将线条图像转换为坐标集

Question

假设我有一个如上图所示的平面图的图像。是否有一种简单的处理方法将此图像转换为由坐标[（x1，y1），（x2，y2）]，[（x2，y2），（x3，y3）]，...表示的一组线， [（XN-1，炔1），（XN，YN）]

Answer 1

我想你想找到图像中所有线条的坐标。通过遵循c ++中的简单步骤很容易找到共同符号：

1. 阈值二进制反转
2. 应用Thinning algorithm以减少厚度。
3. 在二进制图像中查找NonZero像素。
4. 查找轮廓＆amp;将近似值应用于近似点。

void thinningIteration(Mat& im, int iter)
{
    Mat marker = Mat::zeros(im.size(), CV_8UC1);
    for (int i = 1; i < im.rows-1; i++)
    {
        for (int j = 1; j < im.cols-1; j++)
        {
            uchar p2 = im.at<uchar>(i-1, j);
            uchar p3 = im.at<uchar>(i-1, j+1);
            uchar p4 = im.at<uchar>(i, j+1);
            uchar p5 = im.at<uchar>(i+1, j+1);
            uchar p6 = im.at<uchar>(i+1, j);
            uchar p7 = im.at<uchar>(i+1, j-1);
            uchar p8 = im.at<uchar>(i, j-1);
            uchar p9 = im.at<uchar>(i-1, j-1);

            int A  = (p2 == 0 && p3 == 1) + (p3 == 0 && p4 == 1) + 
                     (p4 == 0 && p5 == 1) + (p5 == 0 && p6 == 1) + 
                     (p6 == 0 && p7 == 1) + (p7 == 0 && p8 == 1) +
                     (p8 == 0 && p9 == 1) + (p9 == 0 && p2 == 1);
            int B  = p2 + p3 + p4 + p5 + p6 + p7 + p8 + p9;
            int m1 = iter == 0 ? (p2 * p4 * p6) : (p2 * p4 * p8);
            int m2 = iter == 0 ? (p4 * p6 * p8) : (p2 * p6 * p8);

            if (A == 1 && (B >= 2 && B <= 6) && m1 == 0 && m2 == 0)
                marker.at<uchar>(i,j) = 1;
        }
    }
    im &= ~marker;
}

void thinning(Mat& im)
{
    im /= 255;
    Mat prev = Mat::zeros(im.size(), CV_8UC1);
    Mat diff;
    do 
    {
        thinningIteration(im, 0);
        thinningIteration(im, 1);
        absdiff(im, prev, diff);
        im.copyTo(prev);
    } 
    while (countNonZero(diff) > 0);

    im *= 255;
}

void main()
{
    Mat mSource_Bgr,mSource_Gray,mThreshold,mThinning;
    mSource_Bgr= imread(FileName_S.c_str(),IMREAD_COLOR);
    mSource_Gray= imread(FileName_S.c_str(),0);

    threshold(mSource_Gray,mThreshold,120,255,THRESH_BINARY_INV);

    mThinning= mThreshold.clone();

    thinning(mThinning);

    imshow("mThinning",mThinning);



    vector<Point2i> locations;   // output, locations of non-zero pixels 
    findNonZero(mThinning, locations);

    for (int i = 0; i < locations.size(); i++)
    {
        circle(mSource_Bgr,locations[i],2,Scalar(0,255,0),1);
    }

    imshow("mResult",mSource_Bgr);



    //To Find the Lines from the image
    vector<vector<Point> > contours;
    vector<Vec4i> hierarchy;
    /// Find contours
    findContours( mThinning.clone(), contours, hierarchy, CV_RETR_EXTERNAL, CV_CHAIN_APPROX_SIMPLE, Point(0, 0) );
    /// Approximate contours to polygons + get bounding rects and circles
    vector<vector<Point> > contours_poly( contours.size() );
    for( size_t i = 0; i < contours.size(); i++ )
    { 
        approxPolyDP( Mat(contours[i]), contours_poly[i], 2, true );
    }

    for (int i = 0; i < contours_poly.size(); i++)
    {
        drawContours(mSource_Bgr,contours_poly,i,Scalar(255,0,0),2);
    }
    imshow("mResult",mSource_Bgr);



}