使用控制点扭曲图像

Fre*_*man 13 matlab image-processing octave homography projective-geometry

我想根据从这里提取的这个方案使用控制点转换图像:

在此输入图像描述

AB包含源目标顶点的坐标.

我正在将转换矩阵计算为:

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');
Run Code Online (Sandbox Code Playgroud)

到目前为止它似乎正常工作,因为(使用标准化坐标)源顶点产生其目标顶点:

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
Run Code Online (Sandbox Code Playgroud)

问题是会imtransform产生意想不到的结果:

Z = imtransform(X,T,'XData',[1 info.Width], 'YData',[1 info.Height]);
imwrite(Z,'projective.png');
Run Code Online (Sandbox Code Playgroud)

出乎意料的结果

我如何用它imtransform来产生我预期的结果?:

在此输入图像描述

有没有其他方法可以实现它?

Amr*_*mro 15

您必须将控制点"调整"为您正在使用的图像的大小.我这样做的方法是通过计算控制点A的角和源图像的角之间的仿射变换(最好是你想让点按顺时针顺序排列).

我应该指出的一点是,矩阵中点的顺序与A您显示的图像不匹配,所以我在下面的代码中修复了...

这是估算单应性的代码(在MATLAB中测试):

% 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
Run Code Online (Sandbox Code Playgroud)

我明白了:

>> H
H =
   -0.3268    0.6419   -0.0015
   -0.4871    0.4667    0.0009
  324.0851 -221.0565    1.0000
Run Code Online (Sandbox Code Playgroud)

现在让我们想象一下这些要点:

% 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
Run Code Online (Sandbox Code Playgroud)

control_points

最后,我们可以在源图像上应用转换:

% transform input image and show result
J = imtransform(I, T);
figure(2)
subplot(121), imshow(I), title('image')
subplot(122), imshow(J), title('warped')
Run Code Online (Sandbox Code Playgroud)

warped_image