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。

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>

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])

Vincenty距离从GeoPy 1.13版开始就被弃用了-你应该使用geo .distance.distance()来代替!

上面的答案是基于haversine公式，该公式假设地球是一个球体，结果误差高达0.5%(根据help(earth .distance))。Vincenty距离采用更精确的椭球模型，如WGS-84，并在地质学中实现。例如,

import geopy.distance

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

print geopy.distance.geodesic(coords_1, coords_2).km

将使用默认的椭球WGS-84打印279.352901604公里的距离。(你也可以选择。miles或其他距离单位。)

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)