我正在使用适用于Android的OpenCV 3.0。我有一个图像,我想要检测圆形表盘内的手的角度。因为我正在- (void)viewDidLoad
{
[self performSelector:@selector(Fb_Login) withObject:nil];
[super viewDidLoad];
}
-(void)Fb_Login
{
if (!FBSession.activeSession.isOpen)
{
// if the session is closed, then we open it here, and establish a handler for state changes
[FBSession openActiveSessionWithReadPermissions:nil allowLoginUI:YES completionHandler:^(FBSession *session,FBSessionState state, NSError *error)
{
if (error)
{
UIAlertView *alertView = [[UIAlertView alloc] initWithTitle:@"Alert" message:error.localizedDescription delegate:nil cancelButtonTitle:@"OK" otherButtonTitles:nil];
[alertView show];
}
else if(session.isOpen)
{
[self Fb_Login];
}
}];
return;
}
探测手。
这是代码。
HoughLinesP
我需要的是这些圈子中的手的角度。任何有关此问题的帮助都非常感谢。感谢ADvance
修改 我用这个
更新了我的代码Mat imgSource = new Mat(), imgCirclesOut = new Mat(),imgLinesOut=new Mat();
//grey opencv
Imgproc.cvtColor(Image, imgSource, Imgproc.COLOR_BGR2GRAY);
Imgproc.GaussianBlur( imgSource, imgSource, new Size(9, 9), 2, 2 );
int threshold = 0;
int minLineSize = 0;
int lineGap = 0;
Imgproc.HoughLinesP(imgSource, imgLinesOut, 1, Math.PI/180, threshold, minLineSize, lineGap);
for( int j = 0; i < imgLinesOut.cols(); i++ )
{
double[] vec=imgLinesOut.get(0,j);
Point pt1, pt2;
pt1=new Point(vec[0],vec[1]);
pt2=new Point(vec[2],vec[3]);
Imgproc.line( Image, pt1, pt2, new Scalar(0,0,255), 3, Core.LINE_AA,0);
}
正如@Micka所建议的那样,不需要灰化图像(我删除了 Mat imgSource = new Mat(), imgCirclesOut = new Mat(),imgLinesOut=new Mat();
Imgproc.GaussianBlur( Image, imgSource, new Size(5, 5), 2, 2 );
int threshold = 20;
int minLineSize = 0;
int lineGap = 10;
Imgproc.Canny(imgSource, imgSource, 70, 100);
Imgproc.HoughLinesP(imgSource, imgLinesOut, 1, Math.PI/180, threshold, minLineSize, lineGap);
for( int j = 0; j < imgLinesOut.cols(); j++ )
{
double[] vec=imgLinesOut.get(0,j);
Point pt1, pt2;
pt1=new Point(vec[0],vec[1]);
pt2=new Point(vec[2],vec[3]);
Imgproc.line( imgSource, pt1, pt2, new Scalar(0,0,255), 3, Core.LINE_AA,0);
}
)。我还将cvtcolor
尺寸的值减少到GuassianBlur
。我也在图像上添加了Canny for edge。
产生的模糊图像
答案 0 :(得分:3)
在如此小的图像中检测线条可能是个问题,因为你需要很少的点才能正确填充霍夫累加器。
我建议使用不同的方法:
以下是这个想法的简单实现。代码是用C ++编写的,但您可以轻松地移植到Java,或至少用作参考。
#include "opencv2/opencv.hpp"
using namespace cv;
int main(int, char**)
{
Mat1b img = imread("path_to_image", IMREAD_GRAYSCALE);
Mat3b res;
cvtColor(img, res, COLOR_GRAY2BGR);
// Find dials
vector<Vec3f> circles;
HoughCircles(img, circles, CV_HOUGH_GRADIENT, 1, img.cols/10, 400, 40);
// For each dial
for (int i = 0; i < circles.size(); ++i)
{
// Segment the dial
Mat1b dial(img.size(), uchar(255));
Mat1b mask(img.size(), uchar(0));
circle(mask, Point(circles[i][0], circles[i][1]), circles[i][2], Scalar(255), CV_FILLED);
img.copyTo(dial, mask);
// Apply threshold and open
Mat1b bin;
threshold(dial, bin, 127, 255, THRESH_BINARY_INV);
Mat kernel = getStructuringElement(MORPH_ELLIPSE, Size(5,5));
morphologyEx(bin, bin, MORPH_OPEN, kernel);
// Get min area rect
vector<Point> points;
findNonZero(bin, points);
RotatedRect r = minAreaRect(points);
// Draw min area rect
Point2f pts[4];
r.points(pts);
for (int j = 0; j < 4; ++j) {
line(res, pts[j], pts[(j + 1) % 4], Scalar(0, 255, 0), 1);
}
}
imshow("Result", res);
waitKey();
return 0;
}
从这张图片开始:
我在这里找到 hands :
答案 1 :(得分:-1)
for( int j = 0; j < imgLinesOut.size(); j++ )
这将给出vector的大小。迭代该向量