Question

我有几点，并希望确定它们是否在彼此的特定距离内。如果是，我想将它们合并为一个点。我构建了一个搜索树，并从中获得了一个距离矩阵。是否存在优雅（如果可能的话没有慢速循环）方法来确定哪些点在特定距离内而不使用某些复杂的聚类算法（kmeans，层次结构等）？

import numpy as np
from sklearn.neighbors import NearestNeighbors
from sklearn.neighbors import radius_neighbors_graph

RADIUS = 0.025
points = np.array([
    [13.2043373032, 52.3818529896],
    [13.0530692845, 52.3991668707],
    [13.229309674, 52.3840231],
    [13.489018081, 52.4180538095],
    [13.3209738098, 52.6375963146],
    [13.0160362703, 52.4187139243],
    [13.0448485, 52.4143229343],
    [13.32478977, 52.5090253],
    [13.35514839, 52.5219323867],
    [13.1982523828, 52.3592620828]
])

tree = NearestNeighbors(n_neighbors=2, radius=RADIUS, leaf_size=30, algorithm="auto", n_jobs=1).fit(points)
nnGraph = radius_neighbors_graph(tree, RADIUS, mode='distance', include_self=False)

print nnGraph

(0, 9)        0.0233960536484
(1, 6)        0.0172420289306
(6, 1)        0.0172420289306
(9, 0)        0.0233960536484

Answer 1

您可以使用pdist中的squareform和scipy.spatial.distance作为矢量化解决方案，就像这样 -

from scipy.spatial.distance import pdist, squareform

# Get pairwise euclidean distances              
dists = squareform(pdist(points))

# Get valid distances mask and the corresponding indices
mask = dists < RADIUS
np.fill_diagonal(mask,0)
idx = np.argwhere(mask)

# Present indices and corresponding distances as zipped output
out = zip(map(tuple,idx),dists[idx[:,0],idx[:,1]])

示例运行 -

In [91]: RADIUS
Out[91]: 0.025

In [92]: points
Out[92]: 
array([[ 13.2043373 ,  52.38185299],
       [ 13.05306928,  52.39916687],
       [ 13.22930967,  52.3840231 ],
       [ 13.48901808,  52.41805381],
       [ 13.32097381,  52.63759631],
       [ 13.01603627,  52.41871392],
       [ 13.0448485 ,  52.41432293],
       [ 13.32478977,  52.5090253 ],
       [ 13.35514839,  52.52193239],
       [ 13.19825238,  52.35926208]])

In [93]: out
Out[93]: 
[((0, 9), 0.023396053648436933),
 ((1, 6), 0.017242028930573985),
 ((6, 1), 0.017242028930573985),
 ((9, 0), 0.023396053648436933)]

Answer 2

对于小点数（＆lt; 50），使用复数更快一点。在另一篇文章中找到了这个：Efficiently Calculating a Euclidean Distance Matrix Using Numpy

pointsCmplx = np.array([points[...,0] + 1j * points[...,1]])
dists = abs(pointsCmplx.T - pointsCmplx)

我的目标是在半径方面获得非重叠点。我拿了你的代码并删除了下三角矩阵，最后我简单地删除了第二点。这些点按特定观察进行排序。较低的指数意味着更重要。有效合并近集群而不是删除点的任何其他建议？我寻找一个非常快速的解决方案，不想使用一些复杂的聚类算法。

# overlapping points
points = np.array([
    [13.2043373032, 52.3818529896],
    [13.0530692845, 52.3991668707],
    [13.229309674, 52.3840231],
    [13.489018081, 52.4180538095],
    [13.3209738098, 52.6375963146],
    [13.0160362703, 52.4187139243],
    [13.0448485, 52.4143229343],
    [13.32478977, 52.5090253],
    [13.35514839, 52.5219323867],
    [13.1982523828, 52.3592620828],
    [13.1982523828, 52.3592620830]       # nearly identical
])

dists = squareform(pdist(points))
mask = dists < RADIUS
np.fill_diagonal(mask,0)

# delete lower triangular matrix
mask = np.triu(mask)
idx = np.argwhere(mask)

# delete the target ids
idx = idx[:,1]   
points = np.delete(points, idx, 0)

距离矩阵的矢量化解释

2 个答案: