使用 OpenCV 在灰度图像中查找局部最大值

Question

有人知道如何使用 OpenCV 在灰度 IPL_DEPTH_8U 图像中找到局部最大值吗？ HarrisCorner 提到了类似的东西，但我实际上对角落不感兴趣...... 谢谢！

Answer 1

要查找的不仅仅是全局最小值和最大值，请尝试使用skimage 中的此函数：

http://scikit-image.org/docs/dev/api/skimage.feature.html#skimage.feature.peak_local_max

您也可以参数化峰值之间的最小距离。和更多。要找到最小值，请使用否定值（不过请注意数组类型，255-image 可以解决问题）。

Answer 2

这是一个简单的技巧。这个想法是用一个在中心包含一个洞的内核来扩张。在扩张操作之后，每个像素被替换为其最大的邻居（在本例中使用 5 x 5 邻域），排除原始像素。

Mat1b kernelLM(Size(5, 5), 1u);
kernelLM.at<uchar>(2, 2) = 0u;
Mat imageLM;
dilate(image, imageLM, kernelLM);
Mat1b localMaxima = (image > imageLM);

Answer 3

找到了一个简单的解决方案。

在此示例中，如果您尝试查找 matchTemplate 函数的 2 个结果，并且彼此之间的距离最小。

    cv::Mat result;
    matchTemplate(search, target, result, CV_TM_SQDIFF_NORMED);
    float score1;
    cv::Point displacement1 = MinMax(result, score1);
    cv::circle(result, cv::Point(displacement1.x+result.cols/2 , displacement1.y+result.rows/2), 10, cv::Scalar(0), CV_FILLED, 8, 0);
    float score2;
    cv::Point displacement2 = MinMax(result, score2);

在哪里

cv::Point MinMax(cv::Mat &result, float &score)
{
    double minVal, maxVal;
    cv::Point  minLoc, maxLoc, matchLoc;

    minMaxLoc(result, &minVal, &maxVal, &minLoc, &maxLoc, cv::Mat());
    matchLoc.x = minLoc.x - result.cols/2;
    matchLoc.y = minLoc.y - result.rows/2;
    return minVal;
}

流程是：

1. 使用 minMaxLoc 查找全局最小值
2. 使用最小值之间的最小距离作为半径在全局最小值周围绘制一个实心白色圆圈
3. 找到另一个最小值

可以将分数相互比较以确定例如匹配的确定性，

Answer 4

您可以检查每个像素并测试它是否是局部最大值。这是我将如何做到的。 假设输入为 CV_32FC1 类型

#include <vector>//std::vector
#include <algorithm>//std::sort
#include "opencv2/imgproc/imgproc.hpp"
#include "opencv2/core/core.hpp"

//structure for maximal values including position
struct SRegionalMaxPoint
{
    SRegionalMaxPoint():
        values(-FLT_MAX),
        row(-1),
        col(-1)
    {}
    float values;
    int row;
    int col;
    //ascending order
    bool operator()(const SRegionalMaxPoint& a, const SRegionalMaxPoint& b)
    {   
        return a.values < b.values;
    }   
};

//checks if pixel is local max
bool isRegionalMax(const float* im_ptr, const int& cols )
{
    float center = *im_ptr;
    bool is_regional_max = true;
    im_ptr -= (cols + 1);
    for (int ii = 0; ii < 3; ++ii, im_ptr+= (cols-3))
    {
        for (int jj = 0; jj < 3; ++jj, im_ptr++)
        {
            if (ii != 1 || jj != 1)
            {
                is_regional_max &= (center > *im_ptr);
            }
        }
    }
    return is_regional_max;
}

void imregionalmax(
    const cv::Mat& input, 
    std::vector<SRegionalMaxPoint>& buffer)
{
    //find local max - top maxima
    static const int margin = 1;
    const int rows = input.rows;
    const int cols = input.cols;
    for (int i = margin; i < rows - margin; ++i)
    {
        const float* im_ptr = input.ptr<float>(i, margin);
        for (int j = margin; j < cols - margin; ++j, im_ptr++)
        {
            //Check if pixel is local maximum
            if ( isRegionalMax(im_ptr, cols ) )
            {
                cv::Rect roi = cv::Rect(j - margin, i - margin, 3, 3);
                cv::Mat subMat = input(roi);

                float val = *im_ptr;
                //replace smallest value in buffer
                if ( val > buffer[0].values )
                {
                    buffer[0].values = val;
                    buffer[0].row    = i;
                    buffer[0].col    = j;
                    std::sort(buffer.begin(), buffer.end(), SRegionalMaxPoint());
                }

            }
        }
    }

}

为了测试代码，你可以试试这个：

cv::Mat temp = cv::Mat::zeros(15, 15, CV_32FC1);
temp.at<float>(7, 7) = 1;
temp.at<float>(3, 5) = 6;
temp.at<float>(8, 10) = 4;
temp.at<float>(11, 13) = 7;
temp.at<float>(10, 3) = 8;
temp.at<float>(7, 13) = 3;

vector<SRegionalMaxPoint> buffer_(5);
imregionalmax(temp, buffer_);

cv::Mat debug;
cv::cvtColor(temp, debug, cv::COLOR_GRAY2BGR);
for (auto it = buffer_.begin(); it != buffer_.end(); ++it)
{
    circle(debug, cv::Point(it->col, it->row), 1, cv::Scalar(0, 255, 0));
}

这个解决方案没有考虑高原，所以它与matlab的imregionalmax()并不完全相同

Answer 5

下面的清单是一个类似于 Matlab 的“imregionalmax”的函数。它最多在 threshold 之上寻找 nLocMax 个局部最大值，其中找到的局部最大值相距至少 minDistBtwLocMax 个像素。它返回找到的局部最大值的实际数量。请注意，它使用 OpenCV 的 minMaxLoc 来查找全局最大值。它是“opencv-self-contained”的，除了（易于实现）函数vdist，它计算点（r，c）和（row，col）之间的（欧几里德）距离。

input 是单通道 CV_32F 矩阵，locations 是 nLocMax（行）乘 2（列）CV_32S 矩阵。

int imregionalmax(Mat input, int nLocMax, float threshold, float minDistBtwLocMax, Mat locations)
{
    Mat scratch = input.clone();
    int nFoundLocMax = 0;
    for (int i = 0; i < nLocMax; i++) {
        Point location;
        double maxVal;
        minMaxLoc(scratch, NULL, &maxVal, NULL, &location);
        if (maxVal > threshold) {
            nFoundLocMax += 1;
            int row = location.y;
            int col = location.x;
            locations.at<int>(i,0) = row;
            locations.at<int>(i,1) = col;
            int r0 = (row-minDistBtwLocMax > -1 ? row-minDistBtwLocMax : 0);
            int r1 = (row+minDistBtwLocMax < scratch.rows ? row+minDistBtwLocMax : scratch.rows-1);
            int c0 = (col-minDistBtwLocMax > -1 ? col-minDistBtwLocMax : 0);
            int c1 = (col+minDistBtwLocMax < scratch.cols ? col+minDistBtwLocMax : scratch.cols-1);
            for (int r = r0; r <= r1; r++) {
                for (int c = c0; c <= c1; c++) {
                    if (vdist(Point2DMake(r, c),Point2DMake(row, col)) <= minDistBtwLocMax) {
                        scratch.at<float>(r,c) = 0.0;
                    }
                }
            }
        } else {
            break;
        }
    }
    return nFoundLocMax;
}

Answer 6

如果像素等于“局部”邻域中的最大值，则将其视为局部最大值。下面的函数在两行代码中捕获了这个属性。

要处理“高原”上的像素（值等于它们的邻域），可以使用局部最小值属性，因为高原像素等于它们的局部最小值。其余代码过滤掉这些像素。

void non_maxima_suppression(const cv::Mat& image, cv::Mat& mask, bool remove_plateaus) {
    // find pixels that are equal to the local neighborhood not maximum (including 'plateaus')
    cv::dilate(image, mask, cv::Mat());
    cv::compare(image, mask, mask, cv::CMP_GE);

    // optionally filter out pixels that are equal to the local minimum ('plateaus')
    if (remove_plateaus) {
        cv::Mat non_plateau_mask;
        cv::erode(image, non_plateau_mask, cv::Mat());
        cv::compare(image, non_plateau_mask, non_plateau_mask, cv::CMP_GT);
        cv::bitwise_and(mask, non_plateau_mask, mask);
    }
}

Answer 7

这是一种非常快速的方法。它将建立的最大值存储在一个向量中 积分。

vector <Point> GetLocalMaxima(const cv::Mat Src,int MatchingSize, int Threshold, int GaussKernel  )
{  
  vector <Point> vMaxLoc(0); 

  if ((MatchingSize % 2 == 0) || (GaussKernel % 2 == 0)) // MatchingSize and GaussKernel have to be "odd" and > 0
  {
    return vMaxLoc;
  }

  vMaxLoc.reserve(100); // Reserve place for fast access 
  Mat ProcessImg = Src.clone();
  int W = Src.cols;
  int H = Src.rows;
  int SearchWidth  = W - MatchingSize;
  int SearchHeight = H - MatchingSize;
  int MatchingSquareCenter = MatchingSize/2;

  if(GaussKernel > 1) // If You need a smoothing
  {
    GaussianBlur(ProcessImg,ProcessImg,Size(GaussKernel,GaussKernel),0,0,4);
  }
  uchar* pProcess = (uchar *) ProcessImg.data; // The pointer to image Data 

  int Shift = MatchingSquareCenter * ( W + 1);
  int k = 0;

  for(int y=0; y < SearchHeight; ++y)
  { 
    int m = k + Shift;
    for(int x=0;x < SearchWidth ; ++x)
    {
      if (pProcess[m++] >= Threshold)
      {
        Point LocMax;
        Mat mROI(ProcessImg, Rect(x,y,MatchingSize,MatchingSize));
        minMaxLoc(mROI,NULL,NULL,NULL,&LocMax);
        if (LocMax.x == MatchingSquareCenter && LocMax.y == MatchingSquareCenter)
        { 
          vMaxLoc.push_back(Point( x+LocMax.x,y + LocMax.y )); 
          // imshow("W1",mROI);cvWaitKey(0); //For gebug              
        }
      }
    }
    k += W;
  }
  return vMaxLoc; 
}

Answer 8

实际上，在我发布上面的代码之后，我写了一个更好、更快的代码.. 即使是 640x480 的图片，上面的代码也会受到影响。 我对其进行了优化，现在即使是 1600x1200 的图片也非常快。 这是代码：

void localMaxima(cv::Mat src,cv::Mat &dst,int squareSize)
{
if (squareSize==0)
{
    dst = src.clone();
    return;
}

Mat m0;
dst = src.clone();
Point maxLoc(0,0);

//1.Be sure to have at least 3x3 for at least looking at 1 pixel close neighbours
//  Also the window must be <odd>x<odd>
SANITYCHECK(squareSize,3,1);
int sqrCenter = (squareSize-1)/2;

//2.Create the localWindow mask to get things done faster
//  When we find a local maxima we will multiply the subwindow with this MASK
//  So that we will not search for those 0 values again and again
Mat localWindowMask = Mat::zeros(Size(squareSize,squareSize),CV_8U);//boolean
localWindowMask.at<unsigned char>(sqrCenter,sqrCenter)=1;

//3.Find the threshold value to threshold the image
    //this function here returns the peak of histogram of picture
    //the picture is a thresholded picture it will have a lot of zero values in it
    //so that the second boolean variable says :
    //  (boolean) ? "return peak even if it is at 0" : "return peak discarding 0"
int thrshld =  maxUsedValInHistogramData(dst,false);
threshold(dst,m0,thrshld,1,THRESH_BINARY);

//4.Now delete all thresholded values from picture
dst = dst.mul(m0);

//put the src in the middle of the big array
for (int row=sqrCenter;row<dst.size().height-sqrCenter;row++)
    for (int col=sqrCenter;col<dst.size().width-sqrCenter;col++)
    {
        //1.if the value is zero it can not be a local maxima
        if (dst.at<unsigned char>(row,col)==0)
            continue;
        //2.the value at (row,col) is not 0 so it can be a local maxima point
        m0 =  dst.colRange(col-sqrCenter,col+sqrCenter+1).rowRange(row-sqrCenter,row+sqrCenter+1);
        minMaxLoc(m0,NULL,NULL,NULL,&maxLoc);
        //if the maximum location of this subWindow is at center
        //it means we found the local maxima
        //so we should delete the surrounding values which lies in the subWindow area
        //hence we will not try to find if a point is at localMaxima when already found a neighbour was
        if ((maxLoc.x==sqrCenter)&&(maxLoc.y==sqrCenter))
        {
            m0 = m0.mul(localWindowMask);
                            //we can skip the values that we already made 0 by the above function
            col+=sqrCenter;
        }
    }
}

Answer 9

我想你想使用

MinMaxLoc(arr, mask=NULL)-> (minVal, maxVal, minLoc, maxLoc)
Finds global minimum and maximum in array or subarray

作用于你的图像

Answer 10

首先要回答的问题是您认为什么是“本地”。答案很可能是一个方形窗口（比如 3x3 或 5x5）或一定半径的圆形窗口。然后，您可以扫描整个图像，窗口以每个像素为中心，并选择窗口中的最大值。

请参阅this，了解如何在 OpenCV 中访问像素值。