Question

我想按顺序在几个角度旋转图像。我使用cv2.getRotationMatrix2D和cv2.warpAffine来做到这一点。有一个像素坐标[x，y]，其中x = cols，y =行（在这种情况下）我想在旋转的图像中找到它们的新坐标。

我使用了http://www.pyimagesearch.com/2017/01/02/rotate-images-correctly-with-opencv-and-python/提供的以下略微更改的代码和来自仿射变换的解释来尝试映射旋转图像中的点：http://docs.opencv.org/2.4/doc/tutorials/imgproc/imgtrans/warp_affine/warp_affine.html。

问题是我的映射或我的旋转错误，因为转换的计算坐标是错误的。（我试图手动计算角点以进行简单验证）

CODE ：

def rotate_bound(image, angle):
    # grab the dimensions of the image and then determine the
    # center
    (h, w) = image.shape[:2]
    (cX, cY) = ((w-1) // 2.0, (h-1)// 2.0)


# grab the rotation matrix (applying the negative of the
# angle to rotate clockwise), then grab the sine and cosine
# (i.e., the rotation components of the matrix)
M = cv2.getRotationMatrix2D((cX, cY), -angle, 1.0)
cos = np.abs(M[0, 0])
sin = np.abs(M[0, 1])

# compute the new bounding dimensions of the image
nW = int((h * sin) + (w * cos))
nH = int((h * cos) + (w * sin))
print nW, nH

# adjust the rotation matrix to take into account translation
M[0, 2] += ((nW-1) / 2.0) - cX
M[1, 2] += ((nH-1) / 2.0) - cY

# perform the actual rotation and return the image
return M, cv2.warpAffine(image, M, (nW, nH))

#function that calculates the updated locations of the coordinates
#after rotation
def rotated_coord(points,M):
    points = np.array(points)
    ones = np.ones(shape=(len(points),1))
    points_ones = np.concatenate((points,ones), axis=1)
    transformed_pts = M.dot(points_ones.T).T
    return transformed_pts

#READ IMAGE & CALL FCT
img = cv2.imread("Lenna.png")
points = np.array([[511,  511]])
#rotate by 90 angle for example
M, rotated = rotate_bound(img, 90)
#find out the new locations
transformed_pts = rotated_coord(points,M)

如果我有坐标[511,511]，我希望获得[-0.5, 511.50]时获得[0,511]（[col，row]）。

如果我使用w // 2，则会在图像上添加黑色边框，我的旋转更新坐标将会再次关闭。

问题：如何使用Python在旋转的图像中找到一对像素坐标的正确位置（按某个角度）？

Answer 1

对于图像旋转的情况，旋转后图像大小和参考点都会发生变化，必须修改变换矩阵。可以使用以下关系计算新的with和height：

new.width = h * \ sin（\ theta）+ w * \ cos（\ theta）

new.height = h * \ cos（\ theta）+ w * \ sin（\ theta）

由于图像尺寸发生变化，由于您可能会看到黑色边框，因此旋转点（图像中心）的坐标也会发生变化。然后必须在转换矩阵中考虑它。

我在我的博客image rotation bounding box opencv

中解释了一个例子

def rotate_box(bb, cx, cy, h, w):  
    new_bb = list(bb)                                                                                                                                                 
    for i,coord in enumerate(bb):
        # opencv calculates standard transformation matrix                                                                                                            
        M = cv2.getRotationMatrix2D((cx, cy), theta, 1.0)
        # Grab  the rotation components of the matrix)                                                                                                                
        cos = np.abs(M[0, 0])
        sin = np.abs(M[0, 1])                                                                                                                                         
        # compute the new bounding dimensions of the image                                                                                                            
        nW = int((h * sin) + (w * cos))
        nH = int((h * cos) + (w * sin))
        # adjust the rotation matrix to take into account translation
        M[0, 2] += (nW / 2) - cx
        M[1, 2] += (nH / 2) - cy
        # Prepare the vector to be transformed 
        v = [coord[0],coord[1],1]
        # Perform the actual rotation and return the image
        calculated = np.dot(M,v)
        new_bb[i] = (calculated[0],calculated[1]) 
        return new_bb   


 ## Calculate the new bounding box coordinates
 new_bb = {}
 for i in bb1: 
 new_bb[i] = rotate_box(bb1[i], cx, cy, heigth, width)

Answer 2

上面提到的@ cristianpb Python代码的相应C ++代码，如果有人在寻找像我一样的C ++代码：

 // send the original angle i.e. don't transform it in radian
        cv::Point2f rotatePointUsingTransformationMat(const cv::Point2f& inPoint, const cv::Point2f& center, const double& rotAngle)
        {
            cv::Mat rot = cv::getRotationMatrix2D(center, rotAngle, 1.0);
            float cos = rot.at<double>(0,0);
            float sin = rot.at<double>(0,1);
            int newWidth = int( ((center.y*2)*sin) +  ((center.x*2)*cos) );
            int newHeight = int( ((center.y*2)*cos) +  ((center.x*2)*sin) );

            rot.at<double>(0,2) += newWidth/2.0 - center.x;
            rot.at<double>(1,2) += newHeight/2.0 - center.y;

            int v[3] = {static_cast<int>(inPoint.x),static_cast<int>(inPoint.y),1};
            int mat3[2][1] = {{0},{0}};

            for(int i=0; i<rot.rows; i++)
            {
                for(int j=0; j<= 0; j++)
                {
                    int sum=0;
                    for(int k=0; k<3; k++)
                    {
                        sum = sum + rot.at<double>(i,k) * v[k];
                    }
                    mat3[i][j] = sum;
                }
            }
            return Point2f(mat3[0][0],mat3[1][0]);
        }

在图像中找到[x，y]旋转坐标位置[OpenCV / Python]

2 个答案: