Question

使用OpenCV在图像中使用Ransac和一组点拟合一条或多条优质线条的最佳和最有效的方法是什么？

Ransac是适合生产线的最有效方式吗？

Answer 1

看看Least Mean Square metod。它比RANSAC更快更简单。

另请参阅OpenCV的fitLine方法。

当您的数据中存在大量异常值或复杂假设时，RANSAC的表现会更好。

Answer 2

RANSAC效率最高，但对于大量异常值更好。以下是使用opencv的方法：

有用的结构 -

struct SLine
{
    SLine():
        numOfValidPoints(0),
        params(-1.f, -1.f, -1.f, -1.f)
    {}
    cv::Vec4f params;//(cos(t), sin(t), X0, Y0)
    int numOfValidPoints;
};

用于成功配对的总最小二乘

cv::Vec4f TotalLeastSquares(
    std::vector<cv::Point>& nzPoints,
    std::vector<int> ptOnLine)
{
    //if there are enough inliers calculate model
    float x = 0, y = 0, x2 = 0, y2 = 0, xy = 0, w = 0;
    float dx2, dy2, dxy;
    float t;
    for( size_t i = 0; i < nzPoints.size(); ++i )
    {
        x += ptOnLine[i] * nzPoints[i].x;
        y += ptOnLine[i] * nzPoints[i].y;
        x2 += ptOnLine[i] * nzPoints[i].x * nzPoints[i].x;
        y2 += ptOnLine[i] * nzPoints[i].y * nzPoints[i].y;
        xy += ptOnLine[i] * nzPoints[i].x * nzPoints[i].y;
        w += ptOnLine[i];
    }

    x /= w;
    y /= w;
    x2 /= w;
    y2 /= w;
    xy /= w;

    //Covariance matrix
    dx2 = x2 - x * x;
    dy2 = y2 - y * y;
    dxy = xy - x * y;

    t = (float) atan2( 2 * dxy, dx2 - dy2 ) / 2;
    cv::Vec4f line;
    line[0] = (float) cos( t );
    line[1] = (float) sin( t );

    line[2] = (float) x;
    line[3] = (float) y;

    return line;
}

实际的RANSAC

SLine LineFitRANSAC(
    float t,//distance from main line
    float p,//chance of hitting a valid pair
    float e,//percentage of outliers
    int T,//number of expected minimum inliers 
    std::vector<cv::Point>& nzPoints)
{
    int s = 2;//number of points required by the model
    int N = (int)ceilf(log(1-p)/log(1 - pow(1-e, s)));//number of independent trials

    std::vector<SLine> lineCandidates;
    std::vector<int> ptOnLine(nzPoints.size());//is inlier
    RNG rng((uint64)-1);
    SLine line;
    for (int i = 0; i < N; i++)
    {
        //pick two points
        int idx1 = (int)rng.uniform(0, (int)nzPoints.size());
        int idx2 = (int)rng.uniform(0, (int)nzPoints.size());
        cv::Point p1 = nzPoints[idx1];
        cv::Point p2 = nzPoints[idx2];

        //points too close - discard
        if (cv::norm(p1- p2) < t)
        {
            continue;
        }

        //line equation ->  (y1 - y2)X + (x2 - x1)Y + x1y2 - x2y1 = 0 
        float a = static_cast<float>(p1.y - p2.y);
        float b = static_cast<float>(p2.x - p1.x);
        float c = static_cast<float>(p1.x*p2.y - p2.x*p1.y);
        //normalize them
        float scale = 1.f/sqrt(a*a + b*b);
        a *= scale;
        b *= scale;
        c *= scale;

        //count inliers
        int numOfInliers = 0;
        for (size_t i = 0; i < nzPoints.size(); ++i)
        {
            cv::Point& p0 = nzPoints[i];
            float rho      = abs(a*p0.x + b*p0.y + c);
            bool isInlier  = rho  < t;
            if ( isInlier ) numOfInliers++;
            ptOnLine[i]    = isInlier;
        }

        if ( numOfInliers < T)
        {
            continue;
        }

        line.params = TotalLeastSquares( nzPoints, ptOnLine);
        line.numOfValidPoints = numOfInliers;
        lineCandidates.push_back(line);
    }

    int bestLineIdx = 0;
    int bestLineScore = 0;
    for (size_t i = 0; i < lineCandidates.size(); i++)
    {
        if (lineCandidates[i].numOfValidPoints > bestLineScore)
        {
            bestLineIdx = i;
            bestLineScore = lineCandidates[i].numOfValidPoints;
        }
    }

    if ( lineCandidates.empty() )
    {
        return SLine();
    }
    else
    {
        return lineCandidates[bestLineIdx];
    }
}

OpenCV - Ransac装配线

2 个答案: