计算（x，y，z）点列表中最近邻居的（欧几里得）距离的最有效方法是什么？

Question

对于数组中的每个点，最有效的方法是计算最近邻的距离（欧几里得）？

我有一个（X，Y，Z）点的列表，我想计算一个最近邻距离的列表。距离的索引将对应于该点的索引。

我调查了PYOD和sklearn邻居，但是这些邻居似乎需要“教学”。我认为我的问题比这简单。对于每个点：找到最近的邻居，计算距离。

示例数据：

points = [
     (0.003467119 ,0.01422762 ,0.0101960126),
     (0.007279433  ,0.01651597  ,0.0045558849),
     (0.005392258  ,0.02149997  ,0.0177409387),
     (0.017898802  ,0.02790659  ,0.0006487222),
     (0.013564214  ,0.01835688  ,0.0008102952),
     (0.013375397  ,0.02210725 ,0.0286032185)
]

计算k = 1最近的邻居距离

结果格式：

results = [(point, nearest neighbor distance), ...]

示例结果：

results = [
((0.003467119, 0.01422762, 0.0101960126), (0.007279433, 0.01651597, 0.0045558849)), 
((0.007279433, 0.01651597, 0.0045558849), (0.003467119, 0.01422762, 0.0101960126)), 
((0.005392258, 0.02149997, 0.0177409387), (0.003467119, 0.01422762, 0.0101960126)), 
((0.017898802, 0.02790659, 0.0006487222), (0.013564214, 0.01835688, 0.0008102952)), 
((0.013564214, 0.01835688, 0.0008102952), (0.007279433, 0.01651597, 0.0045558849)), 
((0.013375397, 0.02210725, 0.0286032185), (0.005392258, 0.02149997, 0.0177409387))
]

Answer 1

怎么样？

from scipy.spatial import distance

A = (0.003467119 ,0.01422762 ,0.0101960126)
B = (0.007279433  ,0.01651597  ,0.0045558849)
C = (0.005392258  ,0.02149997  ,0.0177409387)
D = (0.017898802  ,0.02790659  ,0.0006487222)
E = (0.013564214  ,0.01835688  ,0.0008102952)
F = (0.013375397  ,0.02210725 ,0.0286032185)

points = [A, B, C, D, E, F]
results = []
for point in points:
    distances = [{'point':point, 'neighbor':p, 'd':distance.euclidean(point, p)} for p in points if p != point]
    results.append(min(distances, key=lambda k:k['d']))

结果将是对象列表，如下所示：

results = [
    {'point':(x1, y1, z1), 'neighbor':(x2, y2, z2), 'd':"distance from point to neighbor"},
...]

point是参考点，neighbor是该点的最近邻居。