Несколько ключевых концепций
- делают декартово произведение между двумя кадрами данных, чтобы получить все комбинации (к этому подходит объединение по одинаковому значению между двумя кадрами данных
foo=1) - как только оба набора данных будут вместе, используйте оба набора широты / долготы для расчета расстояния) геопия была использована для этого
- очистка столбцов, используйте
sort_values(), чтобы найти наименьшее расстояние - наконец
groupby() и agg(), чтобы получить первые значения для кратчайшего расстояния
Есть два фрейма данных для использования
dfdist содержит все комбинации и расстояния dfnearest который содержит результат
dfstat = pd.DataFrame({'STOP_ID': ['19970', '19971', '19972', '19973', '19974'],
'STOP_NAME': ['Royal Park Railway Station (Parkville)',
'Flemington Bridge Railway Station (North Melbo...',
'Macaulay Railway Station (North Melbourne)',
'North Melbourne Railway Station (West Melbourne)',
'Clifton Hill Railway Station (Clifton Hill)'],
'LATITUDE': ['-37.781193',
'-37.788140',
'-37.794267',
'-37.807419',
'-37.788657'],
'LONGITUDE': ['144.952301',
'144.939323',
'144.936166',
'144.942570',
'144.995417'],
'TICKETZONE': ['1', '1', '1', '1', '1'],
'ROUTEUSSP': ['Upfield',
'Upfield',
'Upfield',
'Flemington,Sunbury,Upfield,Werribee,Williamsto...',
'Mernda,Hurstbridge'],
'geometry': ['POINT (144.95230 -37.78119)',
'POINT (144.93932 -37.78814)',
'POINT (144.93617 -37.79427)',
'POINT (144.94257 -37.80742)',
'POINT (144.99542 -37.78866)']})
dfsub = pd.DataFrame({'id': ['4901', '4902', '4903', '4904', '4905'],
'postcode': ['3000', '3002', '3003', '3005', '3006'],
'suburb': ['MELBOURNE',
'EAST MELBOURNE',
'WEST MELBOURNE',
'WORLD TRADE CENTRE',
'SOUTHBANK'],
'state': ['VIC', 'VIC', 'VIC', 'VIC', 'VIC'],
'lat': ['-37.814563', '-37.816640', '-37.806255', '-37.822262', '-37.823258'],
'lon': ['144.970267', '144.987811', '144.941123', '144.954856', '144.965926']})
import geopy.distance
# cartesian product so we get all combinations
dfdist = (dfsub.assign(foo=1).merge(dfstat.assign(foo=1), on="foo")
# calc distance in km between each suburb and each train station
.assign(km=lambda dfa: dfa.apply(lambda r:
geopy.distance.geodesic(
(r["LATITUDE"],r["LONGITUDE"]),
(r["lat"],r["lon"])).km, axis=1))
# reduce number of columns to make it more digestable
.loc[:,["postcode","suburb","STOP_ID","STOP_NAME","km"]]
# sort so shortest distance station from a suburb is first
.sort_values(["postcode","suburb","km"])
# good practice
.reset_index(drop=True)
)
# finally pick out stations nearest to suburb
# this can easily be joined back to source data frames as postcode and STOP_ID have been maintained
dfnearest = dfdist.groupby(["postcode","suburb"])\
.agg({"STOP_ID":"first","STOP_NAME":"first","km":"first"}).reset_index()
print(dfnearest.to_string(index=False))
dfnearest
вывод
postcode suburb STOP_ID STOP_NAME km
3000 MELBOURNE 19973 North Melbourne Railway Station (West Melbourne) 2.564586
3002 EAST MELBOURNE 19974 Clifton Hill Railway Station (Clifton Hill) 3.177320
3003 WEST MELBOURNE 19973 North Melbourne Railway Station (West Melbourne) 0.181463
3005 WORLD TRADE CENTRE 19973 North Melbourne Railway Station (West Melbourne) 1.970909
3006 SOUTHBANK 19973 North Melbourne Railway Station (West Melbourne) 2.705553
подход к уменьшению размера тестируемых комбинаций
# pick nearer places, based on lon/lat then all combinations
dfdist = (dfsub.assign(foo=1, latr=dfsub["lat"].round(1), lonr=dfsub["lon"].round(1))
.merge(dfstat.assign(foo=1, latr=dfstat["LATITUDE"].round(1), lonr=dfstat["LONGITUDE"].round(1)),
on=["foo","latr","lonr"])
# calc distance in km between each suburb and each train station
.assign(km=lambda dfa: dfa.apply(lambda r:
geopy.distance.geodesic(
(r["LATITUDE"],r["LONGITUDE"]),
(r["lat"],r["lon"])).km, axis=1))
# reduce number of columns to make it more digestable
.loc[:,["postcode","suburb","STOP_ID","STOP_NAME","km"]]
# sort so shortest distance station from a suburb is first
.sort_values(["postcode","suburb","km"])
# good practice
.reset_index(drop=True)
)