使用 cv::warpAffine 旋转 cv::Mat 偏移目标图像

2022-01-16 00:00:00 opencv image-processing rotation c++

我正在尝试使用 OpenCV 的 C++ API 将 1296x968 图像90 度旋转我面临一些问题.

I'm trying to rotate a 1296x968 image by 90 degrees using the C++ API of OpenCV and I'm facing a few problems.

输入:

轮换:

如您所见，旋转后的图像存在一些问题.首先，它具有与原始大小相同的大小，尽管我专门创建了目标 Mat 与原始大小相反的大小.结果，目标图像被裁剪.

As you can see, the rotated image has a few problems. First, it has the same size of the original, even though I specifically create the destination Mat with the inverted size of the original. As a result, the destination image gets cropped.

我怀疑这是因为我正在调用 warpAffine() 并传递原始 Mat 的大小而不是目标 Mat 的大小代码>.但我这样做是因为 我关注了 这个答案，但现在我怀疑答案可能是错误的.所以这是我的第一个疑问/问题.

I suspect this is happening because I'm calling warpAffine() and passing the size of the original Mat instead of the size of destination Mat. But I'm doing this because I followed this answer, but now I suspect that the answer may be wrong. So this is my first doubt/problem.

第二个，是 warpAffine() 是 在某个偏移量处写入目的地(可能是将旋转后的数据复制到图像的中间)并且这个操作会在图像周围留下可怕的黑色大边框.

The second, is that warpAffine() is writing to the destination at a certain offset (probably to copy the rotated data to the middle of the image) and this operation leaves a horrible and large black border around the image.

如何解决这些问题?

我在下面分享源代码:

#include <cv.h>
#include <highgui.h>
#include <iostream>

using namespace cv;
using namespace std;

void rotate(Mat& image, double angle)
{
    Point2f src_center(image.cols/2.0F, image.rows/2.0F);

    Mat rot_matrix = getRotationMatrix2D(src_center, angle, 1.0);

    Mat rotated_img(Size(image.size().height, image.size().width), image.type());

    warpAffine(image, rotated_img, rot_matrix, image.size());
    imwrite("rotated.jpg", rotated_img);
}

int main(int argc, char* argv[])
{
    Mat orig_image = imread(argv[1], 1);
    if (orig_image.empty())
    {
        cout << "!!! Couldn't load " << argv[1] << endl;
        return -1;
    }

    rotate(orig_image, 90);

    return 0;
}

推荐答案

我找到了一个解决方案，不涉及 warpAffine().

I've found a solution that doesn't involve warpAffine().

但在此之前，我需要说明(以供将来参考)我的怀疑是正确的，您需要在调用 warpAffine():

But before that, I need to state (for future references) that my suspicion was right, you needed to pass the size of the destination when calling warpAffine():

warpAffine(image, rotated_img, rot_matrix, rotated_img.size());

据我所知，这个函数绘制的黑色边框(由偏移量引起)似乎是它的标准行为.我注意到 C 接口以及在 Mac 和 Linux 上运行的 OpenCV 的 C++ 接口，使用版本 2.3.1a 和 2.3.0.

As far as I can tell, the black border (caused by writing at an offset) drawed by this function seems to be it's standard behavior. I've noticed this with the C interface and also with the C++ interface of OpenCV running on Mac and Linux, using the versions 2.3.1a and 2.3.0.

我最终使用的解决方案比所有这些 扭曲的东西简单得多.您可以使用 cv::transpose() 和 cv::flip() 将图像旋转 90 度.这里是:

The solution I ended up using is much simpler than all this warp thing. You can use cv::transpose() and cv::flip() to rotate an image by 90 degrees. Here it is:

Mat src = imread(argv[1], 1);

cv::Mat dst;
cv::transpose(src, dst);
cv::flip(dst, dst, 1);

imwrite("rotated90.jpg", dst);

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