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我不能使用我需要从头开始构建它的任何工具箱功能。

% load images
img1 = readImage('roadSign.tif');
img2 = readImage('lena.tif');

% call the main function  
mapIntoImage(img1,img2)


function    [newImage] = mapIntoImage(imageA,imageB)
%  Input:      imageA, imageB - a grayscale image in the range [0..255].
% 
%  Output:    newImage – imageA into which image B has been mapped.
% 
showImage(imageA)
hold on
% Initially, the list of points is empty.
xy = [];
% Loop, picking up the points.
disp('Please enter corners of place to insert image in clockwise order.')

for j = 1:4
   [xi,yi] = ginput(1);
   %draw a yellow dot
   plot(xi,yi,'yo')
   xy(:,j) = [xi;yi];
end

% get x1 y1 cordinates  - xy(:, 1)

imgRow = size(imageB,1);
imgCol = size(imageB,2);

[X,Y] = meshgrid(1:imgCol,1:imgRow);

imgBcords = [0 size(imageB, 1) size(imageB,1) 0 ;
        0 0 size(imageB,2) size(imageB,2)];
coefs = findCoefficients(xy, imgBcords);

A = [coefs(1) coefs(2) coefs(5);coefs(3) coefs(4) coefs(6); coefs(7) coefs(8) 1];

temp = zeros(size(X,1), size(X,2), 3);
new = ones(256);
for i = 1:size(X,1)
    for j = 1:size(X,2)
        temp(i,j,:) =A*[X(i,j); Y(i,j); new(i,j)];
end
end

end 

function [ result ] = findCoefficients( imageA, imageB )
% finds coefficients for inverse mapping algorithem 
%   takes 2 X 2d vectors each consists of 4 points x,y
%   and returns the coef accroding to reverse mapping function
%
% x y 0 0 1 0 -xx' -yx'
% 0 0 x y 0 1 -xy' -yy'  
%                       y' and x' are in the destenation picture;


A = [imageB(1,1) imageB(2,1) 0 0 1 0 -imageB(1,1)*imageA(1,1) -imageB(2,1)*imageA(1,1); 
    0 0 imageB(1,1) imageB(2,1) 0 1 -imageB(1,1)*imageA(2,1) -imageB(2,1)*imageA(2,1);
imageB(1,2) imageB(2,2) 0 0 1 0 -imageB(1,2)*imageA(1,2) -imageB(2,2)*imageA(1,2);
0 0 imageB(1,2) imageB(2,2) 0 1 -imageB(1,2)*imageA(2,2) -imageB(2,2)*imageA(2,2);
imageB(1,3) imageB(2,3) 0 0 1 0 -imageB(1,3)*imageA(1,3) -imageB(2,3)*imageA(1,3);
0 0 imageB(1,3) imageB(2,3) 0 1 -imageB(1,3)*imageA(2,3) -imageB(2,3)*imageA(2,3);
imageB(1,4) imageB(2,4) 0 0 1 0 -imageB(1,4)*imageA(1,4) -imageB(2,4)*imageA(1,4);
0 0 imageB(1,4) imageB(2,4) 0 1 -imageB(1,4)*imageA(2,4) -imageB(2,4)*imageA(2,4)];
B = [imageB(1,1); imageB(2,1); imageB(1,2); imageB(2,2); imageB(1,3); imageB(2,3); imageB(1,4); imageB(2,4)];

result = pinv(A)*B;
end

我现在想构建变换 [x' y' 1] = A*[XY 1]; 我发现我需要使用 repmat，但我似乎无法在没有循环的情况下获得真正的语法。最有效的方法是什么？

4

1 回答 1

1

射影变换的形式为

$ x' = \frac {a_{11}x+a_{12}y+a_{13}}{a_{13}x+a_{23}y+a_{33}} \\ y' = \frac { a_{21}x+a_{22}y+a_{23}}{a_{13}x+a_{23}y+a_{33}} $

其中系数被定义为某个比例因子。确保比例因子恒定的方法之一是设置 $a_{33}=1$。考虑它的一种简单方法是使用齐次坐标：

$ \left( \begin{array}{ccc} x'\\y'\\S\end{array} \right) = \left( \begin{array}{ccc} a_{11} & a_{12} & a_{13}\\a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\end{array} \right) \left( \begin {array}{ccc} x\\y\\1\end{array} \right) $

这些坐标是按比例定义的。那是，

$ \left( \begin{array}{ccc} x'/S\\y'/S\\1\end{array} \right) \equiv \left( \begin{array}{ccc} x'\\ y'\\S\end{数组} \right)$

因此，在您的情况下，您应该这样做：（假设xandy是列向量，并且A是我上面描述的矩阵的转置：

  XY = A * [x y ones(size(x))];
  XY(:,1) = XY(:,1)./XY(:,3);
  XY(:,2) = XY(:,2)./XY(:,3);
于 2012-12-01T21:26:19.270 回答