我尝试在基于经纬度查找距离中实现公式。applet对我正在测试的两点很好:
但是我的代码没有工作。
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
它返回距离5447.05546147。为什么?
当前回答
对于像我这样通过搜索引擎来到这里的人，以及正在寻找开箱即用的解决方案的人，我建议安装mpu。通过pip Install mpu——user安装它，然后像这样使用它来获得haversine距离:
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.
另一个包是gpxpy。
如果你不想要依赖，你可以使用:
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
radius = 6371 # km
dlat = math.radians(lat2 - lat1)
dlon = math.radians(lon2 - lon1)
a = (math.sin(dlat / 2) * math.sin(dlat / 2) +
math.cos(math.radians(lat1)) * math.cos(math.radians(lat2)) *
math.sin(dlon / 2) * math.sin(dlon / 2))
c = 2 * math.atan2(math.sqrt(a), math.sqrt(1 - a))
d = radius * c
return d
if __name__ == '__main__':
import doctest
doctest.testmod()
另一种替代方案是haversine:
from haversine import haversine, Unit
lyon = (45.7597, 4.8422) # (latitude, longitude)
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
他们声称对两个向量中所有点之间的距离进行了性能优化:
from haversine import haversine_vector, Unit
lyon = (45.7597, 4.8422) # (latitude, longitude)
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])
其他回答
最简单的方法是用哈弗辛包装。
import haversine as hs
coord_1 = (lat, lon)
coord_2 = (lat, lon)
x = hs.haversine(coord_1, coord_2)
print(f'The distance is {x} km')
在2022年，人们可以发布JavaScript和Python混合代码，使用最新的Python库，即地理库来解决这个问题。总的好处是，用户可以在运行在现代设备上的web页面上看到结果。
async function main(){ let pyodide = await loadPyodide(); await pyodide.loadPackage(["micropip"]); console.log(pyodide.runPythonAsync(` import micropip await micropip.install('geographiclib') from geographiclib.geodesic import Geodesic lat1 = 52.2296756; lon1 = 21.0122287; lat2 = 52.406374; lon2 = 16.9251681; ans = Geodesic.WGS84.Inverse(lat1, lon1, lat2, lon2) dkm = ans["s12"] / 1000 print("Geodesic solution", ans) print(f"Distance = {dkm:.4f} km.") `)); } main(); <script src="https://cdn.jsdelivr.net/pyodide/v0.21.0/full/pyodide.js"></script>
您可以使用Uber的H3,point_dist()函数来计算两个(纬度，经度)点之间的球面距离。我们可以设置返回单位('km'、'm'或'rads')。默认单位为km。
例子:
import h3
coords_1 = (52.2296756, 21.0122287)
coords_2 = (52.406374, 16.9251681)
distance = h3.point_dist(coords_1, coords_2, unit='m') # To get distance in meters
对于像我这样通过搜索引擎来到这里的人，以及正在寻找开箱即用的解决方案的人，我建议安装mpu。通过pip Install mpu——user安装它，然后像这样使用它来获得haversine距离:
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.
另一个包是gpxpy。
如果你不想要依赖，你可以使用:
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
radius = 6371 # km
dlat = math.radians(lat2 - lat1)
dlon = math.radians(lon2 - lon1)
a = (math.sin(dlat / 2) * math.sin(dlat / 2) +
math.cos(math.radians(lat1)) * math.cos(math.radians(lat2)) *
math.sin(dlon / 2) * math.sin(dlon / 2))
c = 2 * math.atan2(math.sqrt(a), math.sqrt(1 - a))
d = radius * c
return d
if __name__ == '__main__':
import doctest
doctest.testmod()
另一种替代方案是haversine:
from haversine import haversine, Unit
lyon = (45.7597, 4.8422) # (latitude, longitude)
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
他们声称对两个向量中所有点之间的距离进行了性能优化:
from haversine import haversine_vector, Unit
lyon = (45.7597, 4.8422) # (latitude, longitude)
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])
我找到了一个更简单、更健壮的解决方案，即使用geogeoy包中的测地线，因为你很可能在你的项目中使用它，所以不需要额外的包安装。
以下是我的解决方案:
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
地珞
