OpenCV 2.4.2 calcOpticalFlowPyrLK找不到任何点

Question

我在Linux上使用OpenCV 2.4.2.我是用C++写的.我想跟踪简单的对象(例如白色背景上的黑色矩形).首先,我使用goodFeaturesToTrack,然后使用calcOpticalFlowPyrLK在另一个图像上找到这些点.问题是calcOpticalFlowPyrLK找不到这些点.

我找到了用C语言编写的代码,在我的例子中不起作用:http://dasl.mem.drexel.edu/~noahKuntz/openCVTut9.html

我已将其转换为C++:

int main(int, char**) {
    Mat imgAgray = imread("ImageA.png", CV_LOAD_IMAGE_GRAYSCALE);
    Mat imgBgray = imread("ImageB.png", CV_LOAD_IMAGE_GRAYSCALE);
    Mat imgC = imread("ImageC.png", CV_LOAD_IMAGE_UNCHANGED);

    vector<Point2f> cornersA;

    goodFeaturesToTrack(imgAgray, cornersA, 30, 0.01, 30);

    for (unsigned int i = 0; i < cornersA.size(); i++) {
        drawPixel(cornersA[i], &imgC, 2, blue);
    }

    // I have no idea what does it do
//    cornerSubPix(imgAgray, cornersA, Size(15, 15), Size(-1, -1),
//            TermCriteria(TermCriteria::COUNT + TermCriteria::EPS, 20, 0.03));

    vector<Point2f> cornersB;
    vector<uchar> status;
    vector<float> error;

    // winsize has to be 11 or 13, otherwise nothing is found
    int winsize = 11;
    int maxlvl = 5;

    calcOpticalFlowPyrLK(imgAgray, imgBgray, cornersA, cornersB, status, error,
            Size(winsize, winsize), maxlvl);

    for (unsigned int i = 0; i < cornersB.size(); i++) {
        if (status[i] == 0 || error[i] > 0) {
            drawPixel(cornersB[i], &imgC, 2, red);
            continue;
        }
        drawPixel(cornersB[i], &imgC, 2, green);
        line(imgC, cornersA[i], cornersB[i], Scalar(255, 0, 0));
    }

    namedWindow("window", 1);
    moveWindow("window", 50, 50);
    imshow("window", imgC);

    cvWaitKey(0);

    return 0;
}

ImageA:http://oi50.tinypic.com/14kv05v.jpg

ImageB:http://oi46.tinypic.com/4l3xom.jpg

ImageC:http://oi47.tinypic.com/35n3uox.jpg

我发现它只适用于winsize = 11.我尝试在移动的矩形上使用它来检查它与原点的距离.它几乎没有检测到所有四个角落.

int main(int, char**) {
    std::cout << "Compiled at " << __TIME__ << std::endl;

    Scalar white = Scalar(255, 255, 255);
    Scalar black = Scalar(0, 0, 0);
    Scalar red = Scalar(0, 0, 255);
    Rect rect = Rect(50, 100, 100, 150);

    Mat org = Mat(Size(640, 480), CV_8UC1, white);
    rectangle(org, rect, black, -1, 0, 0);

    vector<Point2f> features;
    goodFeaturesToTrack(org, features, 30, 0.01, 30);
    std::cout << "POINTS FOUND:" << std::endl;
    for (unsigned int i = 0; i < features.size(); i++) {
        std::cout << "Point found: " << features[i].x;
        std::cout << " " << features[i].y << std::endl;
    }

    bool goRight = 1;

    while (1) {

        if (goRight) {
            rect.x += 30;
            rect.y += 30;
            if (rect.x >= 250) {
                goRight = 0;
            }
        } else {
            rect.x -= 30;
            rect.y -= 30;
            if (rect.x <= 50) {
                goRight = 1;
            }
        }

        Mat frame = Mat(Size(640, 480), CV_8UC1, white);
        rectangle(frame, rect, black, -1, 0, 0);

        vector<Point2f> found;
        vector<uchar> status;
        vector<float> error;
        calcOpticalFlowPyrLK(org, frame, features, found, status, error,
                    Size(11, 11), 5);

        Mat display;
        cvtColor(frame, display, CV_GRAY2BGR);

        for (unsigned int i = 0; i < found.size(); i++) {
            if (status[i]  == 0 || error[i] > 0) {
                continue;
            } else {
                line(display, features[i], found[i], red);
            }
        }

        namedWindow("window", 1);
        moveWindow("window", 50, 50);
        imshow("window", display);

        if (cvWaitKey(300) > 0) {
            break;
        }
    }

}

Lucas-Kanade的OpenCV实现似乎无法跟踪二进制图像上的矩形.我做错了什么或者这个功能不起作用？

Answer 1

Lucas Kanade方法通过使用该区域中的梯度来估计区域的运动.在一种情况下,梯度下降方法.因此,如果在x和y方向上没有渐变,则方法将失败.第二个重要的注意事项是Lucas Kanade方程式

E = sum_ {winsize}(Ix*u + Iy*v*It)²

是强度恒定约束的一阶泰勒近似.

I(x,y,t)= I(x + u,y + v,t + 1)

所以没有水平(图像金字塔)的方法的限制是图像需要是线性函数.在实践中,这意味着只能估计小动作,取决于你选择的胜利.这就是为什么你使用水平,线性化图像(It).所以5的水平有点高到3应该足够了.在您的情况下,顶级图像的大小为640x480/2 ^ 5 = 20 x 15.

最后你的代码中的问题是:

 if (status[i]  == 0 || error[i] > 0) {

你从lucas kanade方法得到的错误是产生的SSD意味着:

error = sum(winSize)(I(x,y,0) - I(x + u,y + u,1)^ 2)/(winsize*winsize)

错误不太可能是0.所以最后你跳过所有功能.我通过忽略错误获得了很好的经验,这只是一种信心度量.作为前瞻/后退信心,有非常好的替代信心措施.如果丢弃太多的女性,你也可以通过忽略状态标志来开始实验