c – 用于跟踪对象的Meanshift算法发布计算搜索窗口的质心更新

2023年10月29日 239次阅读

我一直在尝试实现用于跟踪对象的meanshift算法,并且已经完成了所涉及的概念.

到目前为止,我已经成功地从我的相机生成了一个反投影流,单个通道色调roi直方图和单通道色调视频流看起来很好,我知道opencv库中有一个meanshift函数,但我正在尝试实现我自己使用opencv中提供的数据结构,计算时刻并计算搜索窗口的平均质心.

但由于某些原因,我无法在我的代码中找到问题,因为它会继续收敛到我的视频流的左上角,以便跟踪任何输入roi(感兴趣的区域).以下是计算搜索窗口质心的函数的代码片段,我觉得问题在于但不确定它是什么,如果有人能指出我正确的方向,我真的很感激：

void moment(Mat &backproj, Rect &win){

    int x_c, y_c, x_c_new, y_c_new;    
    int idx_row, idx_col;
    double m00 = 0.0 , m01 = 0.0 , m10 = 0.0 ;
    double res = 1.0, TOL = 0.003 ;

    //Set the center of search window as the center of the probabilistic image:
    y_c =  (int) backproj.rows / 2 ; 
    x_c =  (int) backproj.cols / 2 ; 

    //Centroid search solver until residual below certain tolerance:
    while (res > TOL){

        win.width = (int) 80; 
        win.height = (int) 60; 

        //First array element at position (x,y) "lower left corner" of the search window:
        win.x = (int) (x_c - win.width / 2) ;
        win.y = (int) (y_c - win.height / 2); 

        //Modulo correction since modulo of negative integer is negative in C:
        if (win.x < 0)
                win.x = win.x % backproj.cols + backproj.cols ;

        if (win.y < 0)
                win.y = win.y % backproj.rows + backproj.rows ;   

        for (int i = 0; i < win.height; i++ ){  

                //Traverse along y-axis (height) i.e. rows ensuring wrap around top/bottom boundaries:                  
                idx_row = (win.y + i) % (int)backproj.rows ;

                for (int j = 0; j < win.width; j++ ){

                        //Traverse along x-axis (width) i.e. cols ensuring wrap around left/right boundaries:
                        idx_col = (win.x + j) % (int)backproj.cols ;    
                        //Compute Moments:                            
                        m00 += (double) backproj.at<uchar>(idx_row, idx_col) ;
                        m10 += (double) backproj.at<uchar>(idx_row, idx_col) * i ;
                        m01 += (double) backproj.at<uchar>(idx_row, idx_col) * j ;
                }
        }

        //Compute new centroid coordinates of the search window:
        x_c_new = (int) ( m10 / m00 ) ;
        y_c_new = (int) ( m01 / m00 );

        //Compute the residual:
        res = sqrt( pow((x_c_new - x_c), 2.0) + pow((y_c_new - y_c), 2.0) ) ;

        //Set new search window centroid coordinates:
        x_c = x_c_new;
        y_c = y_c_new;
    }
}

这是我对stackoverflow的第二次查询,所以请原谅我忘记遵循的任何指导.

编辑

将m00,m01,m10更改为WHILE-LOOP中的块级变量而不是函数级变量,感谢Daniel Strul指出它但问题仍然存在.现在搜索窗口围绕框架边界跳跃而不是关注roi.

    void moment(Mat &backproj, Rect &win){

    int x_c, y_c, x_c_new, y_c_new;    
    int idx_row, idx_col;
    double m00 , m01 , m10 ;
    double res = 1.0, TOL = 0.003 ;

    //Set the center of search window as the center of the probabilistic image:
    y_c =  (int) backproj.rows / 2 ; 
    x_c =  (int) backproj.cols / 2 ; 

    //Centroid search solver until residual below certain tolerance:
    while (res > TOL){

        m00 = 0.0 , m01 = 0.0 , m10 = 0.0
        win.width = (int) 80; 
        win.height = (int) 60; 

        //First array element at position (x,y) "lower left corner" of the search window:
        win.x = (int) (x_c - win.width / 2) ;
        win.y = (int) (y_c - win.height / 2); 

        //Modulo correction since modulo of negative integer is negative in C:
        if (win.x < 0)
                win.x = win.x % backproj.cols + backproj.cols ;

        if (win.y < 0)
                win.y = win.y % backproj.rows + backproj.rows ;   

        for (int i = 0; i < win.height; i++ ){  

                //Traverse along y-axis (height) i.e. rows ensuring wrap around top/bottom boundaries:                  
                idx_row = (win.y + i) % (int)backproj.rows ;

                for (int j = 0; j < win.width; j++ ){

                        //Traverse along x-axis (width) i.e. cols ensuring wrap around left/right boundaries:
                        idx_col = (win.x + j) % (int)backproj.cols ;    
                        //Compute Moments:                            
                        m00 += (double) backproj.at<uchar>(idx_row, idx_col) ;
                        m10 += (double) backproj.at<uchar>(idx_row, idx_col) * i ;
                        m01 += (double) backproj.at<uchar>(idx_row, idx_col) * j ;
                }
        }

        //Compute new centroid coordinates of the search window:
        x_c_new = (int) ( m10 / m00 ) ;
        y_c_new = (int) ( m01 / m00 );

        //Compute the residual:
        res = sqrt( pow((x_c_new - x_c), 2.0) + pow((y_c_new - y_c), 2.0) ) ;

        //Set new search window centroid coordinates:
        x_c = x_c_new;
        y_c = y_c_new;
    }
}

最佳答案 您的算法总是收敛到左上角而不依赖于输入数据的原因是m00,m10和m01永远不会重置为零：

>在迭代0,对于每个时刻变量m00,m10和m01,计算正确的值m0
>在迭代0和迭代1之间,矩量变量不会重置并保持其先前的值
>因此,在迭代1中,对于每个时刻变量m00,m10和m01,您实际上将新时刻与旧时刻相加并获得(m0 m1)
>在迭代2中,您继续将前一个时刻的新时刻相加并获得(m0 m1 m2)
>依此类推,迭代迭代.

至少,应该在每次迭代开始时重置力矩变量.

理想情况下,它们不应该是函数级变量,而应该是块级变量,因为它们在循环迭代之外没有用处(除了调试目的)：

while (res > TOL){
    ...
    double m00 = 0.0, m01 = 0.0, m10 = 0.0;
    for (int i = 0; i < win.height; i++ ){
        ...

编辑1

您遇到的第二个问题(ROI在该处周围跳跃)的原因是矩的计算基于相对坐标i和j.

因此,你计算的是[avg(j),avg(i)],你真正想要的是[avg(y),avg(x)].为了解决这个问题,我提出了第一个解决方案.我已经用下面一个更简单的解决方案取代了它.

编辑2
最简单的解决方案是在每次迭代结束时添加ROI角的坐标：

    x_c_new = win.x + (int) ( m10 / m00 ) ;
    y_c_new = win.y + (int) ( m01 / m00 );
点赞