问题
I want to convert an image using control points according to this scheme extracted from here:
A
and B
contains the coordinates of the source an target vertices.
I am computing the transformation matrix as:
A = [51 228; 51 127; 191 127; 191 228];
B = [152 57; 219 191; 62 240; 92 109];
X = imread('rectangle.png');
info = imfinfo('rectangle.png');
T = cp2tform(A,B,'projective');
Up to here it seems to properly work, because (using normalized coordinates) a source vertex produces its target vertex:
H = T.tdata.T;
> [51 228 1]*H
ans =
-248.2186 -93.0820 -1.6330
> [51 228 1]*H/ -1.6330
ans =
152.0016 57.0006 1.0000
The problem is that imtransform
produces an unexpected result:
Z = imtransform(X,T,'XData',[1 info.Width], 'YData',[1 info.Height]);
imwrite(Z,'projective.png');
How can I use imtransform
to produce this my expected result?:
Is there an alternative way to achieve it?
回答1:
You have to "adapt" the control points to the size of the image you're working with. The way I did this is by computing an affine transformation between the corners of the control points in A
and the corners of the source image (preferrably you want to make the points are in the same clockwise order).
One thing I should point out is that the order of points in your matrix A
does not match the picture you've shown, so I fixed that in the code below...
Here is the code to estimate the homography (tested in MATLAB):
% initial control points
A = [51 228; 51 127; 191 127; 191 228];
B = [152 57; 219 191; 62 240; 92 109];
A = circshift(A, [-1 0]); % fix the order of points to match the picture
% input image
%I = imread('peppers.png');
I = im2uint8(checkerboard(32,5,7));
[h,w,~] = size(I);
% adapt control points to image size
% (basically we estimate an affine transform from 3 corner points)
aff = cp2tform(A(1:3,:), [1 1; w 1; w h], 'affine');
A = tformfwd(aff, A);
B = tformfwd(aff, B);
% estimate homography between A and B
T = cp2tform(B, A, 'projective');
T = fliptform(T);
H = T.tdata.Tinv
I get:
>> H
H =
-0.3268 0.6419 -0.0015
-0.4871 0.4667 0.0009
324.0851 -221.0565 1.0000
Now let's visualize the points:
% check by transforming A points into B
%{
BB = [A ones(size(A,1),1)] * H; % convert to homogeneous coords
BB = bsxfun(@rdivide, BB, BB(:,end)); % convert from homogeneous coords
%}
BB = tformfwd(T, A(:,1), A(:,2));
fprintf('error = %g\n', norm(B-BB));
% visually check by plotting control points and transformed A
figure(1)
subplot(121)
plot(A([1:end 1],1), A([1:end 1],2), '.-', 'MarkerSize',20, 'LineWidth',2)
line(BB([1:end 1],1), BB([1:end 1],2), 'Color','r', 'Marker','o')
text(A(:,1), A(:,2), num2str((1:4)','a%d'), ...
'VerticalAlign','top', 'HorizontalAlign','left')
title('A'); legend({'A', 'A*H'}); axis equal ij
subplot(122)
plot(B([1:end 1],1), B([1:end 1],2), '.-', 'MarkerSize',20, 'LineWidth',2)
text(B(:,1), B(:,2), num2str((1:4)','b%d'), ...
'VerticalAlign','top', 'HorizontalAlign','left')
title('B'); legend('B'); axis equal ij
Finally we can apply the transformation on the source image:
% transform input image and show result
J = imtransform(I, T);
figure(2)
subplot(121), imshow(I), title('image')
subplot(122), imshow(J), title('warped')
回答2:
Your problem is that you accidentally cropped the output image when you specified your XData
and YData
in imtransform
. One option would be to use tformfwd
with to transform A
to compute the valid XData
and YData
ranges.
[U,V] = tformfwd(T, A(:,1), A(:,2));
Z = imtransform(X,T,'XData',[min(U) max(U)], 'YData', [min(V) max(V)]);
来源:https://stackoverflow.com/questions/32266408/warping-an-image-using-control-points