我正在尝试从头开始实现openCV方法warpPerspective()，我编写了下面的代码，它可以处理y和x的变化，但是，当我将单应矩阵从findHomography()传递给函数时，我让它总是给出空白图像warpPerspective() 输出。
我按照这个定义来找到像素的新位置：
s*x' h1 h2 h3 x
s*y' = h4 h5 h6 * y
s h7 h8 1 1
我的映射适用于简单的转变，例如 { 1, 0.5,-51,0,1,50,0,0,1}
但是当矩阵是这样
[1.0340946, 0.032195676, -6.419126;
0.00302419, 1.0487343, -96.520393;
3.7013847e-06, 0.00010837225, 1]
的：输出是这样的：
我的实现： - 给定 H 和图像 A， - 查找 A 中像素的新位置并将它们保存在 TransArry 中。数组的索引是 A 的线性化索引。 - 将 A 的像素重新映射到 tranImg。
Mat transform(Mat A, Mat H)
{
// allocate array of all locations
int Numrows = A.rows;
int Numcols = A.cols;
int channels = A.channels();
cout << "rows " << Numrows << "col " << Numcols << "channels " << channels <<endl;
int size = Numrows*Numcols;
int MaxX,MaxY = -1000;
int MinX,MinY = 1000;
int *TransArry = (int *)malloc(sizeof(int)*size);
int Idx;
int homeX=Idx % Numcols;
int homeY=Idx / Numcols;
cout << H << endl;
waitKey();
for (Idx=0; Idx < size; ++Idx ){
homeX=Idx % Numcols;
homeY=Idx / Numcols;
float x = (H.at<float>(0,0) * (homeX)) +( H.at<float>(0,1) * (homeY)) + ( H.at<float>(0,2) * 1) ;
float y = (H.at<float>(1,0) * (homeX)) +( H.at<float>(1,1) * (homeY)) + ( H.at<float>(1,2) * 1) ;
float s = (H.at<float>(2,0) * (homeX)) +( H.at<float>(2,1) * (homeY)) + ( H.at<float>(2,2) * 1) ;
cout << " x = " << x << " y= " << y << " s= " << s;
x = (x/s);
y = y/s;
// for the first col in TransMatrix
if (homeX ==0){
if (x > MaxX) MaxX = x;
if (x < MinX) MinX = x;
}
//for thee first row in TransMatrix
if (homeY ==0){
if (y > MaxY) MaxY = y;
if (y < MinY) MinY = y;
}
if((y)>=A.rows || (y)<0 || (x)>=A.cols || (x)<0){
TransArry[Idx] = -1;
cout << "x= " << x << "y= "<< y << endl;
}else{
TransArry[Idx] = (y * Numcols + x);
}
//cout << Numcols << endl;
cout << "New index of " << Idx << "is " << TransArry[Idx] << endl;
}
Mat tranImg ;
A.copyTo(tranImg);
tranImg = tranImg - tranImg;
cout << "Rows" << tranImg.rows << "cols" << tranImg.cols << "cha" << A.channels() << endl;
waitKey();
// Remap Image
for (Idx=0; Idx < size; Idx ++ ){
homeX=Idx % Numcols;
homeY=Idx / Numcols;
//tranImg.at<uchar>(homeY, homeX) =0;
if(TransArry[Idx] != -1){
//cout << "Index " << Idx << "Passed " << endl;
int newhomeX=TransArry[Idx] % Numcols; // Col ID
int newhomeY=TransArry[Idx] / Numcols; // Row ID
cout << "Index is " << Idx << endl;
cout << "HomeX is " << homeX << " and HomeY is " << homeY << endl;
cout << "New Index is " << TransArry[Idx] << endl;
cout << "New HomeX is " << newhomeX << " and New HomeY is " << newhomeY << endl;
cout << "*****************************************"<< endl;
// if (!(Idx%100)) sleep(20);
tranImg.at<uchar>(newhomeY, (newhomeX*channels)) = A.at<uchar>(homeY, homeX*channels);
if(channels>1)
tranImg.at<uchar>(newhomeY, newhomeX*channels+1) = A.at<uchar>(homeY, homeX*channels+1);
if(channels>2)
tranImg.at<uchar>(newhomeY, newhomeX*channels+2) = A.at<uchar>(homeY, homeX*channels+2);
// if (!(Idx%100)){
// imshow("inside", tranImg);
// waitKey(1);
// }
}
}
//cout << tranImg << endl;
return tranImg;
}
计算和验证 H。
那么，我访问矩阵 H 和 A 的方式有问题吗？