我是Open-CV的新手。我试图在相当简单的图像中检测90度角。我需要检测围绕物体的那个矩形的角落。我正在使用shi-Thomasi功能。以下是我的代码:
for x in range(0, 50):
ret, frame = cap.read()
# make image gray scale
im = cv2.cvtColor(frame, cv2.COLOR_BGR2GRAY)
#finding corners
corners = cv2.goodFeaturesToTrack(im, 1, 0.01, 10)
corners = np.int0(corners)
for i in corners:
x, y = i.ravel()
cv2.circle(frame, (x, y), 3, 255,-1)
cv2.imwrite("DetectedCorners.png", frame)
问题:始终检测到该对象中的某些角落。我需要一种方法,完全删除该对象,然后检测角落。
我不知道如何删除该对象。 有什么建议 ?照片显示了我的结果。有些时候检测到周围矩形的角,有时是该复杂对象中的一些随机点。 在检测角落之前我也使用了Canny,但结果差了10倍。
答案 0 :(得分:4)
嗯,快速而肮脏的C ++解决方案,只是为了证明使用Hough变换来检测线条,然后计算它们的交集。
如果需要,您最终可以将代码移植到Python。
#include <opencv2\opencv.hpp>
#include <iostream>
using namespace cv;
using namespace std;
int main()
{
Mat3b img = imread("path_to_image");
// Convert to grayscale
Mat1b gray;
cvtColor(img, gray, COLOR_BGR2GRAY);
// Compute edges
Mat1b edges;
Canny(gray, edges, 400, 100);
// Create the output result image
Mat3b res;
cvtColor(edges, res, COLOR_GRAY2BGR);
// Call hough
vector<Vec2f> lines;
HoughLines(edges, lines, 1, CV_PI / 180, 200, 0, 0);
vector<pair<Point,Point>> pts;
vector<Point> intersections;
for (size_t i = 0; i < lines.size(); i++)
{
float rho = lines[i][0], theta = lines[i][1];
// Get 2 points on each line
Point pt1, pt2;
double a = cos(theta), b = sin(theta);
double x0 = a*rho, y0 = b*rho;
pt1.x = cvRound(x0 + 1000 * (-b));
pt1.y = cvRound(y0 + 1000 * (a));
pt2.x = cvRound(x0 - 1000 * (-b));
pt2.y = cvRound(y0 - 1000 * (a));
// Save the pair of points
pts.push_back(make_pair(pt1, pt2));
// Draw lines
line(res, pt1, pt2, Scalar(0, 0, 255), 3, CV_AA);
}
// for each couple of lines
for (int i = 0; i < pts.size() - 1; ++i)
{
// get the two points of the first line
const Point& p1 = pts[i].first;
const Point& p2 = pts[i].second;
for (int j = i + 1; j < pts.size(); ++j)
{
// Get the two points of the second line
const Point& p3 = pts[j].first;
const Point& p4 = pts[j].second;
// Compute intersection
Point p;
float den = (p1.x - p2.x) * (p3.y - p4.y) - (p1.y - p2.y) * (p3.x - p4.x);
if (den != 0) // if not parallel lines
{
p.x = ((p1.x*p2.y - p1.y*p2.x)*(p3.x - p4.x) - (p1.x - p2.x)*(p3.x*p4.y - p3.y*p4.x)) / den;
p.y = ((p1.x*p2.y - p1.y*p2.x)*(p3.y - p4.y) - (p1.y - p2.y)*(p3.x*p4.y - p3.y*p4.x)) / den;
// Draw intersection
circle(res, p, 7, Scalar(0, 255, 0), 2);
}
// Save intersections
intersections.push_back(p);
}
}
return 0;
}
结果: