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

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
    lat1,lon1,lat2,lon2 = map(np.radians, [lat1,lon1,lat2,lon2])

    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)

其他回答

(2022年,JavaScript版本)下面是使用最新的JavaScript库解决这个问题的代码。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。

// Using the WGS84 ellipsoid model for computation var geod84 = geodesic.Geodesic.WGS84; // Input data lat1 = 52.2296756; lon1 = 21.0122287; lat2 = 52.406374; lon2 = 16.9251681; // Do the classic `geodetic inversion` computation geod84inv = geod84.Inverse(lat1, lon1, lat2, lon2); // Present the solution (only the geodetic distance) console.log("The distance is " + (geod84inv.s12/1000).toFixed(5) + " km."); <script type="text/javascript" src="https://cdn.jsdelivr.net/npm/geographiclib-geodesic@2.0.0/geographiclib-geodesic.min.js"> </script>

注意,如果你只是需要一种快速简单的方法来找到两点之间的距离,我强烈建议使用Kurt回答中描述的方法,而不是重新实现haversine,查看他的帖子来了解基本原理。

这个答案只关注OP遇到的特定bug。


这是因为在Python中,所有的三角函数都使用弧度,而不是度。

您可以手动将数字转换为弧度,或使用数学模块中的radians函数:

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

# Approximate radius of earth in km
R = 6373.0

lat1 = radians(52.2296756)
lon1 = radians(21.0122287)
lat2 = radians(52.406374)
lon2 = radians(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, "km")

距离现在返回正确的值278.545589351 km。

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
    lat1,lon1,lat2,lon2 = map(np.radians, [lat1,lon1,lat2,lon2])

    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)

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>

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')