haversine 2.8.1

Calculate the distance between 2 points on Earth.

Haversine

Calculate the distance (in various units) between two points on Earth using their latitude and longitude.

Installation

pip install haversine

Usage

Calculate the distance between Lyon and Paris

from haversine import haversine, Unit

lyon = (45.7597, 4.8422) # (lat, lon)
paris = (48.8567, 2.3508)

haversine(lyon, paris)
>> 392.2172595594006  # in kilometers

haversine(lyon, paris, unit=Unit.MILES)
>> 243.71250609539814  # in miles

# you can also use the string abbreviation for units:
haversine(lyon, paris, unit='mi')
>> 243.71250609539814  # in miles

haversine(lyon, paris, unit=Unit.NAUTICAL_MILES)
>> 211.78037755311516  # in nautical miles

The lat/lon values need to be provided in degrees of the ranges [-90,90] (lat) and [-180,180] (lon). If values are outside their ranges, an error will be raised. This can be avoided by automatic normalization via the normalize parameter.

The haversine.Unit enum contains all supported units:

import haversine

print(tuple(haversine.Unit))

outputs

(<Unit.KILOMETERS: 'km'>, <Unit.METERS: 'm'>, <Unit.MILES: 'mi'>,
 <Unit.NAUTICAL_MILES: 'nmi'>, <Unit.FEET: 'ft'>, <Unit.INCHES: 'in'>,
 <Unit.RADIANS: 'rad'>, <Unit.DEGREES: 'deg'>)

Note for radians and degrees

The radian and degrees returns the great circle distance between two points on a sphere.

Notes:

• on a unit-sphere the angular distance in radians equals the distance between the two points on the sphere (definition of radians)
• When using "degree", this angle is just converted from radians to degrees

Inverse Haversine Formula

Calculates a point from a given vector (distance and direction) and start point. Currently explicitly supports both cardinal (north, east, south, west) and intercardinal (northeast, southeast, southwest, northwest) directions. But also allows for explicit angles expressed in Radians.

Example: Finding arbitary point from Paris

from haversine import inverse_haversine, Direction
from math import pi
paris = (48.8567, 2.3508) # (lat, lon)
# Finding 32 km west of Paris
inverse_haversine(paris, 32, Direction.WEST)
# returns tuple (48.85587279023947, 1.9134085092836945)
# Finding 32 km southwest of Paris
inverse_haversine(paris, 32, pi * 1.25)
# returns tuple (48.65279552300661, 2.0427666779658806)
# Finding 50 miles north of Paris
inverse_haversine(paris, 50, Direction.NORTH, unit=Unit.MILES)
# returns tuple (49.58035791599536, 2.3508)
# Finding 10 nautical miles south of Paris
inverse_haversine(paris, 10, Direction.SOUTH, unit=Unit.NAUTICAL_MILES)
# returns tuple (48.690145868497645, 2.3508)

Performance optimisation for distances between all points in two vectors

You will need to install numpy in order to gain performance with vectors. For optimal performance, you can turn off coordinate checking by adding check=False and install the optional packages numba and icc_rt.

You can then do this:

from haversine import haversine_vector, Unit

lyon = (45.7597, 4.8422) # (lat, lon)
paris = (48.8567, 2.3508)
new_york = (40.7033962, -74.2351462)

haversine_vector([lyon, lyon], [paris, new_york], Unit.KILOMETERS)

>> array([ 392.21725956, 6163.43638211])

It is generally slower to use haversine_vector to get distance between two points, but can be really fast to compare distances between two vectors.

Combine matrix

You can generate a matrix of all combinations between coordinates in different vectors by setting comb parameter as True.

from haversine import haversine_vector, Unit

lyon = (45.7597, 4.8422) # (lat, lon)
london = (51.509865, -0.118092)
paris = (48.8567, 2.3508)
new_york = (40.7033962, -74.2351462)

haversine_vector([lyon, london], [paris, new_york], Unit.KILOMETERS, comb=True)

>> array([[ 392.21725956,  343.37455271],
 	  [6163.43638211, 5586.48447423]])

The output array from the example above returns the following table:

	Paris	New York
Lyon	Lyon <-> Paris	Lyon <-> New York
London	London <-> Paris	London <-> New York

By definition, if you have a vector a with n elements, and a vector b with m elements. The result matrix M would be $n x m$ and a element M[i,j] from the matrix would be the distance between the ith coordinate from vector a and jth coordinate with vector b.

Contributing

Clone the project.

Install pipenv.

Run pipenv install --dev

Launch test with pipenv run pytest