opencv检测相交点_opencv求解两条直线的交点
假设现在有一个点集，需要拟合出最能够表达点集轮廓的几条直线，并求直线之间的交点。
从点集中拟合直线可以采用的方法：随机抽样一致性(RANSAC),霍夫变换(though transform)
c++ 程序代码
/** @brief 计算直线的交点
@param lines 直线：Vec4d=(vx, vy, x0, y0), where (vx, vy) is a normalized vector collinear to the line and (x0, y0) is a point on the line.
@param crossPoints 保存直线的交点
@param mask 掩膜
*/
void crossPointsOfLines(std::vector<:vec4d>& lines, std::vector<:point2f> &crossPoints, int nPoints, cv::Mat& mask)
{
crossPoints.clear();
for (int i = 0; i < lines.size() && crossPoints.size() < nPoints; i++)
{
for (int j = i + 1; j < lines.size() && crossPoints.size() < nPoints; j++)
{
float ka = (float)lines.at(i)[1] / float(lines.at(i)[0] + 0.000001f);//slope of LineA
float kb = (float)lines.at(j)[1] / float(lines.at(j)[0] + 0.000001f);//slope of LineB
//if (std::abs(std::abs(ka) - std::abs(kb)) > 1.0) //two lines are not probably parallel
if ((std::abs(ka) > 1) && (std::abs(kb) < 1) || (std::abs(ka) < 1) && (std::abs(kb) > 1))//two lines are not probably parallel
{
cv::Point2d ptA(lines.at(i)[2], lines.at(i)[3]);
cv::Point2d ptB(lines.at(j)[2], lines.at(j)[3]);
cv::Point2f crossPoint;
crossPoint.x = float(ka*ptA.x - ptA.y - kb*ptB.x + ptB.y) / float(ka - kb);
crossPoint.y = float(ka*kb*(ptA.x - ptB.x) - kb*ptA.y + ka*ptB.y) / float(ka - kb);
crossPoints.push_back(crossPoint);
#ifdef _DEBUG
cv::circle(mask, crossPoint, 2, cv::Scalar(0, 0, 255), -1, cv::FILLED);
#endif
}
}
}
if (crossPoints.size() < nPoints){
LOG(ERROR) << type() << "::crossPointsOfLines: " << "cross points is less than parameter nPoints.";
}
}
参考文献
OpenCV找任意两条直线的交点
OpenCV找直线及直线的交点
随机抽样一致性算法(ransac)