Find local maxima in grayscale image using OpenCV

Question

Does anybody know how to find the local maxima in a grayscale IPL_DEPTH_8U image using OpenCV? HarrisCorner mentions something like that but I'm actually not interested in corners ... Thanks!

Answer 1

A pixel is considered a local maximum if it is equal to the maximum value in a 'local' neighborhood. The function below captures this property in two lines of code.

To deal with pixels on 'plateaus' (value equal to their neighborhood) one can use the local minimum property, since plateaus pixels are equal to their local minimum. The rest of the code filters out those pixels.

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 2

Here's a simple trick. The idea is to dilate with a kernel that contains a hole in the center. After the dilate operation, each pixel is replaced with the maximum of it's neighbors (using a 5 by 5 neighborhood in this example), excluding the original pixel.

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

Answer 3

Actually after I posted the code above I wrote a better and very very faster one .. The code above suffers even for a 640x480 picture.. I optimized it and now it is very very fast even for 1600x1200 pic. Here is the code :

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 4

The following listing is a function similar to Matlab's "imregionalmax". It looks for at most nLocMax local maxima above threshold, where the found local maxima are at least minDistBtwLocMax pixels apart. It returns the actual number of local maxima found. Notice that it uses OpenCV's minMaxLoc to find global maxima. It is "opencv-self-contained" except for the (easy to implement) function vdist, which computes the (euclidian) distance between points (r,c) and (row,col).

input is one-channel CV_32F matrix, and locations is nLocMax (rows) by 2 (columns) CV_32S matrix.

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 5

The first question to answer would be what is "local" in your opinion. The answer may well be a square window (say 3x3 or 5x5) or circular window of a certain radius. You can then scan over the entire image with the window centered at each pixel and pick the highest value in the window.

See this for how to access pixel values in OpenCV.

Answer 6

This is very fast method. It stored founded maxima in a vector of Points.

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 7

Found a simple solution.

In this example, if you are trying to find 2 results of a matchTemplate function with a minimum distance from each other.

    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);

where

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;
}

The process is:

1. Find global Minimum using minMaxLoc
2. Draw a filled white circle around global minimum using min distance between minima as radius
3. Find another minimum

The the scores can be compared to each other to determine, for example, the certainty of the match,

Answer 8

To find more than just the global minimum and maximum try using this function from skimage:

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

You can parameterize the minimum distance between peaks, too. And more. To find minima, use negated values (take care of the array type though, 255-image could do the trick).

Answer 9

You can go over each pixel and test if it is a local maxima. Here is how I would do it. The input is assumed to be type CV_32FC1

#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());
                }

            }
        }
    }

}

For testing the code you can try this:

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));
}

This solution does not take plateaus into account so it is not exactly the same as matlab's imregionalmax()

Answer 10

I think you want to use the

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

function on you image