如何根据点坐标和所需的 bin 网格大小有效地 bin 3D 点云数据

Question

我在 open3D 中有一个大点云，我想基本上制作一个 3D 网格并根据它们所在的立方体对点进行分类。其他人称之为“在 3D 空间中进行分类”。

仅在一个方向上带有网格的示例图像（我想拆分为 3D 卷）

更好地了解我正在尝试做的事情

例子：

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]  

谢谢你们！

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，沿每个坐标轴应用np.digitize ：

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

这给出了：

[[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]]

该数组的行描述了A的相应行在 bin 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]在沿 X 轴的第一个 bin 中（因为0 < 1 <= 4 ），在沿 Y 轴的第二个 bin 中（因为-5 < -2 <= -1 ），在沿 Z 轴的第一个 bin 中（因为10 < 11 <= 14 ）。 然后你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)] 

这给出了：

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