Question

我已经花了一些时间尝试从植物叶子的轮廓中实现链码，直到我遇到this question声称可以使用OpenCV方法findContours找到链码。到目前为止，我有这段代码：

public static List<MatOfPoint> chainCode;

public static void main(String[] args) {
    // load the Core OpenCV library by name
    System.loadLibrary(Core.NATIVE_LIBRARY_NAME);

    // create a display window
    Imshow window1 = new Imshow("My Image - original");

    // load an image from file (read and decode JPEG file)
    Mat inputImg = Highgui.imread("files/11. Acer palmaturu/iPAD2_C11_EX01.JPG");
    Mat grayImg = new Mat(inputImg.height(), inputImg.width(), CvType.CV_8UC1);

    //turn into binary image and invert
    Imgproc.cvtColor(inputImg, grayImg, Imgproc.COLOR_RGB2GRAY);
    Imgproc.threshold(grayImg, grayImg, 0, 255, Imgproc.THRESH_BINARY | Imgproc.THRESH_OTSU);
    Mat invertcolormatrix= new Mat(grayImg.rows(),grayImg.cols(), grayImg.type(), new Scalar(255,255,255));
    Core.subtract(invertcolormatrix, grayImg, grayImg);

    //get chain code
    chainCode = new Vector<MatOfPoint>();
    Imgproc.findContours(grayImg, chainCode, new Mat(), Imgproc.RETR_EXTERNAL, Imgproc.CHAIN_APPROX_NONE);

    //loop through indices of individual contours to find largest contour
    double largest_area = 0;
    int largest_contour_index = 0;
    for(int i = 0; i< chainCode.size(); i++){
        //find the area of the contour
        double a = Imgproc.contourArea(chainCode.get(i),false); 
        //find largest contour and it's index
        if(a > largest_area){
            largest_area= a;
            largest_contour_index = i;    
        }
     }

    //draw largest contour
    if(chainCode.size() > 0){
        Imgproc.drawContours(inputImg, chainCode, largest_contour_index, new Scalar (0, 0, 255), 1);
        System.out.println("chain code at " + largest_contour_index + "; " + chainCode.get(largest_contour_index));
    }

    //show image
    window1.showImage(inputImg);
}

现在我被困在可以获得原始链码的地方，如V = 000567454 .... 或者我应该采用完全不同的方法尝试？因为我真的被卡住了，请有人给我一些帮助！

我希望稍后使用链式代码进行进一步处理和图像识别。

如何从OpenCVs findContours方法获取Chain Code

0 个答案: