使用两个数据框计算多个点的半正弦最近距离答案

【问题标题】：Calculate the haversine nearest distance for multiple points ,using two dataframes使用两个数据框计算多个点的半正弦最近距离
【发布时间】：2021-01-14 03:48:41
【问题描述】：

我想知道数据框二中最近位置到数据框一中每一行的距离是多少。（我的数据框一中每个坐标的最近距离是多少）

查看我的所有代码

我有两个数据框：（在原始数据框中我有数千行）

名为“place_locations”的DataFrame 1：*

|CLUSTER|     |CLIENT|     |LATITUDE|           |LENGHT|

0            X1        19.45685402  -70.68645898
1            X1        19.39320504  -70.52567322
2            X1        18.614736        -68.71711383
3            X2        18.47977644  -69.93177289
4            X2        19.76546997  -70.51085451
5            X3        18.55835346  -68.38226906
6            X3        19.79037017  -70.68748243
7            X4        19.2232559   -70.52629188
8            X4        18.42865751  -68.9703434
9            X5        19.37935119  -70.51440314
10           X5        18.68743273  -68.45068029
11           X6        19.44126162  -70.73159162
12           X6        19.6678557   -70.36758867
13           X7        18.7816069   -70.2598325
14           X8        19.48708304  -70.74375908
15           X8        18.93720371  -70.40746487
16           X9        19.299298        -69.5559162
17          X10        18.60044506  -68.41991221
18          X10        19.30702896  -69.54500792
19          X11        19.3783253   -70.618205

名为“Coordinates_coords”的 DataFrame 2：

 | PLACE|               | LATITUDE|           | LENGHT| 
  supermarket       18.63609095 -68.39650565
   school           19.44512055 -70.66851055
   restarant        18.48377033 -69.93910793
   spa              18.46608496 -69.92713481
   supermarket      18.45646778 -69.9395694
   restaurant       18.4845644  -69.9300583
   school           18.47284417 -69.9345797

def haversine_np(lon1, lat1, lon2, lat2):
    
   
    lon1, lat1, lon2, lat2 = map(np.radians, [lon1, lat1, lon2, lat2])

    dlon = lon2 - lon1
    dlat = lat2 - lat1

    a = np.sin(dlat/2.0)**2 + np.cos(lat1) * np.cos(lat2) * np.sin(dlon/2.0)**2

    c = 2 * np.arcsin(np.sqrt(a))
    km = 6371 * c
    return km

def Top_nearest(distancia,distancias,todos = False,limit= 1.0):
    results = []
    for d in distancias:
        results.append(haversine_np(distancia[0],distancia[1],d[0],d[1]))
    results= np.array(results)
    if not todos:
        print(results.argmin())
        indexes = np.where(results < limit)
    else:
        indexes = np.where(results>= 0)
    
    return list(indexes[0]),results[indexes]

nearest_coordinates = list()

for index,row in place_locations.iterrows():
    indexes,distances=Top_nearest(row[['LATITUDE', 'LENGHT']].values,
                                     Coordinates_coords[['LATITUDE', 'LENGHT']].reset_index(drop=True).values,
                                            todos=True)
    nearest_coordinates.append(distances[0])

nearest_coordinates [:5]

place_locations['Distance_locations'] = nearest_coordinates 
place_locations

我得到的结果不正确，微积分中有一些我无法识别的东西。我得到的距离位置列，不要给我最近的距离位置

【问题讨论】：

您实际上在哪里对距离进行排序？看起来您正在计算与每个点的距离，然后只是将它们留在数组中而不对它们进行排序（尽管您说当结果> = 0时，当我认为结果不能低于 0 时）。也许做一个 argsort 而不是 np.where 并取第一个（或前 N 个）值？

标签： python python-3.x pandas

【解决方案1】：

我之前已经发布过这个。 location, distance, nearest

使用库来计算距离是最简单的。 geopy 对我来说效果很好

import geopy.distance
import pandas as pd
import io
df1 = pd.read_csv(io.StringIO("""CLUSTER     CLIENT     LATITUDE           LENGHT

0            X1        19.45685402  -70.68645898
1            X1        19.39320504  -70.52567322
2            X1        18.614736        -68.71711383
3            X2        18.47977644  -69.93177289
4            X2        19.76546997  -70.51085451
5            X3        18.55835346  -68.38226906
6            X3        19.79037017  -70.68748243
7            X4        19.2232559   -70.52629188
8            X4        18.42865751  -68.9703434
9            X5        19.37935119  -70.51440314
10           X5        18.68743273  -68.45068029
11           X6        19.44126162  -70.73159162
12           X6        19.6678557   -70.36758867
13           X7        18.7816069   -70.2598325
14           X8        19.48708304  -70.74375908
15           X8        18.93720371  -70.40746487
16           X9        19.299298        -69.5559162
17          X10        18.60044506  -68.41991221
18          X10        19.30702896  -69.54500792
19          X11        19.3783253   -70.618205"""), sep="\s+")

df2 = pd.read_csv(io.StringIO(""" PLACE                LATITUDE            LENGHT 
  supermarket       18.63609095 -68.39650565
   school           19.44512055 -70.66851055
   restarant        18.48377033 -69.93910793
   spa              18.46608496 -69.92713481
   supermarket      18.45646778 -69.9395694
   restaurant       18.4845644  -69.9300583
   school           18.47284417 -69.9345797"""), sep="\s+")

# no need to calc distance in miles and kms,  there for reference
df3 = (df1
 .assign(foo=1)
 .merge(df2.assign(foo=1), on="foo")
 .assign(distance_km=lambda dfa: dfa.apply(lambda r: 
                                      geopy.distance.geodesic(
                                          (r["LATITUDE_x"],r["LENGHT_y"]), 
                                          (r["LATITUDE_y"],r["LENGHT_y"])).km, axis=1))
 .assign(distance_miles=lambda dfa: dfa.apply(lambda r: 
                                      geopy.distance.geodesic(
                                          (r["LATITUDE_x"],r["LENGHT_y"]), 
                                          (r["LATITUDE_y"],r["LENGHT_y"])).miles, axis=1))

 
)

# now find nearest PLACE to a CLIENT and count
(df3.sort_values(["CLIENT","distance_km"])
 .groupby(["CLIENT"]).agg({"PLACE":"first","distance_km":"first"})
 .reset_index()
 .groupby("PLACE")["CLIENT"].count()
 .to_frame().reset_index().sort_values("CLIENT",ascending=False)
)

输出

         PLACE  CLIENT
2       school       5
3  supermarket       4
0    restarant       1
1   restaurant       1

【讨论】：