我有一个纬度和经度对的数据帧.

这是我的数据帧外观.

    order_lat  order_long
0   19.111841   72.910729
1   19.111342   72.908387
2   19.111342   72.908387
3   19.137815   72.914085
4   19.119677   72.905081
5   19.119677   72.905081
6   19.119677   72.905081
7   19.120217   72.907121
8   19.120217   72.907121
9   19.119677   72.905081
10  19.119677   72.905081
11  19.119677   72.905081
12  19.111860   72.911346
13  19.111860   72.911346
14  19.119677   72.905081
15  19.119677   72.905081
16  19.119677   72.905081
17  19.137815   72.914085
18  19.115380   72.909144
19  19.115380   72.909144
20  19.116168   72.909573
21  19.119677   72.905081
22  19.137815   72.914085
23  19.137815   72.914085
24  19.112955   72.910102
25  19.112955   72.910102
26  19.112955   72.910102
27  19.119677   72.905081
28  19.119677   72.905081
29  19.115380   72.909144
30  19.119677   72.905081
31  19.119677   72.905081
32  19.119677   72.905081
33  19.119677   72.905081
34  19.119677   72.905081
35  19.111860   72.911346
36  19.111841   72.910729
37  19.131674   72.918510
38  19.119677   72.905081
39  19.111860   72.911346
40  19.111860   72.911346
41  19.111841   72.910729
42  19.111841   72.910729
43  19.111841   72.910729
44  19.115380   72.909144
45  19.116625   72.909185
46  19.115671   72.908985
47  19.119677   72.905081
48  19.119677   72.905081
49  19.119677   72.905081
50  19.116183   72.909646
51  19.113827   72.893833
52  19.119677   72.905081
53  19.114100   72.894985
54  19.107491   72.901760
55  19.119677   72.905081

我想聚集这些彼此最近的点(距离200米)以下是我的距离矩阵.

from scipy.spatial.distance import pdist, squareform
distance_matrix = squareform(pdist(X, (lambda u,v: haversine(u,v))))

array([[ 0.        ,  0.2522482 ,  0.2522482 , ...,  1.67313071,
     1.05925366,  1.05420922],
   [ 0.2522482 ,  0.        ,  0.        , ...,  1.44111548,
     0.81742536,  0.98978355],
   [ 0.2522482 ,  0.        ,  0.        , ...,  1.44111548,
     0.81742536,  0.98978355],
   ..., 
   [ 1.67313071,  1.44111548,  1.44111548, ...,  0.        ,
     1.02310118,  1.22871515],
   [ 1.05925366,  0.81742536,  0.81742536, ...,  1.02310118,
     0.        ,  1.39923529],
   [ 1.05420922,  0.98978355,  0.98978355, ...,  1.22871515,
     1.39923529,  0.        ]])

然后我在距离矩阵上应用DBSCAN聚类算法.

 from sklearn.cluster import DBSCAN

 db = DBSCAN(eps=2,min_samples=5)
 y_db = db.fit_predict(distance_matrix)

我不知道如何选择eps& min_samples值.它在一个星团中聚集了太远的点.(距离约2公里)是因为它在聚类时计算欧氏距离？请帮忙.

解决方法:

DBSCAN旨在用于原始数据,具有加速的空间索引.我知道加速地理距离的唯一工具是ELKI(Java) – 不幸的是,scikit-learn只支持像欧几里德距离这样的一些距离(参见sklearn.neighbors.NearestNeighbors).
但显然,你可以预先计算成对距离,因此这不是一个问题.

但是,您没有仔细阅读文档,并且您认为DBSCAN使用距离矩阵是错误的：

from sklearn.cluster import DBSCAN
db = DBSCAN(eps=2,min_samples=5)
db.fit_predict(distance_matrix)

在距离矩阵行上使用欧几里德距离,这显然没有任何意义.

请参阅DBSCAN的文档(重点已添加)：

class sklearn.cluster.DBSCAN(eps=0.5, min_samples=5, metric=’euclidean’, algorithm=’auto’, leaf_size=30, p=None, random_state=None)

metric : string, or callable

The metric to use when calculating distance between instances in a feature array. If metric is a string or callable, it must be one of the options allowed by metrics.pairwise.calculate_distance for its metric parameter. If metric is “precomputed”, X is assumed to be a distance matrix and must be square. X may be a sparse matrix, in which case only “nonzero” elements may be considered neighbors for DBSCAN.

类似于fit_predict：

X : array or sparse (CSR) matrix of shape (n_samples, n_features), or array of shape (n_samples, n_samples)

A feature array, or array of distances between samples if metric=’precomputed’.

换句话说,你需要这样做

db = DBSCAN(eps=2, min_samples=5, metric="precomputed")