我是数据框中的一个位置,位于纬度和经度列名称下方。我想展示它与另一个数据框中最近火车站的距离。
例如,假设我有一个纬度和经度为(37.814563 144.970267)的点,以及以下其他地理空间点的列表。我想找到最接近的点,然后在郊区数据框中作为额外的一列计算这些点之间的距离。
这是火车数据集的示例。
<bound method NDFrame.to_clipboard of STOP_ID STOP_NAME LATITUDE \
0 19970 Royal Park Railway Station (Parkville) -37.781193
1 19971 Flemington Bridge Railway Station (North Melbo... -37.788140
2 19972 Macaulay Railway Station (North Melbourne) -37.794267
3 19973 North Melbourne Railway Station (West Melbourne) -37.807419
4 19974 Clifton Hill Railway Station (Clifton Hill) -37.788657
LONGITUDE TICKETZONE ROUTEUSSP \
0 144.952301 1 Upfield
1 144.939323 1 Upfield
2 144.936166 1 Upfield
3 144.942570 1 Flemington,Sunbury,Upfield,Werribee,Williamsto...
4 144.995417 1 Mernda,Hurstbridge
geometry
0 POINT (144.95230 -37.78119)
1 POINT (144.93932 -37.78814)
2 POINT (144.93617 -37.79427)
3 POINT (144.94257 -37.80742)
4 POINT (144.99542 -37.78866) >
这是郊区的一个例子
<bound method NDFrame.to_clipboard of postcode suburb state lat lon
4901 3000 MELBOURNE VIC -37.814563 144.970267
4902 3002 EAST MELBOURNE VIC -37.816640 144.987811
4903 3003 WEST MELBOURNE VIC -37.806255 144.941123
4904 3005 WORLD TRADE CENTRE VIC -37.822262 144.954856
4905 3006 SOUTHBANK VIC -37.823258 144.965926>
我想要展示的是,在郊区列表中新增一列,其中包含从纬度和经度到最近火车站的距离。
使用一种解决方案得到了奇怪的输出,不确定是否正确?
两个解决方案都被展示了,
from sklearn.neighbors import NearestNeighbors
from haversine import haversine
NN = NearestNeighbors(n_neighbors=1, metric='haversine')
NN.fit(trains_shape[['LATITUDE', 'LONGITUDE']])
indices = NN.kneighbors(df_complete[['lat', 'lon']])[1]
indices = [index[0] for index in indices]
distances = NN.kneighbors(df_complete[['lat', 'lon']])[0]
df_complete['closest_station'] = trains_shape.iloc[indices]['STOP_NAME'].reset_index(drop=True)
df_complete['closest_station_distances'] = distances
print(df_complete)
这里的输出,
<bound method NDFrame.to_clipboard of postcode suburb state lat lon Venues Cluster \
1 3040 aberfeldie VIC -37.756690 144.896259 4.0
2 3042 airport west VIC -37.711698 144.887037 1.0
4 3206 albert park VIC -37.840705 144.955710 0.0
5 3020 albion VIC -37.775954 144.819395 2.0
6 3078 alphington VIC -37.780767 145.031160 4.0
#1 #2 #3 \
1 Café Electronics Store Grocery Store
2 Fast Food Restaurant Café Supermarket
4 Café Pub Coffee Shop
5 Café Fast Food Restaurant Grocery Store
6 Café Park Bar
#4 ... #6 \
1 Coffee Shop ... Bakery
2 Grocery Store ... Italian Restaurant
4 Breakfast Spot ... Burger Joint
5 Vietnamese Restaurant ... Pub
6 Pizza Place ... Vegetarian / Vegan Restaurant
#7 #8 #9 \
1 Shopping Mall Japanese Restaurant Indian Restaurant
2 Portuguese Restaurant Electronics Store Middle Eastern Restaurant
4 Bar Bakery Gastropub
5 Chinese Restaurant Gym Bakery
6 Italian Restaurant Gastropub Bakery
#10 Ancestry Cluster ClosestStopId \
1 Greek Restaurant 8.0 20037
2 Convenience Store 5.0 20032
4 Beach 6.0 22180
5 Convenience Store 5.0 20004
6 Coffee Shop 5.0 19931
ClosestStopName \
1 Essendon Railway Station (Essendon)
2 Glenroy Railway Station (Glenroy)
4 Southern Cross Railway Station (Melbourne City)
5 Albion Railway Station (Sunshine North)
6 Alphington Railway Station (Alphington)
closest_station closest_station_distances
1 Glenroy Railway Station (Glenroy) 0.019918
2 Southern Cross Railway Station (Melbourne City) 0.031020
4 Alphington Railway Station (Alphington) 0.023165
5 Altona Railway Station (Altona) 0.005559
6 Newport Railway Station (Newport) 0.002375
还有第二个功能。
def ClosestStop(r):
# Cartesin Distance: square root of (x2-x2)^2 + (y2-y1)^2
distances = ((r['lat']-StationDf['LATITUDE'])**2 + (r['lon']-StationDf['LONGITUDE'])**2)**0.5
# Stop with minimum Distance from the Suburb
closestStationId = distances[distances == distances.min()].index.to_list()[0]
return StationDf.loc[closestStationId, ['STOP_ID', 'STOP_NAME']]
df_complete[['ClosestStopId', 'ClosestStopName']] = df_complete.apply(ClosestStop, axis=1)
奇怪的是,这个代码给出了不同的答案,这让我想到代码存在问题。 KM 的结果似乎也不对。
完全不知道如何解决这个问题 - 希望能在这里得到一些指导,谢谢!
distance(lat1, lon1, lat2, lon2)
,2. 对每个郊区和车站的组合应用该函数,3. 选择每个郊区距离最近的车站并将其添加到数据框中。(或者使用 sklearn 中的 NearestNeighbor 分类器) - Niklas Mertsch