Function to determine all local maxima of a histogram - c++

Is there an OpenCV function that can give me a list of all the local maxima for a histogram? Maybe there is a function that lets me specify a minimum peak/threshold and will tell me the bins of all those local maxima above that threshold.
If not, is there a function that can sort the bins from highest(most frequent) to lowest (least frequent). I can then grab all the first 20 or so bins and I have my 20 biggest local maxima.

Opencv minMaxLoc can be used in this context with a sliding window. If the location of the maxima is on an edge then ignore the maxima, otherwise record as maxima. You can use something like the function below (Note: this code is more like psuedocode it has not been tested)
/**
* Assumes a 1 channel histogram
*/
vector<int> findMaxima(Mat histogram, int windowsize, int histbins){
vector<int> maximas;
int lastmaxima;
for(int i = 0; i < histbins - windowsize; i++){
//Just some Local variables, only maxloc and maxval are used.
int maxval,minval;
Point* maxloc, maxloc;
//Crop the windows
Rect window(i,0,windowsize,1);
//Get the maxima
minMaxLoc(histogram(window), minval,maxval,maxloc,minloc);
//Check if its not on the side
if(maxloc.x != 0&&maxloc.x != windowsize-1){
//Translate from cropped window into real position
int originalposition = maxloc.x+i;
//Check that this is a new maxima and not already recorded
if(lastmaxima != originalposition){
maximas.push(originalposition);
lastmaxima = originalposition;
}
}
}
return maximas;
}
Of course this is a very simplistic system. You might want to use a multiscale approach with different sliding window sizes. You may also need to apply gaussian smoothing depending on your data. Another approach could be to run this for a small window size like 3 or 4 (you need a mimimum of 3). Then you could use something else for non maxima-suppression.
For your approach in which you suggested
Maybe there is a function that lets me specify a minimum peak/threshold and will tell me the bins of all those local maxima above that threshold.
You could simply perform a threshold before finding the maxima with the above function.
threshold(hist,res ...aditional parameters...);
vector<int> maximas = findMaximas(hist, ...other parameters...);

AFAIK OpenCV doesn't have such functionality, but it is possible do implement something similar yourself.
In order to sort histogram bins you can possibly use sortIdx, but as a result you will obtain list of largest bins, which is different than local maxima (those should be "surrounded" by smaller values).
To obtain local maxima you can compare each bin with its neighbors (2 in 1D case). A bin should be larger than neighbors with some margin to be considered a local maximum.
Depending on the size of the bins, you may want to filter the histogram before this step (for example convolve it with Gaussian kernel), since otherwise you'd obtain too much of these maxima, especially for small bin sizes. If you've used Gaussian kernel - it's sigma would be related to the size of the neighborhood in which detected local maxima are "global".
Once you detect those points - you may want to perform non-maximal suppression, to replace groups of points that lie very close together with a single point. A simple strategy for that would be to sort those maxima according to some criteria (for example difference with neighbors), then take one maximum and remove all the points in its neighborhood (its size can be related the the Gaussian kernel sigma), take next remaining maximum and again remove points in its neighborhood and so on until you run out of points or go below some meaningful difference values.
Finally, you may want to sort remaining candidate points by their absolute values (to get "largest" local maxima), or by their differences with neighbors (to get "sharpest" ones).
You may try another approach. We can use this definition of local maximum to implement a simpler algorithm: just move a sliding window of size S along the histogram and pick maximum in each position. This will have some problems:
in locations with prominent maximum multiple window positions will generate points that correspond to the same maximum (can be fixed with non maximum suppression),
in locations with no or small variation it will return
semi-random maxima (can be fixed with threshold on variance in
window or difference between maximum and neighborhood),
in regions with monotonic histogram it will return a largest value (which is not necessarily a maximum).
Once you perform all the "special case" handling - those 2 approaches would be quite similar I believe.
Another thing to implement may be "multi scale" approach, which can be considered as an extension if those 2. Basically it boils down to detecting local maxima for different neighborhood sizes, and then storing them all along with corresponding neighborhood size, which can be helpful for some purposes.
As you can see, this is a quite vague guide, and there's a reason for that: the type and amount of local maximas you want to get will most likely depend on the problem you have in mind. There's no hard and easy rule to decide if the point should be considered a local maxima, so you should probably start with some simple approach and then refine it for your specific case.

Related

Histogram Binning of Gradient Vectors

I am working on a project that has a small component requiring the comparison of distributions over image gradients. Assume I have computed the image gradients in the x and y directions using a Sobel filter and have for each pixel a 2-vector. Obviously getting the magnitude and direction is reasonably trivial and is as follows:
However, what is not clear to me is how to bin these two components in to a two dimensional histogram for an arbitrary number of bins.
I had considered something along these lines(written in browser):
//Assuming normalised magnitudes.
//Histogram dimensions are bins * bins.
int getHistIdx(float mag, float dir, int bins) {
const int magInt = reinterpret_cast<int>(mag);
const int dirInt = reinterpret_cast<int>(dir);
const int magMod = reinterpret_cast<int>(static_cast<float>(1.0));
const int dirMod = reinterpret_cast<int>(static_cast<float>(TWO_PI));
const int idxMag = (magInt % magMod) & bins
const int idxDir = (dirInt % dirMod) & bins;
return idxMag * bins + idxDir;
}
However, I suspect that the mod operation will introduce a lot of incorrect overlap, i.e. completely different gradients getting placed in to the same bin.
Any insight in to this problem would be very much appreciated.
I would like to avoid using any off the shelf libraries as I want to keep this project as dependency light as possible. Also I intend to implement this in CUDA.
This is more of a what is an histogram question? rather than one of your tags. Two things:
In a 2D plain two directions equal by modulation of 2pi are in fact the same - so it makes sense to modulate.
I see no practical or logical reason of modulating the norms.
Next, you say you want a "two dimensional histogram", but return a single number. A 2D histogram, and what would make sense in your context, is a 3D plot - the plane is theta/R, 2 indexed, while the 3D axis is the "count".
So first suggestion, return
return Pair<int,int>(idxMag,idxDir);
Then you can make a 2D histogram, or 2 2D histograms.
Regarding the "number of bins"
this is use case dependent. You need to define the number of bins you want (maybe different for theta and R). Maybe just some constant 10 bins? Maybe it should depend on the amount of vectors? In any case, you need a function that receives either the number of vectors, or the total set of vectors, and returns the number of bins for each axis. This could be a constant (10 bins) initially, and you can play with it. Once you decide on the number of bins:
Determine the bins
For a bounded case such as 0<theta<2 pi, this is easy. Divide the interval equally into the number of bins, assuming a flat distribution. Your modulation actually handles this well - if you would have actually modulated by 2*pi, which you didn't. You would still need to determine the bin bounds though.
For R this gets trickier, as this is unbounded. Two options here, but both rely on the same tactic - choose a maximal bin. Either arbitrarily (Say R=10), so any vector longer than that is placed in the "longer than max" bin. The rest is divided equally (for example, though you could choose other distributions). Another option is for the longest vector to determine the edge of the maximal bin.
Getting the index
Once you have the bins, you need to search the magnitude/direction of the current vector in your bins. If bins are pairs representing min/max of bin (and maybe an index), say in a linked list, then it would be something like (for mag for example):
bin = histogram.first;
while ( mag > bin.min ) bin = bin.next;
magIdx = bin.index;
If the bin does not hold the index you can just use a counter and increase it in the while. Also, for the magnitude the final bin should hold "infinity" or some large number as a limit. Note this has nothing to do with modulation, though that would work for your direction - as you have coded. I don't see how this makes sense for the norm.
Bottom line though, you have to think a bit about what you want. In any case all the "objects" here are trivial enough to write yourself, or even use small arrays.
I think you should arrange your bins in a square array, and then bin by vx and vy independently.
If your gradients are reasonably even you just need to scan the data first to accumulate the min and max in x and y, and then split the gradients evenly.
If the gradients are very unevenly distributed, you might want to sort the (eg) vx first and arrange that the boundaries between each bin exactly evenly divides the values.
An intermediate solution might be to obtain the min and max ignoring the (eg) 10% most extreme values.

Error in calculating exact nearest neighbors in radius with FLANN

I am trying to find the exact number of neighbour nodes in a big 3D points dataset. The goal is for each point of the dataset to retrieve all the possible neighbours in a region with a given radius. FLANN ensures that for lower dimensional data can retrieve the exact neighbors while comparing with brute force search it seems to not be the case. The neighbors are essential for further calculations and therefore I need the exact number. I tested increasing the radius a little bit but doesn't seem to be this the problem. Is anyone aware how to calculate the exact neighbors with FLANN or other C++ library?
The code:
// All nodes to be tested for inclusion in support domain.
flann::Matrix<double> query_nodes = flann::Matrix<double>(&nodes_pos[0].x, nodes_pos.size(), 3);
// Set default search parameters
flann::SearchParams search_parameters = flann::SearchParams();
search_parameters.checks = -1;
search_parameters.sorted = false;
search_parameters.use_heap = flann::FLANN_True;
flann::KDTreeSingleIndexParams index_parameters = flann::KDTreeSingleIndexParams();
flann::KDTreeSingleIndex<flann::L2_3D<double> > index(query_nodes, index_parameters);
index.buildIndex();
//FLANN uses L2 for radius search.
double l2_radius = (this->support_layer_*grid.spacing)*(this->support_layer_*grid.spacing);
double extension = l2_radius/10.;
l2_radius+= extension;
index.radiusSearch(query_nodes, indices, dists, l2_radius, search_parameters);
Try nanoflann. It is designed for low dimensional spaces and gives exact nearest neighbors. Furthermore, it is just one header file that you can either "install" or just copy to your project.
You should check page 6+ from the flann-manual, to fine-tune your search parameters, such as target_precision, which should be set to 1, for "maximum" accuracy.
That parameter is often found as epsilon (ε) in Approximate Nearest Neighbor Search (ANNS), which is used in high dimensional spaces, in order to (try) to beat the curse of dimensionality. FLANN is usually used in 128 dimensions, not 3, as far as I can tell, which may explain the bad performance you are experiencing.
A c++ library that works well in 3 dimensions is CGAL. However, it's much larger than FLANN, because it is a library for computational geometry, thus it provides functionality for many problems, not just NNS.

Detect clusters of circular objects by iterative adaptive thresholding and shape analysis

I have been developing an application to count circular objects such as bacterial colonies from pictures.
What make it easy is the fact that the objects are generally well distinct from the background.
However, few difficulties make the analysis tricky:
The background will present gradual as well as rapid intensity change.
In the edges of the container, the object will be elliptic rather than circular.
The edges of the objects are sometimes rather fuzzy.
The objects will cluster.
The object can be very small (6px of diameter)
Ultimately, the algorithms will be used (via GUI) by people that do not have deep understanding of image analysis, so the parameters must be intuitive and very few.
The problem has been address many times in the scientific literature and "solved", for instance, using circular Hough transform or watershed approaches, but I have never been satisfied by the results.
One simple approach that was described is to get the foreground by adaptive thresholding and split (as I described in this post) the clustered objects using distance transform.
I have successfully implemented this method, but it could not always deal with sudden change in intensity. Also, I have been asked by peers to come out with a more "novel" approach.
I therefore was looking for a new method to extract foreground.
I therefore investigated other thresholding/blob detection methods.
I tried MSERs but found out that they were not very robust and quite slow in my case.
I eventually came out with an algorithm that, so far, gives me excellent results:
I split the three channels of my image and reduce their noise (blur/median blur). For each channel:
I apply a manual implementation of the first step of adaptive thresholding by calculating the absolute difference between the original channel and a convolved (by a large kernel blur) one. Then, for all the relevant values of threshold:
I apply a threshold on the result of 2)
find contours
validate or invalidate contours on the grant of their shape (size, area, convexity...)
only the valid continuous regions (i.e. delimited by contours) are then redrawn in an accumulator (1 accumulator per channel).
After accumulating continuous regions over values of threshold, I end-up with a map of "scores of regions". The regions with the highest intensity being those that fulfilled the the morphology filter criteria the most often.
The three maps (one per channel) are then converted to grey-scale and thresholded (the threshold is controlled by the user)
Just to show you the kind of image I have to work with:
This picture represents part of 3 sample images in the top and the result of my algorithm (blue = foreground) of the respective parts in the bottom.
Here is my C++ implementation of : 3-7
/*
* cv::Mat dst[3] is the result of the absolute difference between original and convolved channel.
* MCF(std::vector<cv::Point>, int, int) is a filter function that returns an positive int only if the input contour is valid.
*/
/* Allocate 3 matrices (1 per channel)*/
cv::Mat accu[3];
/* We define the maximal threshold to be tried as half of the absolute maximal value in each channel*/
int maxBGR[3];
for(unsigned int i=0; i<3;i++){
double min, max;
cv::minMaxLoc(dst[i],&min,&max);
maxBGR[i] = max/2;
/* In addition, we fill accumulators by zeros*/
accu[i]=cv::Mat(compos[0].rows,compos[0].cols,CV_8U,cv::Scalar(0));
}
/* This loops are intended to be multithreaded using
#pragma omp parallel for collapse(2) schedule(dynamic)
For each channel */
for(unsigned int i=0; i<3;i++){
/* For each value of threshold (m_step can be > 1 in order to save time)*/
for(int j=0;j<maxBGR[i] ;j += m_step ){
/* Temporary matrix*/
cv::Mat tmp;
std::vector<std::vector<cv::Point> > contours;
/* Thresholds dst by j*/
cv::threshold(dst[i],tmp, j, 255, cv::THRESH_BINARY);
/* Finds continous regions*/
cv::findContours(tmp, contours, CV_RETR_LIST, CV_CHAIN_APPROX_TC89_L1);
if(contours.size() > 0){
/* Tests each contours*/
for(unsigned int k=0;k<contours.size();k++){
int valid = MCF(contours[k],m_minRad,m_maxRad);
if(valid>0){
/* I found that redrawing was very much faster if the given contour was copied in a smaller container.
* I do not really understand why though. For instance,
cv::drawContours(miniTmp,contours,k,cv::Scalar(1),-1,8,cv::noArray(), INT_MAX, cv::Point(-rect.x,-rect.y));
is slower especially if contours is very long.
*/
std::vector<std::vector<cv::Point> > tpv(1);
std::copy(contours.begin()+k, contours.begin()+k+1, tpv.begin());
/* We make a Roi here*/
cv::Rect rect = cv::boundingRect(tpv[0]);
cv::Mat miniTmp(rect.height,rect.width,CV_8U,cv::Scalar(0));
cv::drawContours(miniTmp,tpv,0,cv::Scalar(1),-1,8,cv::noArray(), INT_MAX, cv::Point(-rect.x,-rect.y));
accu[i](rect) = miniTmp + accu[i](rect);
}
}
}
}
}
/* Make the global scoreMap*/
cv::merge(accu,3,scoreMap);
/* Conditional noise removal*/
if(m_minRad>2)
cv::medianBlur(scoreMap,scoreMap,3);
cvtColor(scoreMap,scoreMap,CV_BGR2GRAY);
I have two questions:
What is the name of such foreground extraction approach and do you see any reason for which it could be improper to use it in this case ?
Since recursively finding and drawing contours is quite intensive, I would like to make my algorithm faster. Can you indicate me any way to achieve this goal ?
Thank you very much for you help,
Several years ago I wrote an aplication that detects cells in a microscope image. The code is written in Matlab, and I think now that is more complicated than it should be (it was my first CV project), so I will only outline tricks that will actually be helpful for you. Btw, it was deadly slow, but it was really good at separating large groups of twin cells.
I defined a metric by which to evaluate the chance that a given point is the center of a cell:
- Luminosity decreases in a circular pattern around it
- The variance of the texture luminosity follows a given pattern
- a cell will not cover more than % of a neighboring cell
With it, I started to iteratively find the best cell, mark it as found, then look for the next one. Because such a search is expensive, I employed genetic algorithms to search faster in my feature space.
Some results are given below:

Robustly find N circles with the same diameter: alternative to bruteforcing Hough transform threshold

I am developing application to track small animals in Petri dishes (or other circular containers).
Before any tracking takes place, the first few frames are used to define areas.
Each dish will match an circular independent static area (i.e. will not be updated during tracking).
The user can request the program to try to find dishes from the original image and use them as areas.
Here are examples:
In order to perform this task, I am using Hough Circle Transform.
But in practice, different users will have very different settings and images and I do not want to ask the user to manually define the parameters.
I cannot just guess all the parameters either.
However, I have got additional informations that I would like to use:
I know the exact number of circles to be detected.
All the circles have the almost same dimensions.
The circles cannot overlap.
I have a rough idea of the minimal and maximal size of the circles.
The circles must be entirely in the picture.
I can therefore narrow down the number of parameters to define to one: the threshold.
Using these informations and considering that I have got N circles to find, my current solution is to
test many values of threshold and keep the circles between which the standard deviation is the smallest (since all the circles should have a similar size):
//at this point, minRad and maxRad were calculated from the size of the image and the number of circles to find.
//assuming circles should altogether fill more than 1/3 of the images but cannot be altogether larger than the image.
//N is the integer number of circles to find.
//img is the picture of the scene (filtered).
//the vectors containing the detected circles and the --so far-- best circles found.
std::vector<cv::Vec3f> circles, bestCircles;
//the score of the --so far-- best set of circles
double bestSsem = 0;
for(int t=5; t<400 ; t=t+2){
//Apply Hough Circles with the threshold t
cv::HoughCircles(img, circles, CV_HOUGH_GRADIENT, 3, minRad*2, t,3, minRad, maxRad );
if(circles.size() >= N){
//call a routine to give a score to this set of circles according to the similarity of their radii
double ssem = scoreSetOfCircles(circles,N);
//if no circles are recorded yet, or if the score of this set of circles is higher than the former best
if( bestCircles.size() < N || ssem > bestSsem){
//this set become the temporary best set of circles
bestCircles=circles;
bestSsem=ssem;
}
}
}
With:
//the methods to assess how good is a set of circle (the more similar the circles are, the higher is ssem)
double scoreSetOfCircles(std::vector<cv::Vec3f> circles, int N){
double ssem=0, sum = 0;
double mean;
for(unsigned int j=0;j<N;j++){
sum = sum + circles[j][2];
}
mean = sum/N;
for(unsigned int j=0;j<N;j++){
double em = mean - circles[j][2];
ssem = 1/(ssem + em*em);
}
return ssem;
}
I have reached a higher accuracy by performing a second pass in which I repeated this algorithm narrowing the [minRad:maxRad] interval using the result of the first pass.
For instance minRad2 = 0.95 * average radius of best circles and maxRad2 = 1.05 * average radius of best circles.
I had fairly good results using this method so far. However, it is slow and rather dirty.
My questions are:
Can you thing of any alternative algorithm to solve this problem in a cleaner/faster manner ?
Or what would you suggest to improve this algorithm?
Do you think I should investigate generalised Hough transform ?
Thank you for your answers and suggestions.
The following approach should work pretty well for your case:
Binarize your image (you might need to do this on several levels of threshold to make algorithm independent of the lighting conditions)
Find contours
For each contour calculate the moments
Filter them by area to remove too small contours
Filter contours by circularity:
double area = moms.m00;
double perimeter = arcLength(Mat(contours[contourIdx]), true);
double ratio = 4 * CV_PI * area / (perimeter * perimeter);
ratio close to 1 will give you circles.
Calculate radius and center of each circle
center = Point2d(moms.m10 / moms.m00, moms.m01 / moms.m00);
And you can add more filters to improve the robustness.
Actually you can find an implementation of the whole procedure in OpenCV. Look how the SimpleBlobDetector class and findCirclesGrid function are implemented.
Within the current algorithm, the biggest thing that sticks out is the for(int t=5; t<400; t=t+2) loop. Trying recording score values for some test images. Graph score(t) versus t. With any luck, it will either suggest a smaller range for t or be a smoothish curve with a single maximum. In the latter case you can change your loop over all t values into a smarter search using Hill Climbing methods.
Even if it's fairly noisy, you can first loop over multiples of, say, 30, and for the best 1 or 2 of those loop over nearby multiples of 2.
Also, in your score function, you should disqualify any results with overlapping circles and maybe penalize overly spaced out circles.
You don't explain why you are using a black background. Unless you are using a telecentric lens (which seems unlikely, given the apparent field of view), and ignoring radial distortion for the moment, the images of the dishes will be ellipses, so estimating them as circles may lead to significant errors.
All and all, it doesn't seem to me that you are following a good approach. If the goals is simply to remove the background, so you can track the bugs inside the dishes, then your goal should be just that: find which pixels are background and mark them. The easiest way to do that is to take a picture of the background without dishes, under the same illumination and camera, and directly detect differences with the picture with the images. A colored background would be preferable to do that, with a color unlikely to appear in the dishes (e.g. green or blue velvet). So you'd have reduced the problem to bluescreening (or chroma keying), a classic technique in machine vision as applied to visual effects. Do a google search for "matte petro vlahos assumption" to find classic algorithms for solving this problem.

openCV filter image - replace kernel with local maximum

Some details about my problem:
I'm trying to realize corner detector in openCV (another algorithm, that are built-in: Canny, Harris, etc).
I've got a matrix filled with the response values. The biggest response value is - the biggest probability of corner detected is.
I have a problem, that in neighborhood of a point there are few corners detected (but there is only one). I need to reduce number of false-detected corners.
Exact problem:
I need to walk through the matrix with a kernel, calculate maximum value of every kernel, leave max value, but others values in kernel make equal zero.
Are there build-in openCV functions to do this?
This is how I would do it:
Create a kernel, it defines a pixels neighbourhood.
Create a new image by dilating your image using this kernel. This dilated image contains the maximum neighbourhood value for every point.
Do an equality comparison between these two arrays. Wherever they are equal is a valid neighbourhood maximum, and is set to 255 in the comparison array.
Multiply the comparison array, and the original array together (scaling appropriately).
This is your final array, containing only neighbourhood maxima.
This is illustrated by these zoomed in images:
9 pixel by 9 pixel original image:
After processing with a 5 by 5 pixel kernel, only the local neighbourhood maxima remain (ie. maxima seperated by more than 2 pixels from a pixel with a greater value):
There is one caveat. If two nearby maxima have the same value then they will both be present in the final image.
Here is some Python code that does it, it should be very easy to convert to c++:
import cv
im = cv.LoadImage('fish2.png',cv.CV_LOAD_IMAGE_GRAYSCALE)
maxed = cv.CreateImage((im.width, im.height), cv.IPL_DEPTH_8U, 1)
comp = cv.CreateImage((im.width, im.height), cv.IPL_DEPTH_8U, 1)
#Create a 5*5 kernel anchored at 2,2
kernel = cv.CreateStructuringElementEx(5, 5, 2, 2, cv.CV_SHAPE_RECT)
cv.Dilate(im, maxed, element=kernel, iterations=1)
cv.Cmp(im, maxed, comp, cv.CV_CMP_EQ)
cv.Mul(im, comp, im, 1/255.0)
cv.ShowImage("local max only", im)
cv.WaitKey(0)
I didn't realise until now, but this is what #sansuiso suggested in his/her answer.
This is possibly better illustrated with this image, before:
after processing with a 5 by 5 kernel:
solid regions are due to the shared local maxima values.
I would suggest an original 2-step procedure (there may exist more efficient approaches), that uses opencv built-in functions :
Step 1 : morphological dilation with a square kernel (corresponding to your neighborhood). This step gives you another image, after replacing each pixel value by the maximum value inside the kernel.
Step 2 : test if the cornerness value of each pixel of the original response image is equal to the max value given by the dilation step. If not, then obviously there exists a better corner in the neighborhood.
If you are looking for some built-in functionality, FilterEngine will help you make a custom filter (kernel).
http://docs.opencv.org/modules/imgproc/doc/filtering.html#filterengine
Also, I would recommend some kind of noise reduction, usually blur, before all processing. That is unless you really want the image raw.