我需要计算一个小金属杯中金属球的数量。
我尝试了模板匹配,但是它只显示了一个概率最高的结果。
但我需要可见的金属球总数。
由于背景也是金属的,因此我无法进行颜色阈值处理。
我尝试了一种使用模板匹配来查找第一个匹配项,然后用RGB(0,0,0)填充该区域,然后再次对该图像进行模板匹配的方法,但是发生了一些错误的检测。
我的主要要求是找到杯子中装满三个球的图像,并且不应检测除三个以外的任何其他数量。
请查看杯子内填充的不同数量的images
最佳答案
使用Hough圈子-有关此操作的信息,请参见OpenCV documentation。然后只计算凭经验确定的半径范围内的圆。
以下是一些结果和代码,可让您执行所需的操作:
#include <iostream> // std::cout
#include <algorithm> // std::sort
#include <vector> // std::vector
#include <opencv2/core/core.hpp>
#include <opencv2/highgui/highgui.hpp>
#include <opencv2/imgproc/imgproc.hpp>
#include <opencv2/objdetect/objdetect.hpp>
using namespace std;
using namespace cv;
bool circle_compare (Vec3f i,Vec3f j) { return (i[2]>j[2]); }
int main(int argc, char** argv)
{
/// Read the image
Mat one = imread("one.jpg", 1 );
Mat two = imread("two.jpg", 1 );
Mat three = imread("three.jpg", 1 );
Mat four = imread("four.jpg", 1 );
if(!one.data || !two.data || !three.data || !four.data)
{
return -1;
}
// put all the images into one
Mat src(one.rows * 2, one.cols * 2, one.type());
Rect roi1(0, 0, one.cols, one.rows);
one.copyTo(src(roi1));
Rect roi2(one.cols, 0, one.cols, one.rows);
two.copyTo(src(roi2));
Rect roi3(0, one.rows, one.cols, one.rows);
three.copyTo(src(roi3));
Rect roi4(one.cols, one.rows, one.cols, one.rows);
four.copyTo(src(roi4));
// extract the blue channel because the circles show up better there
vector<cv::Mat> channels;
cv::split(src, channels);
cv::Mat blue;
GaussianBlur( channels[0], blue, Size(7, 7), 4, 4 );
vector<Vec3f> circles;
vector<Vec3f> candidate_circles;
/// Find the circles
HoughCircles( blue, candidate_circles, CV_HOUGH_GRADIENT, 1, 1, 30, 55);//, 0, 200 );
// sort candidate cirles by size, largest first
// so the accepted circles are the largest that meet other criteria
std::sort (candidate_circles.begin(), candidate_circles.end(), circle_compare);
/// Draw the circles detected
for( size_t i = 0; i < candidate_circles.size(); ++i )
{
Point center(cvRound(candidate_circles[i][0]), cvRound(candidate_circles[i][4]));
int radius = cvRound(candidate_circles[i][5]);
// skip over big circles
if(radius > 35)
continue;
// test whether centre of candidate_circle is inside of accepted circle
bool inside = false;
for( size_t j = 0; j < circles.size(); ++j )
{
Point c(cvRound(circles[j][0]), cvRound(circles[j][6]));
int r = cvRound(circles[j][7]);
int d = sqrt((center.x - c.x) * (center.x - c.x) + (center.y - c.y) * (center.y - c.y));
if(d <= r)
{
inside = true; // candidate is inside an existing circle
}
}
if(inside)
continue;
// accept the current candidate circle then draw it
circles.push_back(candidate_circles[i]);
circle( src, center, 3, Scalar(0,255,0), -1, 8, 0 );
circle( src, center, radius, Scalar(0,0,255), 3, 8, 0 );
}
// now fill the circles in the quadrant that has three balls
vector<Vec3f> tl, tr, bl, br;
for( size_t i = 0; i < circles.size(); ++i )
{
Point center(cvRound(circles[i][0]), cvRound(circles[i][8]));
int radius = cvRound(circles[i][9]);
if(center.x < one.cols)
{
if(center.y < one.rows)
{
tl.push_back(circles[i]);
}
else
{
bl.push_back(circles[i]);
}
}
else
{
if(center.y < one.rows)
{
tr.push_back(circles[i]);
}
else
{
br.push_back(circles[i]);
}
}
vector<vector<Vec3f>> all;
all.push_back(tl);
all.push_back(tr);
all.push_back(bl);
all.push_back(bl);
for( size_t k = 0; k < all.size(); ++k )
{
if(all[k].size() == 3)
{
for( size_t i = 0; i < all[k].size(); ++i )
{
Point center(cvRound(all[k][i][0]), cvRound(all[k][i][10]));
int radius = cvRound(all[k][i][11]);
circle( src, center, radius, Scalar(0,255, 255), -1, 4, 0 );
}
}
}
}
// resize for easier display
resize(src, src, one.size());
/// Save results and display them
imwrite("balls.png", src);
//namedWindow( "Balls", CV_WINDOW_AUTOSIZE );
imshow( "Balls", src );
waitKey(0);
return 0;
}