

R = [cos(theta) -sin(theta); sin(theta) cos(theta)];
T = [w/2; h/2];
RT = [inv(R) T; 0 0 1];
for z = 1:p
    for x = 1:ww
        for y = 1:hh
            % Using matrix multiplication
            i = zeros(3,1);
            i = RT*[x-ww/2; y-hh/2; 1];

            %Using explicit equations
            %i(1) = ( (x-ww/2)*cos(theta)+(y-hh/2)*sin(theta)+w/2);
            %i(2) = (-(x-ww/2)*sin(theta)+(y-hh/2)*cos(theta)+h/2);

            %% Nearest Neighbour
            i = round(i);
            if i(1)>0 && i(2)>0 && i(1)<=w && i(2)<=h
                im2(y,x,z) = im1(i(2),i(1),z);
            end
        end
    end
end

%Revised code
im1 = imread('file.jpg');
[h, w, p] = size(im1);
theta = (pi)/3;
hh = round(h*abs(cos(theta)) + w*abs(sin(theta)));
ww = round(w*abs(cos(theta)) + h*abs(sin(theta)));
im2 = zeros([hh,ww,p], class(im1));

R = [cos(theta) -sin(theta); sin(theta) cos(theta)];
T = [w/2; h/2];
RT = [inv(R) T; 0 0 1];

x=1:ww;
y=1:hh;

[X, Y] = meshgrid(x,y);
orig_pos = [X(:)' ; Y(:)' ; ones(1,numel(X))];
orig_pos_2 = [X(:)'-(ww/2) ; Y(:)'-(hh/2) ; ones(1,numel(X))];

new_pos = round(RT*orig_pos_2); % Round to nearest neighbour

% Check if new positions fall from map:
valid_pos = new_pos(1,:)>=1 & new_pos(1,:)<=w & new_pos(2,:)>=1 & new_pos(2,:)<=h;

orig_pos = orig_pos(:,valid_pos);
new_pos = new_pos(:,valid_pos);

siz = size(im1);
siz2 = size(im2);

% Expand the 2D indices to include the third dimension.
ind_orig_pos = sub2ind(siz2,orig_pos(2*ones(p,1),:),orig_pos(ones(p,1),:), (1:p)'*ones(1,length(orig_pos)));
ind_new_pos  = sub2ind(siz, new_pos(2*ones(p,1),:), new_pos(ones(p,1),:), (1:p)'*ones(1,length(new_pos)));

im2(ind_orig_pos) = im1(ind_new_pos);
imshow(im2);

Gunther Struyf Gunther Struyf · Accepted Answer · 2012-10-10T08:21:48

You should vectorize the for loops, it doesn't seem that hard. It'll gain you a lot. Trickiest is the position calculation and ignoring positions that fall off the image after rotation.

Solution:

x=1:ww
y=1:hh

[X, Y] = meshgrid(x,y);
orig_pos = [X(:)' ; Y(:)' ; ones(1,numel(X))];

new_pos = round(RT*orig_pos); % round to nearest neighbour

% check if new positions fall from map:
valid_pos = new_pos(1,:)>=1 & new_pos(1,:)<=w & new_pos(2,:)>=1 & new_pos(2,:)<=h;

You can do the assignment to the resulting matrix in a for loop to deal with positions to ignore, or just remove them from the source matrix and do the assignment in one blow:

orig_pos = orig_pos(:,valid_pos);
new_pos = new_pos(:,valid_pos);

siz = size(im1);
im2 = zeros(siz,class(im1));

% expand the 2d indices to include the third dimension
ind_orig_pos = sub2ind(siz,orig_pos(2*ones(siz(3),1),:),orig_pos(ones(siz(3),1),:), (1:siz(3))'*ones(1,N));
ind_new_pos  = sub2ind(siz, new_pos(2*ones(siz(3),1),:), new_pos(ones(siz(3),1),:), (1:siz(3))'*ones(1,N));

im2(ind_orig_pos) = im1(ind_new_pos);

Your code is probably also slow, because you don't initialize im2 so it is expanded in runtime as needed.

As reference: using 'peppers.png' as source image, this whole block of code took 0.12s on my pc, your code took several minutes.. end result was the same.

Matrix multiplication vs equations for image transformations in MATLAB

1 Answers