使用python numpy在3d空间中找到点的k个最近邻居 [英] find the k nearest neighbours of a point in 3d space with python numpy

问题描述

我有n个点的3d点云，格式为np.array((n，3)).例如，可能是这样的:

I have a 3d point cloud of n points in the format np.array((n,3)). e.g This could be something like:

P = [[x1,y1,z1],[x2,y2,z2],[x3,y3,z3],[x4,y4,z4],[x5,y5,z5],.....[xn,yn,zn]]

我希望能够获得每个点的K个最近邻居.

I would like to be able to get the K-nearest neighbors of each point.

例如，P1的k个最近邻居可能是P2，P3，P4，P5，P6，而P2的KNN可能是P100，P150，P2等.

so for example the k nearest neighbors of P1 might be P2,P3,P4,P5,P6 and the KNN of P2 might be P100,P150,P2 etc etc.

如何在python中做到这一点?

how does one go about doing that in python?

推荐答案

可以使用首先，让我们创建一个示例数组，在3D空间中存储点:

First, let's create an example array that stores points in 3D space:

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]])

我们为每个点计算到所有其他点的距离:

We compute for each point, the distance to all other points:

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. ,  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. ]])

您可以这样读取距离矩阵:点1和5之间的距离为distance[0, 4].您还可以看到每个点与其自身之间的距离为0，例如distance[6, 6] == 0

You read this distance matrix like this: the distance between points 1 and 5 is distance[0, 4]. You can also see that the distance between each point and itself is 0, for example distance[6, 6] == 0

我们argsort距离矩阵的每一行，以获得每个点的最接近点的列表:

We argsort each row of the distance matrix to get for each point a list of which points are closest:

closest = np.argsort(D, axis=1)
print(closest)

输出:

[[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]]

同样，我们看到每个点都离自己最近.因此，不用考虑，我们现在可以选择k个最接近的点:

Again, we see that each point is closest to itself. So, disregarding that, we can now select the k closest points:

k = 3  # For each point, find the 3 closest points
print(closest[:, 1:k+1])

输出:

[[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，我们看到对于点4，k = 3个最接近的点是1、3和2.

For example, we see that for point 4, the k=3 closest points are 1, 3 and 2.

这篇关于使用python numpy在3d空间中找到点的k个最近邻居的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！