Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
611 views
in Technique[技术] by (71.8m points)

python - DBSCAN for clustering of geographic location data

I have a dataframe with latitude and longitude pairs.

Here is my dataframe look like.

    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

I want to cluster this points which are nearest to each other(200 meters distance) following is my distance matrix.

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.        ]])

Then I am applying DBSCAN clustering algorithm on distance matrix.

 from sklearn.cluster import DBSCAN

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

I don't know how to choose eps & min_samples value. It clusters the points which are way too far, in one cluster.(approx 2 km in distance) Is it because it calculates euclidean distance while clustering? please help.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

You can cluster spatial latitude-longitude data with scikit-learn's DBSCAN without precomputing a distance matrix.

db = DBSCAN(eps=2/6371., min_samples=5, algorithm='ball_tree', metric='haversine').fit(np.radians(coordinates))

This comes from this tutorial on clustering spatial data with scikit-learn DBSCAN. In particular, notice that the eps value is still 2km, but it's divided by 6371 to convert it to radians. Also, notice that .fit() takes the coordinates in radian units for the haversine metric.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...