Question

整个下午一直困扰着我。下面是一个代码段，我正在尝试将代码分类为具有一定公差角的垂直线（linesY_acc）和水平线（linesX_acc）。

我不知道为什么仅前面的lineX_acc会被填充，而linesY_acc却什么也没有？也许这是一个小错误，但我看不出为什么

需要您的帮助！预先感谢

std::vector<cv::Vec4i> linesX_acc, linesY_acc;
int boxWidth_threshold  = 35;
double threshold_anglesX = 20.0 /180 * CV_PI;
double threshold_anglesY = 20.0 /180 * CV_PI;

std::vector<cv::Vec4i>::iterator itlines;

for(itlines = lines.begin(); itlines != lines.end(); itlines++)
{
    // distlength --- calculate the line length;
    //                discard line if it is smaller than boxWidth_threshold
    if(distlength(cv::Point2f((*itlines)[0], (*itlines)[1]),
                           cv::Point2f((*itlines)[2], (*itlines)[3])) < boxWidth_threshold )
    {
        continue;
    }

    // filtering the angle of line
    // myAngle - function to caluclate angle
    double angle =std::abs(myAngle((*itlines)));

    if( (angle < threshold_anglesX ) || (angle > (CV_PI - threshold_anglesX )) )
    {
        linesX_acc.push_back(lines[i]);
    }
    else if ( (angle > (CV_PI/2 -threshold_anglesY ) ) &&
            (angle < (CV_PI/2 + threshold_anglesY ) ))
    {
        linesY_acc.push_back((*itlines));
    }

    else
    {
        continue;
    }

}

我的角度代码是

double myAngle(cv::Vec4i lines)
{
    cv::Point p1, p2;
    p1=cv::Point(lines[0], lines[1]);
    p2=cv::Point(lines[2], lines[3]);

   //calculate angle in radian,  if you need it in degrees just do angle * 180 / PI
   return atan2(p2.y - p1.y, p2.x - p1.x);
  }

Answer 1

错误在这里：

linesX_acc.push_back(lines[i]); <--- i?!! Whats is i?

但是这里您使用的是迭代器：

linesY_acc.push_back((*itlines));

但是您的代码非常困难，请尝试以下操作：

std::vector<cv::Vec4i> linesX_acc, linesY_acc;
int boxWidth_threshold  = 35;
double threshold_anglesX = 20.0 /180 * CV_PI;
double threshold_anglesY = 20.0 /180 * CV_PI;

for(const cv::Vec4i& itlines : lines)
{
    cv::Point2f p1(itlines[0], itlines[1]);
    cv::Point2f p2(itlines[2], itlines[3]);

    if (cv::norm(p1 - p2) < boxWidth_threshold)
    {
        continue;
    }

    auto angle = fabs(atan2(p2.y - p1.y, p2.x - p1.x));
    if ( (angle < threshold_anglesX ) || (angle > (CV_PI - threshold_anglesX )) )
    {
        linesX_acc.push_back(itlines);
    }
    else if ( (angle > (CV_PI/2 -threshold_anglesY ) ) &&
              (angle < (CV_PI/2 + threshold_anglesY ) ))
    {
        linesY_acc.push_back(itlines);
    }
    else
    {
        continue;
    }
}

需要一些破译的opencv线角

1 个答案: