python - 如何在大型点云中有效计算3D法线的方向

我正在使用Python库（用于管理Python中的点云）（更像边做边学）。

我已经编写了一个函数来计算存储为numpy结构化数组的点云中每个法线的方向，但是我对最终函数不满意（认为它可以工作并且足够快），我想知道是否存在在大型点云中计算方位的另一种更高效/ Python方法。

这是pointcloud的结构：

esfera = PyntCloud.from_ply('Sphere.ply')

esfera.vertex
Out[3]:
array([ (0.2515081465244293, 0.05602749437093735, 1.9830318689346313, 0.12660565972328186, 0.02801010198891163, 0.9915575981140137, 7.450349807739258, 77.52488708496094),
       (0.09723527729511261, 0.02066999115049839, 1.9934484958648682, 0.048643846064805984, 0.011384730227291584, 0.9987513422966003, 2.863548517227173, 76.82744598388672),
       (0.17640848457813263, 0.028193067759275436, 1.9881943464279175, 0.08916780352592468, 0.01611466333270073, 0.9958862066268921, 5.198856830596924, 79.75591278076172),
       ...,
       (0.17817874252796173, -0.046098098158836365, -1.9879237413406372, 0.08992616087198257, -0.02275240235030651, -0.9956884980201721, 5.322407245635986, 284.19854736328125),
       (0.2002459168434143, -0.002330917865037918, -1.986855149269104, 0.09960971027612686, -0.0010710721835494041, -0.9950260519981384, 5.717002868652344, 270.6160583496094),
       (0.12885123491287231, -0.03245270624756813, -1.9912745952606201, 0.06637085974216461, -0.01580258458852768, -0.9976698756217957, 3.912114381790161, 283.3924865722656)],
      dtype=[('x', '<f4'), ('y', '<f4'), ('z', '<f4'), ('nx', '<f4'), ('ny', '<f4'), ('nz', '<f4'), ('scalar_Dip_(degrees)', '<f4'), ('scalar_Dip_direction_(degrees)', '<f4')])

esfera.vertex['nx']
Out[4]:
array([ 0.12660566,  0.04864385,  0.0891678 , ...,  0.08992616,
        0.09960971,  0.06637086], dtype=float32)

esfera.vertex[-1]['nx']
Out[5]: 0.06637086

这是方向函数：

def add_orientation(self, degrees=True):

    """ Adds orientation (with respect to y-axis) values to PyntCloud.vertex

    This function expects the PyntCloud to have a numpy structured array
    with normals x,y,z values (correctly named) as the corresponding vertex
    atribute.

     Args:
        degrees (Optional[bool]): Set the oputput orientation units.
            If True(Default) set units to degrees.
            If False set units to radians.
    """

    #: set copy to False for efficience in large pointclouds
    nx = self.vertex['nx'].astype(np.float64, copy=False)
    ny = self.vertex['ny'].astype(np.float64, copy=False)

    #: get orientations
    angle = np.arctan(np.absolute(nx / ny))

    #: mask for every quadrant
    q2 = np.logical_and((self.vertex['nx']>0),(self.vertex['ny']<0))
    q3 = np.logical_and((self.vertex['nx']<0),(self.vertex['ny']<0))
    q4 = np.logical_and((self.vertex['nx']<0),(self.vertex['ny']>0))

    #: apply modification for every quadrant
    angle[q2] = np.pi - angle[q2]
    angle[q3] = np.pi + angle[q3]
    angle[q4] = (2*np.pi) - angle[q4]

    if degrees == False:
        orientation = np.array(angle, dtype=[('orir', 'f4')])
    else:
        orientation = np.array((180 * angle / np.pi), dtype=[('orid', 'f4')])

    #: merge the structured arrays and replace the old vertex attribute
    self.vertex = join_struct_arrays([self.vertex, orientation])

并且结果在CloudCompare中可视化（不要表示后期图像）：

https://raw.githubusercontent.com/daavoo/sa/master/Captura%20de%20pantalla%20de%202016-03-21%2013%3A28%3A39.png

感谢您的帮助。

最佳答案

好吧，我为自己感到羞耻。 xD

那些numpy内置函数正是我想要的。

谢谢@丹。

这是新功能：

 def add_orientation(self, degrees=True):

        """ Adds orientation (with respect to y-axis) values to PyntCloud.vertex

        This function expects the PyntCloud to have a numpy structured array
        with normals x,y,z values (correctly named) as the corresponding vertex
        atribute.

         Args:
            degrees (Optional[bool]): Set the oputput orientation units.
                If True(Default) set units to degrees.
                If False set units to radians.
        """

        #: set copy to False for efficience in large pointclouds
        nx = self.vertex['nx'].astype(np.float64, copy=False)
        ny = self.vertex['ny'].astype(np.float64, copy=False)

        #: get orientations
        angle = np.arctan2(nx,ny)

        #: convert (-180 , 180) to (0 , 360)
        angle[(np.where(angle < 0))] = (2*np.pi) + angle[(np.where(angle < 0))]

        if degrees:
            orientation = np.array(np.rad2deg(angle), dtype=[("orid2",'f4')])
        else:
            orientation = np.array(angle, dtype=[("orir2",'f4')])

        self.vertex = join_struct_arrays([self.vertex, orientation])

Wich更简单，更快捷。

t0 = t.time()
esfera.add_orientation()
t1 = t.time()
dif = t1-t0
dif
Out[18]: 0.34514379501342773

t0 = t.time()
esfera.add_orientation2()
t1 = t.time()
dif = t1-t0
dif
Out[20]: 0.291456937789917

现在，我感到as愧。

下次，我将在发布问题之前对numpy文档进行更深入的研究。

谢谢。

comp = esfera.vertex['orid'] == esfera.vertex['orid2']

np.all(comp)
Out[15]: True

关于python - 如何在大型点云中有效计算3D法线的方向，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/36119783/