使用python numpy在三维空间中查找点的k个最近邻
我有一个由n个点组成的3d点云,格式为np.arrayn,3。e、 g这可能类似于:使用python numpy在三维空间中查找点的k个最近邻,python,numpy,knn,Python,Numpy,Knn,我有一个由n个点组成的3d点云,格式为np.arrayn,3。e、 g这可能类似于: P = [[x1,y1,z1],[x2,y2,z2],[x3,y3,z3],[x4,y4,z4],[x5,y5,z5],.....[xn,yn,zn]] 我希望能够得到每个点的K近邻 例如,P1的k近邻可能是P2、P3、P4、P5、P6,P2的KNN可能是P100、P150、P2等 如何在python中实现这一点 这可以通过以下方法巧妙地解决 首先,让我们创建一个在三维空间中存储点的示例阵列: import
P = [[x1,y1,z1],[x2,y2,z2],[x3,y3,z3],[x4,y4,z4],[x5,y5,z5],.....[xn,yn,zn]]
我希望能够得到每个点的K近邻
例如,P1的k近邻可能是P2、P3、P4、P5、P6,P2的KNN可能是P100、P150、P2等
如何在python中实现这一点 这可以通过以下方法巧妙地解决 首先,让我们创建一个在三维空间中存储点的示例阵列:
import numpy as np
N = 10 # The number of points
points = np.random.rand(N, 3)
print(points)
输出:
array([[ 0.23087546, 0.56051787, 0.52412935],
[ 0.42379506, 0.19105237, 0.51566572],
[ 0.21961949, 0.14250733, 0.61098618],
[ 0.18798019, 0.39126363, 0.44501143],
[ 0.24576538, 0.08229354, 0.73466956],
[ 0.26736447, 0.78367342, 0.91844028],
[ 0.76650234, 0.40901879, 0.61249828],
[ 0.68905082, 0.45289896, 0.69096152],
[ 0.8358694 , 0.61297944, 0.51879837],
[ 0.80963247, 0.1680279 , 0.87744732]])
array([[ 0. , 0.4, 0.4, 0.2, 0.5, 0.5, 0.6, 0.5, 0.6, 0.8],
[ 0.4, 0. , 0.2, 0.3, 0.3, 0.7, 0.4, 0.4, 0.6, 0.5],
[ 0.4, 0.2, 0. , 0.3, 0.1, 0.7, 0.6, 0.6, 0.8, 0.6],
[ 0.2, 0.3, 0.3, 0. , 0.4, 0.6, 0.6, 0.6, 0.7, 0.8],
[ 0.5, 0.3, 0.1, 0.4, 0. , 0.7, 0.6, 0.6, 0.8, 0.6],
[ 0.5, 0.7, 0.7, 0.6, 0.7, 0. , 0.7, 0.6, 0.7, 0.8],
[ 0.6, 0.4, 0.6, 0.6, 0.6, 0.7, 0. , 0.1, 0.2, 0.4],
[ 0.5, 0.4, 0.6, 0.6, 0.6, 0.6, 0.1, 0. , 0.3, 0.4],
[ 0.6, 0.6, 0.8, 0.7, 0.8, 0.7, 0.2, 0.3, 0. , 0.6],
[ 0.8, 0.5, 0.6, 0.8, 0.6, 0.8, 0.4, 0.4, 0.6, 0. ]])
[[0 3 1 2 5 7 4 6 8 9]
[1 2 4 3 7 0 6 9 8 5]
[2 4 1 3 0 7 6 9 5 8]
[3 0 2 1 4 7 6 5 8 9]
[4 2 1 3 0 7 9 6 5 8]
[5 0 7 3 6 2 8 4 1 9]
[6 7 8 9 1 0 3 2 4 5]
[7 6 8 9 1 0 3 2 4 5]
[8 6 7 9 1 0 3 5 2 4]
[9 6 7 1 8 4 2 0 3 5]]
[[3 1 2]
[2 4 3]
[4 1 3]
[0 2 1]
[2 1 3]
[0 7 3]
[7 8 9]
[6 8 9]
[6 7 9]
[6 7 1]]
我们计算每个点到所有其他点的距离:
from scipy.spatial import distance
D = distance.squareform(distance.pdist(points))
print(np.round(D, 1)) # Rounding to fit the array on screen
输出:
array([[ 0.23087546, 0.56051787, 0.52412935],
[ 0.42379506, 0.19105237, 0.51566572],
[ 0.21961949, 0.14250733, 0.61098618],
[ 0.18798019, 0.39126363, 0.44501143],
[ 0.24576538, 0.08229354, 0.73466956],
[ 0.26736447, 0.78367342, 0.91844028],
[ 0.76650234, 0.40901879, 0.61249828],
[ 0.68905082, 0.45289896, 0.69096152],
[ 0.8358694 , 0.61297944, 0.51879837],
[ 0.80963247, 0.1680279 , 0.87744732]])
array([[ 0. , 0.4, 0.4, 0.2, 0.5, 0.5, 0.6, 0.5, 0.6, 0.8],
[ 0.4, 0. , 0.2, 0.3, 0.3, 0.7, 0.4, 0.4, 0.6, 0.5],
[ 0.4, 0.2, 0. , 0.3, 0.1, 0.7, 0.6, 0.6, 0.8, 0.6],
[ 0.2, 0.3, 0.3, 0. , 0.4, 0.6, 0.6, 0.6, 0.7, 0.8],
[ 0.5, 0.3, 0.1, 0.4, 0. , 0.7, 0.6, 0.6, 0.8, 0.6],
[ 0.5, 0.7, 0.7, 0.6, 0.7, 0. , 0.7, 0.6, 0.7, 0.8],
[ 0.6, 0.4, 0.6, 0.6, 0.6, 0.7, 0. , 0.1, 0.2, 0.4],
[ 0.5, 0.4, 0.6, 0.6, 0.6, 0.6, 0.1, 0. , 0.3, 0.4],
[ 0.6, 0.6, 0.8, 0.7, 0.8, 0.7, 0.2, 0.3, 0. , 0.6],
[ 0.8, 0.5, 0.6, 0.8, 0.6, 0.8, 0.4, 0.4, 0.6, 0. ]])
[[0 3 1 2 5 7 4 6 8 9]
[1 2 4 3 7 0 6 9 8 5]
[2 4 1 3 0 7 6 9 5 8]
[3 0 2 1 4 7 6 5 8 9]
[4 2 1 3 0 7 9 6 5 8]
[5 0 7 3 6 2 8 4 1 9]
[6 7 8 9 1 0 3 2 4 5]
[7 6 8 9 1 0 3 2 4 5]
[8 6 7 9 1 0 3 5 2 4]
[9 6 7 1 8 4 2 0 3 5]]
[[3 1 2]
[2 4 3]
[4 1 3]
[0 2 1]
[2 1 3]
[0 7 3]
[7 8 9]
[6 8 9]
[6 7 9]
[6 7 1]]
你可以这样读这个距离矩阵:点1和点5之间的距离是距离[0,4]。您还可以看到每个点与自身之间的距离为0,例如距离[6,6]==0
我们对距离矩阵的每一行进行argsort排序,以获得每个点的最近点列表:
closest = np.argsort(D, axis=1)
print(closest)
输出:
array([[ 0.23087546, 0.56051787, 0.52412935],
[ 0.42379506, 0.19105237, 0.51566572],
[ 0.21961949, 0.14250733, 0.61098618],
[ 0.18798019, 0.39126363, 0.44501143],
[ 0.24576538, 0.08229354, 0.73466956],
[ 0.26736447, 0.78367342, 0.91844028],
[ 0.76650234, 0.40901879, 0.61249828],
[ 0.68905082, 0.45289896, 0.69096152],
[ 0.8358694 , 0.61297944, 0.51879837],
[ 0.80963247, 0.1680279 , 0.87744732]])
array([[ 0. , 0.4, 0.4, 0.2, 0.5, 0.5, 0.6, 0.5, 0.6, 0.8],
[ 0.4, 0. , 0.2, 0.3, 0.3, 0.7, 0.4, 0.4, 0.6, 0.5],
[ 0.4, 0.2, 0. , 0.3, 0.1, 0.7, 0.6, 0.6, 0.8, 0.6],
[ 0.2, 0.3, 0.3, 0. , 0.4, 0.6, 0.6, 0.6, 0.7, 0.8],
[ 0.5, 0.3, 0.1, 0.4, 0. , 0.7, 0.6, 0.6, 0.8, 0.6],
[ 0.5, 0.7, 0.7, 0.6, 0.7, 0. , 0.7, 0.6, 0.7, 0.8],
[ 0.6, 0.4, 0.6, 0.6, 0.6, 0.7, 0. , 0.1, 0.2, 0.4],
[ 0.5, 0.4, 0.6, 0.6, 0.6, 0.6, 0.1, 0. , 0.3, 0.4],
[ 0.6, 0.6, 0.8, 0.7, 0.8, 0.7, 0.2, 0.3, 0. , 0.6],
[ 0.8, 0.5, 0.6, 0.8, 0.6, 0.8, 0.4, 0.4, 0.6, 0. ]])
[[0 3 1 2 5 7 4 6 8 9]
[1 2 4 3 7 0 6 9 8 5]
[2 4 1 3 0 7 6 9 5 8]
[3 0 2 1 4 7 6 5 8 9]
[4 2 1 3 0 7 9 6 5 8]
[5 0 7 3 6 2 8 4 1 9]
[6 7 8 9 1 0 3 2 4 5]
[7 6 8 9 1 0 3 2 4 5]
[8 6 7 9 1 0 3 5 2 4]
[9 6 7 1 8 4 2 0 3 5]]
[[3 1 2]
[2 4 3]
[4 1 3]
[0 2 1]
[2 1 3]
[0 7 3]
[7 8 9]
[6 8 9]
[6 7 9]
[6 7 1]]
我们再次看到,每个点都离自己最近。因此,忽略这一点,我们现在可以选择k个最近点:
k = 3 # For each point, find the 3 closest points
print(closest[:, 1:k+1])
输出:
array([[ 0.23087546, 0.56051787, 0.52412935],
[ 0.42379506, 0.19105237, 0.51566572],
[ 0.21961949, 0.14250733, 0.61098618],
[ 0.18798019, 0.39126363, 0.44501143],
[ 0.24576538, 0.08229354, 0.73466956],
[ 0.26736447, 0.78367342, 0.91844028],
[ 0.76650234, 0.40901879, 0.61249828],
[ 0.68905082, 0.45289896, 0.69096152],
[ 0.8358694 , 0.61297944, 0.51879837],
[ 0.80963247, 0.1680279 , 0.87744732]])
array([[ 0. , 0.4, 0.4, 0.2, 0.5, 0.5, 0.6, 0.5, 0.6, 0.8],
[ 0.4, 0. , 0.2, 0.3, 0.3, 0.7, 0.4, 0.4, 0.6, 0.5],
[ 0.4, 0.2, 0. , 0.3, 0.1, 0.7, 0.6, 0.6, 0.8, 0.6],
[ 0.2, 0.3, 0.3, 0. , 0.4, 0.6, 0.6, 0.6, 0.7, 0.8],
[ 0.5, 0.3, 0.1, 0.4, 0. , 0.7, 0.6, 0.6, 0.8, 0.6],
[ 0.5, 0.7, 0.7, 0.6, 0.7, 0. , 0.7, 0.6, 0.7, 0.8],
[ 0.6, 0.4, 0.6, 0.6, 0.6, 0.7, 0. , 0.1, 0.2, 0.4],
[ 0.5, 0.4, 0.6, 0.6, 0.6, 0.6, 0.1, 0. , 0.3, 0.4],
[ 0.6, 0.6, 0.8, 0.7, 0.8, 0.7, 0.2, 0.3, 0. , 0.6],
[ 0.8, 0.5, 0.6, 0.8, 0.6, 0.8, 0.4, 0.4, 0.6, 0. ]])
[[0 3 1 2 5 7 4 6 8 9]
[1 2 4 3 7 0 6 9 8 5]
[2 4 1 3 0 7 6 9 5 8]
[3 0 2 1 4 7 6 5 8 9]
[4 2 1 3 0 7 9 6 5 8]
[5 0 7 3 6 2 8 4 1 9]
[6 7 8 9 1 0 3 2 4 5]
[7 6 8 9 1 0 3 2 4 5]
[8 6 7 9 1 0 3 5 2 4]
[9 6 7 1 8 4 2 0 3 5]]
[[3 1 2]
[2 4 3]
[4 1 3]
[0 2 1]
[2 1 3]
[0 7 3]
[7 8 9]
[6 8 9]
[6 7 9]
[6 7 1]]
例如,我们看到,对于第4点,k=3最近点是1、3和2。@marijn van vliet的解决方案在大多数情况下都能满足要求。但是,它被称为蛮力方法,如果点云相对较大或存在计算/时间限制,则可能需要考虑构建KD树,以快速检索点的K近邻 在python中,sklearn库提供了一个易于使用的实现: 从sklearn.KDTree导入 tree=KDTreepcloud 用于查找形状为1,3的P1的K个邻居 索引,距离=tree.queryP1,K 有关更详细的用法和输出,请参见另一篇文章中的以下答案:
许多其他库都有基于KD树的KNN检索的实现,包括和。可能重复的numpy.linalg.norm和numpy.argsort可能会有所帮助。看,我昨晚自己也得出了同样的解决方案,但非常感谢你的详细解决方案。我相信将来会有人觉得这很有帮助。