如何有效地基于点坐标和所需的bin网格大小来存储三维点云数据？

Question

我在open3D中有一个很大的点云，我想基本上建立一个三维网格，并根据它们在哪个立方体上存储点。其他人称它为“三维空间中的二进制”。

只有一个方向的网格的例子图像(我想分裂成3D卷)

我想要做的更好的形象

示例：

import numpy as np

A = np.array([[ 0, -1, 10],
 [ 1, -2 ,11],
 [ 2, -3 ,12],
 [ 3, -4 ,13],
 [ 4, -5 ,14],
 [ 5, -6 ,15],
 [ 6, -7 ,16],
 [ 7, -8 ,17],
 [ 8, -9 ,18]])

#point 1: X,Y,Z
#point 2: X,Y,Z

print(A)

X_segments = np.linspace(0,8,3) #plane at beginning, middle and end - this creates 2 sections where data can be
Y_segments = np.linspace(-9,-1,3)
Z_segments = np.linspace(10,18,3)

#all of these combined form 4 cuboids where data can be 
#its also possible for the data to be outside these cuboids but we can ignore that

bin1 =  A where A[0,:] is > X_segments [0] and < X_segments[1]
    and A where A[1,:] is > Y_segments [0] and < Y_segments[1] 
    and A where A[2,:] is > Z_segments [0] and < Z_segments[1] 

bin2 =  A where A[0,:] is > X_segments [1] and < X_segments[2]
    and A where A[1,:] is > Y_segments [0] and < Y_segments[1] 
    and A where A[2,:] is > Z_segments [0] and < Z_segments[1] 

bin3 =  A where A[0,:] is > X_segments [1] and < X_segments[2]
    and A where A[1,:] is > Y_segments [1] and < Y_segments[2] 
    and A where A[2,:] is > Z_segments [0] and < Z_segments[1] 

bin4 =  A where A[0,:] is > X_segments [1] and < X_segments[2]
    and A where A[1,:] is > Y_segments [1] and < Y_segments[2] 
    and A where A[2,:] is > Z_segments [1] and < Z_segments[2]  

谢谢大家！

Answer 1

您可以尝试以下方法：

import numpy as np

A = np.array([[ 0, -1, 10],
              [ 1, -2 ,11],
              [ 2, -3 ,12],
              [ 3, -4 ,13],
              [ 4, -5 ,14],
              [ 5, -6 ,15],
              [ 6, -7 ,16],
              [ 7, -8 ,17],
              [ 8, -9 ,18]])

X_segments = np.linspace(0,8,3)
Y_segments = np.linspace(-9,-1,3)
Z_segments = np.linspace(10,18,3)

edges = [X_segments, Y_segments, Z_segments]

print(edges) # just to show edges of the bins

我们得到：

[[ 0.  4.  8.]
 [-9. -5. -1.]
 [10. 14. 18.]]

接下来，沿着每个坐标系应用np.digitize：

coords = np.vstack([np.digitize(A.T[i], b, right=True) for i, b in enumerate(edges)]).T
print(coords)

这意味着：

[[0 2 0]
 [1 2 1]
 [1 2 1]
 [1 2 1]
 [1 1 1]
 [2 1 2]
 [2 1 2]
 [2 1 2]
 [2 0 2]]

此数组的行描述A的相应行在回收箱网格中的位置。例如，第二行[1, 2, 1]表示A的第二行，即[1, -2, 11]位于沿X轴的第一行(自0 < 1 <= 4)、沿Y轴的第二行(自-5 < -2 <= -1)，以及沿Z轴的第一行(自10 < 11 <= 14)。然后您可以选择属于每个长方体的元素：

# select rows of A that are in the cuboid with coordinates [1, 2, 1]
A[np.all(coords == [1, 2, 1], axis=1)] 

这意味着：

array([[ 1, -2, 11],
       [ 2, -3, 12],
       [ 3, -4, 13]])