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I have a list of stations in List 1 and List 2 How do I find the closest stations to List 1 from List 2 ?

list 1 and 2

List 1

SS No Latitude Longtitude 977 23.141747 53.796469 946 23.398398 55.422916 742 23.615732 53.717952 980 23.633077 55.567046 660 23.6504 54.4007

List 2

SS No Latitude Longtitude 962 23.657571 53.703683 745 23.671971 52.955976 743 23.766849 53.770344 978 23.847163 52.809653 748 23.942166 52.16236 744 23.955817 52.790424 760 23.984592 55.55764 945 24.030256 55.844842 894 24.03511 53.891547 856 24.741601 55.80063 893 24.04123 53.899958 387 24.059988 51.748138 675 24.061578 53.417912 664 24.063978 51.76195

I can do this manually by mapping them on PowerBI but I am looking for a scalable solution and Prefer Python.

Raul V
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2 Answers2

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This is quite similar to the previous question
Getting distance between two points based on latitude/longitude
so, it could be claimed as duplicate.
Anyway, following Kurt Peek's answer, you could do:

import geopy.distance

def get_distnace(coords_1, coords_2):
    return geopy.distance.vincenty(coords_1, coords_2).km

to get a function that returns the distance in 'km'. Then, given a couple of lists like yours, in the form

list1 = [[stat_name_1, lat_1, lon_1], [stat_name_1, lat_1, lon_1], ... ]

list2 = [[...], ... ]

you could probably do:

min_stat = get_distance(list1[0][1:], list2[0][1:])
for stat_1 in list1:
    coord_1 = stat_1[1:]
    for stat_2 in list2:
        coord_2 = stat_2[1:]
        min_stat = min(min_stat, get_distance(coord_1, coord_2)

Take it as a draft, an idea, to be debugged and tested before any application.

Francesco Pinna
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geopy is your friend. It has functions that calculate the distance between coordinate pairs. Here's one approach:

from geopy import distance

s = {
    977: (23.141747, 53.796469),
    946: (23.398398, 55.422916),
    # etc etc
}

d = {
    962: (23.657571, 53.703683),
    745: (23.671971, 52.955976),
    743: (23.766849, 53.770344),
    # etc etc
}

for (ss, a) in s.items():
    best = None
    dist = None
    for (dd, b) in d.items():
        km = distance.distance(a, b).km
        if dist is None or km < dist:
            best = dd
            dist = km

    print(f'{ss} is nearest {best}: {dist} km')

If I run it with your example data I get:

977 is nearest 962: 57.909274476145846 km
946 is nearest 760: 66.3613771288571 km
742 is nearest 962: 4.857141728990575 km
980 is nearest 760: 38.94400553289674 km
660 is nearest 743: 65.56437761273963 km

If you need the distances from each of list 1 to each of list 2, you could try

sorted([[ss, dd, distance.distance(a, b).km] for (ss, a) in s.items()
        for (dd, b) in d.items()])

This gives a list of lists, ordered by the list 1 item first:

[[660, 387, 273.98088337893], 
 [660, 664, 272.6633222300461], 
 [660, 675, 109.98235440892797], 
 [660, 743, 65.56437761273963],

 # etc

Further ordering or grouping left as an exercice for the reader.

xpqz
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