我尝试使用OpenCV函数cvtColorCannyHoughLinesP在下面的代码中,但是在某些情况下无法获得准确的结果或不起作用。

private boolean opencvProcessCount(Uri picFileUri) {
hairCount = 0;
totalC = 0;
//Log.e(">>>>>>>>","count " + picFileUri);
try {
    InputStream iStream = getContentResolver().openInputStream(picFileUri);
    byte[] im = getBytes(iStream);
    BitmapFactory.Options opt = new BitmapFactory.Options();
    opt.inDither = true;
    opt.inPreferredConfig = Bitmap.Config.ARGB_8888;
    Bitmap image = BitmapFactory.decodeByteArray(im, 0, im.length);

    Mat mYuv = new Mat();
    Utils.bitmapToMat(image, mYuv);
    Mat mRgba = new Mat();
    Imgproc.cvtColor(mYuv, mRgba, Imgproc.COLOR_RGB2GRAY, 4);
    Imgproc.Canny(mRgba, mRgba, 80, 90);
    Mat lines = new Mat();
    int threshold = 80;
    int minLineSize = 30;
    int lineGap = 100;

    Imgproc.HoughLinesP(mRgba, lines, 1, Math.PI/180, threshold, minLineSize, lineGap);

    for (int x = 0; x < lines.rows(); x++)
    {
        double[] vec = lines.get(x, 0);
        double x1 = vec[0],
                y1 = vec[1],
                x2 = vec[2],
                y2 = vec[3];
        Point start = new Point(x1, y1);
        Point end = new Point(x2, y2);
        double dx = x1 - x2;
        double dy = y1 - y2;

        double dist = Math.sqrt (dx*dx + dy*dy);
        totalC ++;
        Log.e(">>>>>>>>","dist " + dist);
        if(dist>300.d)
        {
            hairCount ++;
            // Log.e(">>>>>>>>","count " + x);
            Imgproc.line(mRgba, start, end, new Scalar(0,255, 0, 255),5);// here initimg is the original image.
        }// show those lines that have length greater than 300


    }

    Log.e(">>>>>>>>",totalC+" out hairCount " + hairCount);

    // Imgproc.
} catch (Throwable e) {
    // Log.e(">>>>>>>>","count " + e.getMessage());
    e.printStackTrace();
}
return false;
}

以下是用于计数头发的示例图像:

android - 如何使用OpenCV从图像中检测(计数)头发?-LMLPHP

android - 如何使用OpenCV从图像中检测(计数)头发?-LMLPHP

最佳答案

我认为您会发现这篇文章很有趣:

http://www.cs.ubc.ca/~lowe/papers/aij87.pdf

他们采用2D位图,应用Canny边缘检测器,然后根据它们属于同一对象的可能性重新组合不同边缘的 fragment -在这种情况下为头发(并提供进行此类重新组合的标准)。

我认为您可以使用它来了解图像上有多少个对象,如果图像仅包含头发,那么您就可以算出头发了。

10-08 14:58