将数学函数从 Python 转换为 C++
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【中文标题】将数学函数从 Python 转换为 C++【英文标题】:Converting Math functions from Python to c++ 【发布时间】:2018-08-22 15:14:06 【问题描述】:我正在尝试将一些 opencv 代码从 python 转换为 c++,但我有点迷茫。蟒蛇是:
if 0 < R[1,1] < 1:
# If it gets here, the pose is flipped.
# Flip the axes. E.g., Y axis becomes [-y0, -y1, y2].
R *= np.array([
[ 1, -1, 1],
[ 1, -1, 1],
[-1, 1, -1],
])
# Fixup: rotate along the plane spanned by camera's forward (Z) axis and vector to marker's position
forward = np.array([0, 0, 1])
tnorm = T / np.linalg.norm(T)
axis = np.cross(tnorm, forward)
angle = -2*math.acos(tnorm @ forward)
R = cv2.Rodrigues(angle * axis)[0] @ R
到目前为止我有:
cv::Mat R;
if (0 < R.at<double>(1, 1) < 1)
// Flip the axes.E.g., Y axis becomes[-y0, -y1, y2].
float mult[9] = 1, -1, 1, 1, -1, 1,-1, 1, -1 ;
cv::Mat FlipAxes = cv::Mat(3, 3, CV_32F, mult);
R *= FlipAxes;
//# Fixup: rotate along the plane spanned by camera's forward (Z) axis and vector to marker's position
cv::Vec3d forward(0, 0, 1);
double tnorm = tvecBest / np.linalg.norm(T)
axis = np.cross(tnorm, forward)
angle = -2 * math.acos(tnorm @ forward)
R = cv2.Rodrigues(angle * axis)[0] * R
我迷路了:
double tnorm = tvecBest / np.linalg.norm(T)
axis = np.cross(tnorm, forward)
angle = -2 * math.acos(tnorm @ forward)
c++ opencv 中np.linalg.norm
的等价物是什么?
【问题讨论】:
没有一个可以直接使用 cv::sqrt() 和 cv::pow() 你能给我一个代码示例吗? 用norm
怎么样,还是直接用normalize
?
谢谢,我可以这样做吗?:cv::Vec3d tvecNorm(tvecBest); cv::normalize(tvecNorm, tvecNorm);
@anti 是的,应该可以。
【参考方案1】:
np.linalg.norm(T) 只是 L2 范数:
sqrt(T[0]*T[0]+T[1]*T[1]+T[2]*T[2])
np.cross(tnorm, forward) 是叉积:
axis[0] = tnorm[1]*forward[2]-tnorm[2]*forward[1]
axis[1] = -tnorm[0]*forward[2]+tnorm[2]*forward[0]
axis[2] = tnorm[0]*forward[1]-tnorm[1]*forward[0]
【讨论】:
谢谢!所以这给了我:cv::Vec3d forward(0, 0, 1); cv::Vec3d T(tvecBest); cv::Vec3d tnorm = sqrt(T[0] * T[0] + T[1] * T[1] + T[2] * T[2]); cv::Vec3d axis; axis[0] = tnorm[1] * forward[2] - tnorm[2] * forward[1]; axis[1] = -tnorm[0] * forward[2] + tnorm[2] * forward[0]; axis[2] = tnorm[0] * forward[1] - tnorm[1] * forward[0];
c++ 是干什么用的:cv::Vec3d angle = -2 * math.acos(tnorm @ forward)
?
@ 不是 python 中的数学运算符。是 / 运算符吗?
从我读到的意思是@是一个点积:tnorm @ forward -> tnorm[0]*forward [0]+ tnorm[1]*forward [1]+ tnorm[2] *转发[2]
并且角度必须是双精度:双精度角 = -2 * acos( tnorm[0]*forward [0]+ tnorm[1]*forward [1]+ tnorm[2]*forward [ 2])以上是关于将数学函数从 Python 转换为 C++的主要内容,如果未能解决你的问题,请参考以下文章