Finding distance between two points with GeoSeries.distance

Question

When I want the distance between two points [(117.454361,38.8459879),(117.459880 ,38.846255)] (longitude,latitude) on the earth, I take the GeoSeries.distance method, but the method does not give me the right answer.

The real distance is about 479 meters, however the method give the result far from it, why?

import geopandas as gpd
from shapely.geometry import Point
geom = [Point(xy) for xy in zip([117.454361,117.459880],[38.8459879,38.846255])]
gdf = gpd.GeoDataFrame(geometry=geom, crs={'init':'epsg:4326'})
dis = gdf.distance(gdf.shift())
print(dis)

score 15 · Answer 1 · edited Aug 22 '23 at 06:17

15

import geopandas as gpd
from shapely.geometry import Point

geom = [Point(xy) for xy in zip([117.454361,117.459880], [38.8459879,38.846255])]
gdf = gpd.GeoDataFrame(geometry=geom, crs={'init':'epsg:4326'})
gdf.to_crs(epsg=3310, inplace=True)
l = gdf.distance(gdf.shift())
print(l)

The result is 479.450134 meters.

edited Aug 22 '23 at 06:17

Taras

32,823
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answered Aug 20 '18 at 11:35

CJ Xu

331
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5

What is crs 3310? Why are you using it? – larsks Jun 03 '19 at 14:33
2

I believe it's a conversion to UTM, a rectilinear projection coordinate system where distance can be calculated with pythagorean theorem. But I don't know why it's EPSG 3310 in particular. Is that a general solution for anywhere on Earth? – user2561747 Jul 31 '19 at 20:43
EPSG 32663 will also be a good candidate (WGS 84 / World Equidistant Cylindrical) https://spatialreference.org/ref/epsg/32663/ – Adrian Tofting Aug 11 '20 at 11:42

score 13 · Accepted Answer · answered Aug 20 '18 at 07:13

This is a distance in degrees, the coordinates of your data. I can get this using Pythagoras' theorem from your coordinates:

>>> a = [117.454361, 38.8459879]
>>> b = [117.45988, 38.846255]
>>> math.sqrt((a[0]-b[0])**2 + (a[1]-b[1])**2)
0.005525459565494833

to get a distance in metres, transform to a metric coordinate system, such as the UTM zone for your data.

Possible help here:

Calculate distance between a coordinate and a county in GeoPandas

score 9 · Answer 3 · answered Apr 07 '20 at 23:58

9

Its pretty easy with Geopy

import geopy.distance
dist = geopy.distance.geodesic((38.8459879,117.454361),(38.846255,117.459880))
dist.meters

answered Apr 07 '20 at 23:58

virtuvious

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Are the coordinates indise (lat, lon) or (lon, lat). Thanks! – M.O. Apr 22 '20 at 03:59
1

its (lat, long) – virtuvious Apr 22 '20 at 14:47

score 6 · Answer 4 · answered Dec 27 '19 at 23:41

Pythagoras only works on a flat plane and not an sphere. The distance between two points on the surface of a sphere is found using great-circle distance:

where φ's are latitude and λ's are longitudes. To convert the distance to meter you need to know the radius of the sphere (6371km for Earth) and multiply it by Δσ in radians. Here is a code that does that:

def haversine(coord1, coord2):
    import math
    # Coordinates in decimal degrees (e.g. 2.89078, 12.79797)
    lon1, lat1 = coord1
    lon2, lat2 = coord2
    R = 6371000  # radius of Earth in meters
    phi_1 = math.radians(lat1)
    phi_2 = math.radians(lat2)

    delta_phi = math.radians(lat2 - lat1)
    delta_lambda = math.radians(lon2 - lon1)

    a = math.sin(delta_phi / 2.0) ** 2 + math.cos(phi_1) * math.cos(phi_2) * math.sin(delta_lambda / 2.0) ** 2

    c = 2 * math.atan2(math.sqrt(a), math.sqrt(1 - a))

    meters = R * c  # output distance in meters
    km = meters / 1000.0  # output distance in kilometers

    meters = round(meters)
    km = round(km, 3)
    print(f"Distance: {meters} m")
    print(f"Distance: {km} km")

and run it for the coordinates you get 479m.

haversine(coord1= (117.454361,38.8459879), coord2= (117.459880,38.846255))

The other way to this is to do what @CJ Xu did and convert the coordinates to UTM.

There is a library to do that in a vectorized way (with the possibility of using numba optimizations) :

https://github.com/mapado/haversine — Adrien Pacifico, May 08 '23 at 22:26

score 3 · Answer 5 · answered Aug 21 '23 at 18:40

It's a little unfortunate that this isn't built into GeoPandas. But you can ask Pyproj to do the geodesic calculations for you:

import geopandas as gpd
import pyproj
df = gpd.GeoDataFrame({
    "lon": [117.454361, 117.459880, 117.462374],
    "lat": [38.8459879, 38.846255, 38.847381],
})
df["pt"] = gpd.points_from_xy(df["lon"], df["lat"], crs="epsg:4326")
df.set_geometry("pt", inplace=True)
p1 = df["pt"].iloc[1:]
p2 = df["pt"].shift().dropna()
geod = df.crs.get_geod()
dist = geod.inv(
    p1.x.to_numpy(),
    p1.y.to_numpy(),
    p2.x.to_numpy(),
    p2.y.to_numpy(),
)[2]
print(dist)

Result:

[480.04157936 250.0043709 ]

score 1 · Answer 6 · answered Aug 21 '23 at 23:18

Using only pyproj, calculate the length of a geodesic from lat/long coords:

from pyproj import Geod
geod = Geod(ellps="WGS84")
lons = [117.454361, 117.459880]
lats = [38.8459879, 38.846255]
total_length = geod.line_length(lons, lats)
print(f"total_length = {total_length:.3f} m")
total_length = 480.042 m

Finding distance between two points with GeoSeries.distance

6 Answers6

total_length = 480.042 m

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