Python location, show distance from closest other location

A few key concepts

  1. do a Cartesian product between two data frames to get all combinations (joining on identical value between two data frames is approach to this foo=1)
  2. once both sets of data is together, have both sets of lat/lon to calculate distance) geopy has been used for this
  3. cleanup the columns, use sort_values() to find smallest distance
  4. finally a groupby() and agg() to get first values for shortest distance

There are two data frames for use

  1. dfdist contains all the combinations and distances
  2. dfnearest which contains result
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

output

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

an approach to reducing size of tested combinations

# 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)
)

You can use sklearn.neighbors.NearestNeighbors with a haversine distance.

import pandas as pd
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']})

Let's begin by finding the closest point in a dataframe to some random point, say -37.814563, 144.970267.

NN = NearestNeighbors(n_neighbors=1, metric='haversine')
NN.fit(dfstat[['LATITUDE', 'LONGITUDE']])
NN.kneighbors([[-37.814563, 144.970267]])

The output is (array([[2.55952637]]), array([[3]])), the distance and the index of the closest point in the dataframe. The haversine distance in sklearn is in radius. If you want to compute is in km, you can use haversine.

from haversine import haversine
NN = NearestNeighbors(n_neighbors=1, metric=haversine)
NN.fit(dfstat[['LATITUDE', 'LONGITUDE']])
NN.kneighbors([[-37.814563, 144.970267]])

The output (array([[2.55952637]]), array([[3]])) has the distance in km.

Now you can apply to all points in the dataframe and get closest stations with indices.

indices = NN.kneighbors(dfsub[['lat', 'lon']])[1]
indices = [index[0] for index in indices]
distances = NN.kneighbors(dfsub[['lat', 'lon']])[0]
dfsub['closest_station'] = dfstat.iloc[indices]['STOP_NAME'].reset_index(drop=True)
dfsub['closest_station_distances'] = distances
print(dfsub)
id  postcode    suburb  state   lat lon closest_station closest_station_distances
0   4901    3000    MELBOURNE   VIC -37.814563  144.970267  North Melbourne Railway Station (West Melbourne)    2.559526
1   4902    3002    EAST MELBOURNE  VIC -37.816640  144.987811  Clifton Hill Railway Station (Clifton Hill) 3.182521
2   4903    3003    WEST MELBOURNE  VIC -37.806255  144.941123  North Melbourne Railway Station (West Melbourne)    0.181419
3   4904    3005    WORLD TRADE CENTRE  VIC -37.822262  144.954856  North Melbourne Railway Station (West Melbourne)    1.972010
4   4905    3006    SOUTHBANK   VIC -37.823258  144.965926  North Melbourne Railway Station (West Melbourne)    2.703926

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