Question

获得旋转角度的当前方式。

cv::Rect rRect(pTopLeft.x,pTopLeft.y,pBottomRight.x-pTopLeft.x,pBottomRight.y-pTopLeft.y);

img(rRect);

std::vector<cv::Point> points;
cv::Mat_<uchar>::iterator it = img.begin<uchar>();
cv::Mat_<uchar>::iterator end = img.end<uchar>();
for (; it != end; ++it)
   if (*it)
      points.push_back(it.pos());

cv::RotatedRect box = cv::minAreaRect(cv::Mat(points));
double angle = box.angle;

if (angle < -45.)
    angle += 90.;

有没有办法避免for循环并直接使用rRect获取Rect的角度，或者openCV中是否有任何内置函数使用Rect获取角度？

Answer 1

如果我们有方格点，我们可以执行以下操作。

cv::Mat debugSquares( std::vector<std::vector<cv::Point> > squares, cv::Mat image )
{
    for ( int i = 0; i< squares.size(); i++ ) {
        // draw contour
        cv::drawContours(image, squares, i, cv::Scalar(255,0,0), 1, 8, std::vector<cv::Vec4i>(), 0, cv::Point());

        // draw bounding rect
        cv::Rect rect = boundingRect(cv::Mat(squares[i]));
        cv::rectangle(image, rect.tl(), rect.br(), cv::Scalar(0,255,0), 1, 8, 0);

        // draw rotated rect
        cv::RotatedRect minRect = minAreaRect(cv::Mat(squares[i]));
    cout<<"\nangle"<<minRect.angle;
        cv::Point2f rect_points[4];
        minRect.points( rect_points );
        for ( int j = 0; j < 4; j++ ) {
            cv::line( image, rect_points[j], rect_points[(j+1)%4], cv::Scalar(0,0,255), 3, 8 ); // blue
        }
    }

    imshow("  Image", image);
    return image;
}

`

是否有更好的方法（性能增加/内置函数）来获得矩形的旋转角度？

1 个答案: