为什么这个旋转方法会给图像死空间? OpenCV的 [英] Why does this rotate method give the image dead space? OpenCV
问题描述
我正在使用此方法旋转 cvMat
,每当我运行它时,我都会返回一个旋转的图像,但是下面有很多死区。
I am using this method to rotate a cvMat
, whenever I run it I get back a rotated image however there is a lot of deadspace below it.
void rotate(cv::Mat& src, double angle, cv::Mat& dst)
{
int len = std::max(src.cols, src.rows);
cv::Point2f pt(len/2., len/2.);
cv::Mat r = cv::getRotationMatrix2D(pt, angle, 1.0);
cv::warpAffine(src, dst, r, cv::Size(len, len));
}
给出此图片时:
我得到这张图片:
图像已经旋转但是你可以看到添加了一些额外的像素,我怎么只能旋转原始图像而不添加任何额外的像素?
The image has been rotated but as you can see some extra pixels have been added, how can I only rotate the original image and not add any extra pixels?
方法调用:
rotate(src,skew,res);
res
dst
。
推荐答案
由于 mayank-baddi 表示您必须使用与输入相同的输出图像大小来解决此问题,我的回答基于您的上述评论如何避免在wrapAffine之后添加黑色区域?
,
As mayank-baddi said you have to use output image size same as the input to resolve this, and my answer is based on your comment above How can I avoid adding the black area?
after wrapAffine,
所以你必须这样做,
-
创建比你的更大的白色图像rce,它将取决于你的倾斜角度,这里我使用50像素。
Create white image little bigger than your source, and it will depend on your skew angle, here I used 50 pixel.
int extend=50;
Mat tmp(src.rows+2*extend,src.cols+2*extend,src.type(),Scalar::all(255));
使用投资回报率将来源复制到上面
Copy the source to above using ROI
Rect ROI(extend,extend,src.cols,src.rows);
src.copyTo(tmp(ROI));
现在轮换 tmp
而不是 src
rotate(tmp, skew, res); res being dst.
使用相同的投资回报率从旋转结果中裁剪最终图像。
Crop back the final image from rotated result using the same ROI.
Mat crop=res(ROI);
imshow("crop",crop);
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