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我正在使用 Emgu CV 2.4.2 并希望执行以下算法:

  1. 获取 blob
  2. 设置 ROI 以加快计算速度
  3. 从 blob 中获取局部最小值的像素位置
  4. 划分blob
  5. 将边界矩形绘制成分割的斑点

我已经完成了步骤 1-2 并使用 BGStatModel 提取了 blob。这是我得到的结果:

斑点图片

我想在垂直投影中获得局部最小值的像素位置。得到它之后,我想划分blob并绘制这样的矩形:

垂直投影图片

我试图通过检查 blob 区域中的每个像素来获取局部最小值的像素位置,但这会使我的应用程序运行得非常慢。这是我的代码:

Point minPix = new Point(0,0);

//copy the foreground frame
Image<Gray, Byte> foreFrame_copy = foreFrame.Copy();

//find the contour
Contour<Point> contours = foreFrame.FindContours(
    CHAIN_APPROX_METHOD.CV_CHAIN_APPROX_SIMPLE,
    RETR_TYPE.CV_RETR_EXTERNAL);

//looping every contour
while (contours != null)
{
    double PPixel = contours.Area;
    if (PPixel >= 1400)
    {
        //get the contour width
        WR = contours.BoundingRectangle.Width;
        //divide the contour using estimated pixel position
        Num = Convert.ToInt32(WR / 40);

    if (Num > 1)
    {
        //save the estimated pixel position for ROI in arraylist
        ArrayList XList = new ArrayList();
        for (int i = 1; i <= Num; i++)
        {
            int x = i * WR / Num;
            XList.Add(x);
        }

        //get the estimated pixel position 
        foreach (int pos in Xlist)
        {
            //roiWidth= 10px
            int roiWidth = (pos-5) + (pos+5);
            //roiHeight= 20px
            int roiHeight = 20;

            int pixValue = 0;
            //STEP 2: set the ROI to speed up computation
            foreFrame_copy.ROI = new Rectangle(contours.BoundingRectangle.X, contours.BoundingRectangle.Y, roiWidth, roiHeight);
            for (int i = (pos-5); i < roiWidth; i++)
            {
                for (int j = (pos-5); j < (pos+5); j++)
                {
                    pixValue = foreFrame_copy.Data[i, j, 0];
                    //find the white pixel
                    if (pixValue == 255) {
                        //find the position of minimum pixel
                        if (j < j-1) {
                        minPix.X = i;
                        minPix.Y = j;
                        }
                    }

                }
            }
        }

    }
    //draw the red rectangle
    estimatedFrame.Draw(contours.BoundingRectangle, new Bgr(Color.Red), 1);
    contours = contours.HNext;
}
else
{
    contours = contours.HNext;
}
}
//show frame in framebox
blobBox.Image = foreFrame_copy;
estimatedBox.Image = estimatedFrame;

请帮助我如何使用 Emgu CV 以最快的方式执行步骤 2-5。如果有人详细说明这三个步骤和一些代码,我将不胜感激。

在此先感谢,大卫

4

1 回答 1

0

好的,所以问题现在变得更加清楚。我个人同意 mmgp 用户的观点,因为缺乏通用性,您作为参考的论文并不是那么好。它是针对一个非常特殊的场景进行编码的,并且对受控环境进行了一些强有力的假设。

我为您提供了一种计算垂直和水平投影以及绝对最大值点的方法作为一个很好的起点

   private void ComputeProjections(Image<Bgr, byte> inputImage)
    {                        
        Image<Gray, Byte> inputGrayImage = inputImage.Convert<Gray, Byte>();
        Matrix<float> imgMatHorizontal = new Matrix<float>(inputGrayImage.Height, 1, 1);
        Matrix<float> imgMatVertical = new Matrix<float>(1, inputGrayImage.Width, 1);

        inputGrayImage.Reduce<float>(imgMatHorizontal, REDUCE_DIMENSION.SINGLE_COL, REDUCE_TYPE.CV_REDUCE_AVG);
        inputGrayImage.Reduce<float>(imgMatVertical, REDUCE_DIMENSION.SINGLE_ROW, REDUCE_TYPE.CV_REDUCE_AVG);
        double minH, maxH, minV, maxV;
        Point minLocH, maxLocH, minLocV, maxLocV;
        imgMatHorizontal.MinMax(out minH, out maxH, out minLocH, out maxLocH);
        imgMatVertical.MinMax(out minV, out maxV, out minLocV, out maxLocV);

        Image<Gray, Byte> maskProvaH = new Image<Gray, byte>(new Size((int)(maxH - minH + 1), imgMatHorizontal.Rows));
        Image<Gray, Byte> maskProvaV = new Image<Gray, byte>(new Size(imgMatVertical.Cols, (int)(maxV - minV + 1)));

        for (int i = 0; i < imgMatHorizontal.Rows; i++)
            maskProvaH.Draw(new CircleF(new PointF((float)(imgMatHorizontal[i, 0] - minH), i), 1f), new Gray(255), 1);

        for (int i = 0; i < imgMatVertical.Cols; i++)
            maskProvaV.Draw(new CircleF(new PointF(i, (float)(imgMatVertical[0, i] - minV)), 1f), new Gray(255), 1);

        inputImage.Draw(new CircleF(new PointF(minLocV.X, minLocH.Y), 2), new Bgr(Color.Green), 1);
        //imageBoxProjected.Image = inputGrayImage;
        imageBoxHorizProj.Image = maskProvaH;
        //imageBoxVerticProj.Image = maskProvaV;

    }

如果您有兴趣找到参考论文中的局部最大值,您可以开始实施简单的局部最大值搜索,例如(水平投影代码):

List<Point> localMaxima = new List<Point>();
        imgMatHorizontal.MinMax(out minH, out maxH, out minLocH, out maxLocH);

        Image<Gray, Byte> maskProvaH = new Image<Gray, byte>(new Size((int)(maxH - minH + 1), imgMatHorizontal.Rows));
        for (int i = 0; i < imgMatHorizontal.Rows; i++)
            maskProvaH.Draw(new CircleF(new PointF((float)(imgMatHorizontal[i, 0] - minH), i), 1f), new Gray(255), 1);

        //// Absolute Horizontal Projection Maxima Drawing
        //inputGrayImage.Draw(new Cross2DF(new PointF(maxLocH.X, maxLocH.Y), 2, 2), new Gray(255), 2);

        // Local maximas search
        for (int i = imgMatHorizontal.Rows - 2; i > 1; i--)
        {
            float prev = imgMatHorizontal[i + 1, 0];
            float curr = imgMatHorizontal[i, 0];
            float next = imgMatHorizontal[i - 1, 0];
            if (curr >= prev && curr >= next)
                localMaxima.Add(new Point((int)curr, i));
        }

此代码来自我的一个 youtube 演示:

使用投影的面部特征演示

其余步骤非常简单,您可以根据自己的需要进行调整,或者按照论文中的建议遵循一些交叉比率阈值。

于 2012-12-12T20:38:12.747 回答