另一种有趣的方法是通过Pyodide和WebAssembly实现混合JavaScript和Python，使用Python的库Pandas和geographiclib来获得解决方案，这也是可行的。

我用Pandas做了额外的工作来准备输入数据，当输出可用时，将它们附加到解决方案列中。Pandas为常见需求提供了许多有用的输入/输出特性。它的toHtml方法可以方便地在网页上呈现最终的解决方案。

我发现这个答案中的代码在某些iPhone和iPad设备上执行不成功。但在较新的中端Android设备上，它运行得很好。

async function main(){ let pyodide = await loadPyodide(); await pyodide.loadPackage(["pandas", "micropip"]); console.log(pyodide.runPythonAsync(` import micropip import pandas as pd import js print("Pandas version: " + pd.__version__) await micropip.install('geographiclib') from geographiclib.geodesic import Geodesic import geographiclib as gl print("Geographiclib version: " + gl.__version__) data = {'Description': ['Answer to the question', 'Bangkok to Tokyo'], 'From_long': [21.0122287, 100.6], 'From_lat': [52.2296756, 13.8], 'To_long': [16.9251681, 139.76], 'To_lat': [52.406374, 35.69], 'Distance_km': [0, 0]} df1 = pd.DataFrame(data) collist = ['Description','From_long','From_lat','To_long','To_lat'] div2 = js.document.createElement("div") div2content = df1.to_html(buf=None, columns=collist, col_space=None, header=True, index=True) div2.innerHTML = div2content js.document.body.append(div2) arr="<i>by Swatchai</i>" def dkm(frLat,frLon,toLat,toLon): print("frLon,frLat,toLon,toLat:", frLon, "|", frLat, "|", toLon, "|", toLat) dist = Geodesic.WGS84.Inverse(frLat, frLon, toLat, toLon) return dist["s12"] / 1000 collist = ['Description','From_long','From_lat','To_long','To_lat','Distance_km'] dist = [] for ea in zip(df1['From_lat'].values, df1['From_long'].values, df1['To_lat'].values, df1['To_long'].values): ans = dkm(*ea) print("ans=", ans) dist.append(ans) df1['Distance_km'] = dist # Update content div2content = df1.to_html(buf=None, columns=collist, col_space=None, header=True, index=False) div2.innerHTML = div2content js.document.body.append(div2) # Using the haversine formula from math import sin, cos, sqrt, atan2, radians, asin # Approximate radius of earth in km from Wikipedia R = 6371 lat1 = radians(52.2296756) lon1 = radians(21.0122287) lat2 = radians(52.406374) lon2 = radians(16.9251681) # https://en.wikipedia.org/wiki/Haversine_formula def hav(angrad): return (1-cos(angrad))/2 h = hav(lat2-lat1)+cos(lat2)*cos(lat1)*hav(lon2-lon1) dist2 = 2*R*asin(sqrt(h)) print(f"Distance by haversine formula = {dist2:8.6f} km.") `)); } main(); <script src="https://cdn.jsdelivr.net/pyodide/v0.21.0/full/pyodide.js"></script> Pyodide implementation<br>

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。

有多种方法来计算基于坐标的距离，即纬度和经度

安装和导入

from geopy import distance
from math import sin, cos, sqrt, atan2, radians
from sklearn.neighbors import DistanceMetric
import osrm
import numpy as np

定义坐标

lat1, lon1, lat2, lon2, R = 20.9467,72.9520, 21.1702, 72.8311, 6373.0
coordinates_from = [lat1, lon1]
coordinates_to = [lat2, lon2]

使用半正矢

dlon = radians(lon2) - radians(lon1)
dlat = radians(lat2) - radians(lat1)
    
a = sin(dlat / 2)**2 + cos(lat1) * cos(lat2) * sin(dlon / 2)**2
c = 2 * atan2(sqrt(a), sqrt(1 - a))
    
distance_haversine_formula = R * c
print('distance using haversine formula: ', distance_haversine_formula)

使用哈弗辛和sklearn

dist = DistanceMetric.get_metric('haversine')
    
X = [[radians(lat1), radians(lon1)], [radians(lat2), radians(lon2)]]
distance_sklearn = R * dist.pairwise(X)
print('distance using sklearn: ', np.array(distance_sklearn).item(1))

使用OSRM

osrm_client = osrm.Client(host='http://router.project-osrm.org')
coordinates_osrm = [[lon1, lat1], [lon2, lat2]] # note that order is lon, lat
    
osrm_response = osrm_client.route(coordinates=coordinates_osrm, overview=osrm.overview.full)
dist_osrm = osrm_response.get('routes')[0].get('distance')/1000 # in km
print('distance using OSRM: ', dist_osrm)

使用geopy

distance_geopy = distance.distance(coordinates_from, coordinates_to).km
print('distance using geopy: ', distance_geopy)
    
distance_geopy_great_circle = distance.great_circle(coordinates_from, coordinates_to).km 
print('distance using geopy great circle: ', distance_geopy_great_circle)

输出

distance using haversine formula:  26.07547017310917
distance using sklearn:  27.847882224769783
distance using OSRM:  33.091699999999996
distance using geopy:  27.7528030550408
distance using geopy great circle:  27.839182219511834

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)

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

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')