## 问题：根据纬度/经度获取两点之间的距离

``````from math import sin, cos, sqrt, atan2

R = 6373.0

lat1 = 52.2296756
lon1 = 21.0122287
lat2 = 52.406374
lon2 = 16.9251681

dlon = lon2 - lon1
dlat = lat2 - lat1
a = (sin(dlat/2))**2 + cos(lat1) * cos(lat2) * (sin(dlon/2))**2
c = 2 * atan2(sqrt(a), sqrt(1-a))
distance = R * c

print "Result", distance
print "Should be", 278.546``````

I tried implementing this formula: http://andrew.hedges.name/experiments/haversine/ The aplet does good for the two points I am testing:

Yet my code is not working.

``````from math import sin, cos, sqrt, atan2

R = 6373.0

lat1 = 52.2296756
lon1 = 21.0122287
lat2 = 52.406374
lon2 = 16.9251681

dlon = lon2 - lon1
dlat = lat2 - lat1
a = (sin(dlat/2))**2 + cos(lat1) * cos(lat2) * (sin(dlon/2))**2
c = 2 * atan2(sqrt(a), sqrt(1-a))
distance = R * c

print "Result", distance
print "Should be", 278.546
``````

The distance it returns is 5447.05546147. Why?

## 回答 0

``````from math import sin, cos, sqrt, atan2, radians

# approximate radius of earth in km
R = 6373.0

dlon = lon2 - lon1
dlat = lat2 - lat1

a = sin(dlat / 2)**2 + cos(lat1) * cos(lat2) * sin(dlon / 2)**2
c = 2 * atan2(sqrt(a), sqrt(1 - a))

distance = R * c

print("Result:", distance)
print("Should be:", 278.546, "km")``````

Edit: Just as a note, if you just need a quick and easy way of finding the distance between two points, I strongly recommend using the approach described in Kurt’s answer below instead of re-implementing Haversine — see his post for rationale.

This answer focuses just on answering the specific bug OP ran into.

It’s because in Python, all the trig functions use radians, not degrees.

You can either convert the numbers manually to radians, or use the `radians` function from the math module:

``````from math import sin, cos, sqrt, atan2, radians

# approximate radius of earth in km
R = 6373.0

dlon = lon2 - lon1
dlat = lat2 - lat1

a = sin(dlat / 2)**2 + cos(lat1) * cos(lat2) * sin(dlon / 2)**2
c = 2 * atan2(sqrt(a), sqrt(1 - a))

distance = R * c

print("Result:", distance)
print("Should be:", 278.546, "km")
``````

The distance is now returning the correct value of `278.545589351` km.

## 回答 1

``````import geopy.distance

coords_1 = (52.2296756, 21.0122287)
coords_2 = (52.406374, 16.9251681)

print geopy.distance.vincenty(coords_1, coords_2).km``````

`279.352901604`使用默认的椭球WGS-84 打印公里的距离。（您也可以选择`.miles`或选择其他几种距离单位之一）。

Update: 04/2018: Note that Vincenty distance is deprecated since GeoPy version 1.13 – you should use geopy.distance.distance() instead!

The answers above are based on the Haversine formula, which assumes the earth is a sphere, which results in errors of up to about 0.5% (according to `help(geopy.distance)`). Vincenty distance uses more accurate ellipsoidal models such as WGS-84, and is implemented in geopy. For example,

``````import geopy.distance

coords_1 = (52.2296756, 21.0122287)
coords_2 = (52.406374, 16.9251681)

print geopy.distance.vincenty(coords_1, coords_2).km
``````

will print the distance of `279.352901604` kilometers using the default ellipsoid WGS-84. (You can also choose `.miles` or one of several other distance units).

## 回答 2

``````import mpu

# Point one
lat1 = 52.2296756
lon1 = 21.0122287

# Point two
lat2 = 52.406374
lon2 = 16.9251681

# What you were looking for
dist = mpu.haversine_distance((lat1, lon1), (lat2, lon2))
print(dist)  # gives 278.45817507541943.``````

``````import math

def distance(origin, destination):
"""
Calculate the Haversine distance.

Parameters
----------
origin : tuple of float
(lat, long)
destination : tuple of float
(lat, long)

Returns
-------
distance_in_km : float

Examples
--------
>>> origin = (48.1372, 11.5756)  # Munich
>>> destination = (52.5186, 13.4083)  # Berlin
>>> round(distance(origin, destination), 1)
504.2
"""
lat1, lon1 = origin
lat2, lon2 = destination

a = (math.sin(dlat / 2) * math.sin(dlat / 2) +
math.sin(dlon / 2) * math.sin(dlon / 2))
c = 2 * math.atan2(math.sqrt(a), math.sqrt(1 - a))

return d

if __name__ == '__main__':
import doctest
doctest.testmod()``````

``````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)
>> 243.71201856934454  # in miles

# you can also use the string abbreviation for units:
haversine(lyon, paris, unit='mi')
>> 243.71201856934454  # in miles

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

``````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])``````

For people (like me) coming here via search engine and just looking for a solution which works out of the box, I recommend installing . Install it via `pip install mpu --user` and use it like this to get the haversine distance:

``````import mpu

# Point one
lat1 = 52.2296756
lon1 = 21.0122287

# Point two
lat2 = 52.406374
lon2 = 16.9251681

# What you were looking for
dist = mpu.haversine_distance((lat1, lon1), (lat2, lon2))
print(dist)  # gives 278.45817507541943.
``````

An alternative package is `gpxpy`.

If you don’t want dependencies, you can use:

``````import math

def distance(origin, destination):
"""
Calculate the Haversine distance.

Parameters
----------
origin : tuple of float
(lat, long)
destination : tuple of float
(lat, long)

Returns
-------
distance_in_km : float

Examples
--------
>>> origin = (48.1372, 11.5756)  # Munich
>>> destination = (52.5186, 13.4083)  # Berlin
>>> round(distance(origin, destination), 1)
504.2
"""
lat1, lon1 = origin
lat2, lon2 = destination

a = (math.sin(dlat / 2) * math.sin(dlat / 2) +
math.sin(dlon / 2) * math.sin(dlon / 2))
c = 2 * math.atan2(math.sqrt(a), math.sqrt(1 - a))

return d

if __name__ == '__main__':
import doctest
doctest.testmod()
``````

The other alternative package is `[haversine][1]`

``````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)
>> 243.71201856934454  # in miles

# you can also use the string abbreviation for units:
haversine(lyon, paris, unit='mi')
>> 243.71201856934454  # in miles

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

They claim to have performance optimization for distances between all points in two vectors

``````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])
``````

## 回答 3

``````from geopy.distance import geodesic

origin = (30.172705, 31.526725)  # (latitude, longitude) don't confuse
dist = (30.288281, 31.732326)

print(geodesic(origin, dist).meters)  # 23576.805481751613
print(geodesic(origin, dist).kilometers)  # 23.576805481751613
print(geodesic(origin, dist).miles)  # 14.64994773134371``````

I arrived at a much simpler and robust solution which is using `geodesic` from `geopy` package since you’ll be highly likely using it in your project anyways so no extra package installation needed.

Here is my solution:

``````from geopy.distance import geodesic

origin = (30.172705, 31.526725)  # (latitude, longitude) don't confuse
dist = (30.288281, 31.732326)

print(geodesic(origin, dist).meters)  # 23576.805481751613
print(geodesic(origin, dist).kilometers)  # 23.576805481751613
print(geodesic(origin, dist).miles)  # 14.64994773134371
``````

geopy

## 回答 4

``````import numpy as np

def Haversine(lat1,lon1,lat2,lon2, **kwarg):
"""
This uses the ‘haversine’ formula to calculate the great-circle distance between two points – that is,
the shortest distance over the earth’s surface – giving an ‘as-the-crow-flies’ distance between the points
(ignoring any hills they fly over, of course!).
Haversine
formula:    a = sin²(Δφ/2) + cos φ1 ⋅ cos φ2 ⋅ sin²(Δλ/2)
c = 2 ⋅ atan2( √a, √(1−a) )
d = R ⋅ c
where   φ is latitude, λ is longitude, R is earth’s radius (mean radius = 6,371km);
note that angles need to be in radians to pass to trig functions!
"""
R = 6371.0088

dlat = lat2 - lat1
dlon = lon2 - lon1
a = np.sin(dlat/2)**2 + np.cos(lat1) * np.cos(lat2) * np.sin(dlon/2) **2
c = 2 * np.arctan2(a**0.5, (1-a)**0.5)
d = R * c
return round(d,4)``````
``````import numpy as np

def Haversine(lat1,lon1,lat2,lon2, **kwarg):
"""
This uses the ‘haversine’ formula to calculate the great-circle distance between two points – that is,
the shortest distance over the earth’s surface – giving an ‘as-the-crow-flies’ distance between the points
(ignoring any hills they fly over, of course!).
Haversine
formula:    a = sin²(Δφ/2) + cos φ1 ⋅ cos φ2 ⋅ sin²(Δλ/2)
c = 2 ⋅ atan2( √a, √(1−a) )
d = R ⋅ c
where   φ is latitude, λ is longitude, R is earth’s radius (mean radius = 6,371km);
note that angles need to be in radians to pass to trig functions!
"""
R = 6371.0088