将手机信号塔的估计位置分配给最近的实际手机信号塔

Question

所以我有一个这样的数据集：

Customer_id   Lat        Lon
0.     A          40         12
1.     A          np.nan     np.nan
2.     A          np.nan     np.nan
3.     A          43         12
4.     A          45         13
5.     B          43         14
6.     B          np.nan     np.nan
7.     B          43         16

其中坐标(40,12(、(43,12(、(45,13(、(43,14(和(43,16(是某个网络的手机信号塔。

然后我应用一些插值函数，结果如下：

Customer_id   Lat        Lon
0.     A          40         12
1.     A          41         12
2.     A          42         12
3.     A          43         12
4.     A          45         13
5.     B          43         14
6.     B          43         15
7.     B          43         16

但这些新坐标只是估计值，而不是实际的塔楼。然后，我想将这些估计值分配给最近的实际塔，以便例如将记录 1 分配给塔 (40,12(。

我使用了这个代码

def haversine_closest_changed(towers, row):
all_points= towers
lat2= all_points[:,0] #the actual latitudes of the towers
lon2= all_points[:,1] #the actual longitudes of the towers
l=len(lat2) #how many towers are there
lat1=row['Expected_Lat']  #make a column with the actual latitude my value and all the towers, 
#the point I'm looking at multiple times
lon1=row['Expected_Lon']  #find the min distance and output the minimum 
lat1, lon1, lat2, lon2 = map(np.radians, [lon1, lat1, lon2, lat2])
dlat = lat2 - lat1
dlon = lon2 - lon1
a = np.sin(dlon/2.0)**2 + np.cos(lon1) * np.cos(lon2) * np.sin(dlat/2.0)**2
c = 2 * np.arcsin(np.sqrt(a))
km = 6367 * c
idx=np.argmin(km)
closest_point=towers[idx,]
return closest_point

其中 towers 是一个 pandas 数据集，其中包含网络中存在的所有塔(一列用于纬度，另一列用于经度(，Expected_Lat和Expected_Lon列是我在进行插值后所说的列。

这段代码只返回 1 个纬度值和 1 个经度值，在整个列中重复。如何更改此代码以仅替换我插值的点/最近的塔之前是 NaN 的点？

Answer 1

首先，我们标记要插值的行，然后插值，最后，通过从这个 SO 答案最接近的距离计算，我们找到所有插值条目的最接近的实际塔：

import pandas as pd
import io
from math import cos, asin, sqrt
s="""     Customer_id   Lat        Lon
0.     A          40         12
1.     A          np.nan     np.nan
2.     A          np.nan     np.nan
3.     A          43         12
4.     A          45         13
5.     B          43         14
6.     B          np.nan     np.nan
7.     B          43         16 
"""

df = pd.read_csv(io.StringIO(s), na_values='np.nan', sep='ss+', engine='python')
df['Interpolated'] = df.Lat.isnull()
df = df.interpolate()
towers = df.loc[~df.Interpolated,['Lat','Lon']].drop_duplicates().values
def distance(lat1, lon1, lat2, lon2):
p = 0.017453292519943295
a = 0.5 - cos((lat2-lat1)*p)/2 + cos(lat1*p)*cos(lat2*p) * (1-cos((lon2-lon1)*p)) / 2
return 12742 * asin(sqrt(a))
def closest_tower(row):
return min(towers, key=lambda p: distance(row.Lat, row.Lon, p[0], p[1]))
df.loc[df.Interpolated,['Lat','Lon']] = df.loc[df.Interpolated,['Lat','Lon']].apply(closest_tower ,axis=1)

结果：

Customer_id   Lat   Lon  Interpolated
0.0           A  40.0  12.0         False
1.0           A  40.0  12.0          True
2.0           A  43.0  12.0          True
3.0           A  43.0  12.0         False
4.0           A  45.0  13.0         False
5.0           B  43.0  14.0         False
6.0           B  43.0  14.0          True
7.0           B  43.0  16.0         False