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[英]Matplotlib: How to efficiently plot a large number of line segments in 3D?
[英]How to efficiently compute orientation of 3D normals in large pointclouds
我正在(更像是邊做邊學)使用 Python 庫來管理 Python 中的點雲。
我編寫了一個函數來計算存儲為 numpy 結構化數組的點雲中每個法線的方向,但我對最終函數並不滿意(認為它有效並且足夠快),我想知道是否有另一種更有效/pythonic 的方法來計算大型點雲中的方向。
這就是點雲的結構:
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
這是方向函數:
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])
以及在 CloudCompare 中可視化的結果(對於發布圖像沒有足夠的代表):
感謝您的幫助。
好吧,我為自己感到羞恥。 xD
那些 numpy 內置函數正是我正在尋找的。
謝謝@丹。
這是新功能:
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])
哪個更簡單,更快。
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
現在我既高興又羞愧。
下次我會在發布問題之前更深入地查看 numpy 文檔。
謝謝。
comp = esfera.vertex['orid'] == esfera.vertex['orid2']
np.all(comp)
Out[15]: True
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