问题描述
我正在使用 JavaCV(OpenCV 包装器)进行硬币检测，但是在连接硬币时遇到了一些问题.如果我试图侵蚀它们以分离这些硬币，它们会失去圆形形状，如果我尝试计算每枚硬币内的像素，则可能会出现问题，因此某些硬币可能会被误算为更大的硬币.我想要做的是首先将它们重新整形，使它们像一个圆(等于那个硬币的半径)，然后计算它们内部的像素.
I'm doing a coin detection using JavaCV (OpenCV wrapper) but I have a little problem when the coins are connected. If I try to erode them to separate these coins they loose their circle form and if I try to count pixels inside each coin there can be problems so that some coins can be miscounted as one that bigger. What I want to do is firstly to reshape them and make them like a circle (equal with the radius of that coin) and then count pixels inside them.
这是我的阈值图像:
这里是被侵蚀的图像:
有什么建议吗?或者有什么更好的方法来打破硬币之间的桥梁?
Any suggestions? Or is there any better way to break bridges between coins?
推荐答案
这看起来类似于我最近不得不分离琼脂平板上生长的细菌菌落的问题.我对阈值图像执行了距离变换(在您的情况下，您需要反转它).然后找到距离图的峰值(通过计算扩张距离图和距离图之间的差异并找到零值).然后，我假设每个峰是圆(硬币)的中心，距离图中峰的值是圆的半径.
It looks similar to a problem I recently had to separate bacterial colonies growing on agar plates.I performed a distance transform on the thresholded image (in your case you will need to invert it).Then found the peaks of the distance map (by calculating the difference between a the dilated distance map and the distance map and finding the zero values).Then, I assumed each peak to be the centre of a circle (coin) and the value of the peak in the distance map to be the radius of the circle.
这是您的图像在此管道后的结果:
Here is the result of your image after this pipeline:
我是 OpenCV 和 C++ 的新手，所以我的代码可能很乱，但我做到了:
I am new to OpenCV, and c++ so my code is probably very messy, but I did that:
int main( int argc, char** argv ){
cv::Mat objects, distance,peaks,results;
std::vector<std::vector<cv::Point> > contours;
objects=cv::imread("CUfWj.jpg");
objects.copyTo(results);
cv::cvtColor(objects, objects, CV_BGR2GRAY);
//THIS IS THE LINE TO BLUR THE IMAGE CF COMMENTS OF THIS POST
cv::blur( objects,objects,cv::Size(3,3));
cv::threshold(objects,objects,125,255,cv::THRESH_BINARY_INV);
/*Applies a distance transform to "objects".
* The result is saved in "distance" */
cv::distanceTransform(objects,distance,CV_DIST_L2,CV_DIST_MASK_5);
/* In order to find the local maxima, "distance"
* is subtracted from the result of the dilatation of
* "distance". All the peaks keep the save value */
cv::dilate(distance,peaks,cv::Mat(),cv::Point(-1,-1),3);
cv::dilate(objects,objects,cv::Mat(),cv::Point(-1,-1),3);
/* Now all the peaks should be exactely 0*/
peaks=peaks-distance;
/* And the non-peaks 255*/
cv::threshold(peaks,peaks,0,255,cv::THRESH_BINARY);
peaks.convertTo(peaks,CV_8U);
/* Only the zero values of "peaks" that are non-zero
* in "objects" are the real peaks*/
cv::bitwise_xor(peaks,objects,peaks);
/* The peaks that are distant from less than
* 2 pixels are merged by dilatation */
cv::dilate(peaks,peaks,cv::Mat(),cv::Point(-1,-1),1);
/* In order to map the peaks, findContours() is used.
* The results are stored in "contours" */
cv::findContours(peaks, contours, CV_RETR_CCOMP, CV_CHAIN_APPROX_SIMPLE);
/* The next steps are applied only if, at least,
* one contour exists */
cv::imwrite("CUfWj2.jpg",peaks);
if(contours.size()>0){
/* Defines vectors to store the moments of the peaks, the center
* and the theoritical circles of the object of interest*/
std::vector <cv::Moments> moms(contours.size());
std::vector <cv::Point> centers(contours.size());
std::vector<cv::Vec3f> circles(contours.size());
float rad,x,y;
/* Caculates the moments of each peak and then the center of the peak
* which are approximatively the center of each objects of interest*/
for(unsigned int i=0;i<contours.size();i++) {
moms[i]= cv::moments(contours[i]);
centers[i]= cv::Point(moms[i].m10/moms[i].m00,moms[i].m01/moms[i].m00);
x= (float) (centers[i].x);
y= (float) (centers[i].y);
if(x>0 && y>0){
rad= (float) (distance.at<float>((int)y,(int)x)+1);
circles[i][0]= x;
circles[i][3]= y;
circles[i][2]= rad;
cv::circle(results,centers[i],rad+1,cv::Scalar( 255, 0,0 ), 2, 4, 0 );
}
}
cv::imwrite("CUfWj2.jpg",results);
}
return 1;
}
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