我想按顺序在几个角度旋转图像。我这样做是使用cv2.getRotationMatrix2D和cv2.warpAffine。有一对像素坐标[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。
我使用了以下略有改变的代码，由http://www.pyimagesearch.com/2017/01/02/rotate-images-correctly-with-opencv-and-python/和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。 （我尝试手动计算角落以进行简单验证）
码：
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]我将获得[-0.5, 511.50]（[col，row]）当我期望获得[0,511]。
如果我使用w // 2，将在图像上添加黑色边框，我的旋转更新坐标将再次关闭。
问题：如何使用Python在旋转的图像中（通过某个角度）找到一对像素坐标的正确位置？
对于图像旋转的情况，其中图像尺寸在旋转之后以及参考点改变，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）
由于图像尺寸会发生变化，因为您可能会看到黑色边框，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)
上面提到的@ cristianpb Python代码的相应C ++代码，如果有人像我一样寻找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]);
}