如何有效地计算大型点云中三维法线的方向

Question

我正在开发一个Python库，用于在Python中管理点云。在

我写了一个函数来计算存储在一个numpy结构数组中的每个法线的方向，但是我对最后一个函数不太满意（认为它能工作而且速度很快），我想知道是否有另一种更高效的python方法来计算大型点云中的方向。在

点云的结构如下：

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：

^{pr2}$

结果显示在CloudCompare中（post images没有正确的代表）：

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

谢谢你的帮助。在

Answer 1

嗯，我为自己感到羞耻。xD公司

那些numpy内置的功能正是我想要的。在

谢谢@Dan。在

新功能如下：

 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])

它更简单快捷。在

^{pr2}$

现在我既高兴又羞愧。在

下一次我会在发帖前更深入地看一下numpy文档。在

谢谢。在

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

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