Question

我正在使用具有lat和long数据的数据帧，我需要聚集彼此最近的点（200米）。这就是我在Python中所做的事情。

    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

然后我在数据帧中找到每对lat和long与每对lat和long之间的距离。

lat_array = np.radians(np.array(order_data['order_lat']))
long_array =  np.radians(np.array(order_data['order_long']))

distance = []
pairs_lat1 = []
pairs_long1 = []
pairs_lat2 = []
pairs_long2 = []
for i in range(len(lat_array)):
   for j in range(i+1,len(lat_array)):
      dlon = long_array[j]-long_array[i]
      dlat = lat_array[j]-lat_array[i]
      a = np.sin(dlat / 2)**2 + np.cos(lat_array[i]) * np.cos(lat_array[j])  
          * np.sin(dlon / 2)**2
      c = 2 * 6371 * np.arcsin(np.sqrt(a))
      pairs_lat1.append(lat_array[i])
      pairs_long1.append(long_array[i])
      pairs_lat2.append(lat_array[j])
      pairs_long2.append(long_array[j])
      distance.append(c)

 df_distance = pd.DataFrame()
 df_distance['lat1'] = np.rad2deg(pairs_lat1)
 df_distance['long1'] = np.rad2deg(pairs_long1)  
 df_distance['lat2'] = np.rad2deg(pairs_lat2)
 df_distance['long2'] = np.rad2deg(pairs_long2)     
 df_distance['distance'] = distance


df_distance.head()

         lat1      long1       lat2      long2      distance
0     19.111841  72.910729  19.111342  72.908387  2.522482e-01
1     19.111841  72.910729  19.111342  72.908387  2.522482e-01
2     19.111841  72.910729  19.137815  72.914085  2.909520e+00
3     19.111841  72.910729  19.119677  72.905081  1.054209e+00
4     19.111841  72.910729  19.119677  72.905081  1.054209e+00
5     19.111841  72.910729  19.119677  72.905081  1.054209e+00

这给了我这对（lat1，long1＆amp; lat2，long2）252米之间的距离我怎样才能聚集积分？所以最近的点在一起。让我们说在半径250米范围内。在我的情况下，我可以使用分层聚类吗？

Answer 1

最简单的方法是建立一个包含任意两点之间距离的距离矩阵，然后使用任何经典的聚类算法。 Scikit-learn是最受欢迎的聚类库之一（以及许多其他内容）。您还可以尝试专为地理空间群集设计的GVM。

python中的纬度和经度聚类

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