Python中有哪些工具可用于实现较大的圆距+线创建？

我需要使用Python创建一个很大的圆周距离-既是一个数字，也最好是一种我可以用来在客户端地图中绘制的“曲线”。我不在乎曲线的格式-无论是WKT还是一组坐标对-只是想获取数据。

那里有什么工具？我应该使用什么？

以下是一些可能有用的链接：

http://www.koders.com/python/fid0A930D7924AE856342437CA1F5A9A3EC0CAEACE2.aspx http://code.activestate.com/recipes/393241-calculating-the-distance-between-zip-codes/

其他人提供的答案则稍微优雅一些​​，但这是一个超简单的，有点怪异的Python，提供了基础知识。该函数采用两个坐标对和用户指定的段数。它沿着大圆路径产生了一组中间点。输出：准备写为KML的文本。注意事项：该代码不考虑对映体，并假设其为球形。

Alan Glennon的代码，http： //enj.com，2010年7月（作者将该代码放置在公共领域。使用时需您自担风险）。

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def tweensegs（经度1，纬度1，经度2，纬度2，num_of_segments）：

import math

ptlon1 = longitude1
ptlat1 = latitude1
ptlon2 = longitude2
ptlat2 = latitude2

numberofsegments = num_of_segments
onelessthansegments = numberofsegments - 1
fractionalincrement = (1.0/onelessthansegments)

ptlon1_radians = math.radians(ptlon1)
ptlat1_radians = math.radians(ptlat1)
ptlon2_radians = math.radians(ptlon2)
ptlat2_radians = math.radians(ptlat2)

distance_radians=2*math.asin(math.sqrt(math.pow((math.sin((ptlat1_radians-ptlat2_radians)/2)),2) + math.cos(ptlat1_radians)*math.cos(ptlat2_radians)*math.pow((math.sin((ptlon1_radians-ptlon2_radians)/2)),2)))
# 6371.009 represents the mean radius of the earth
# shortest path distance
distance_km = 6371.009 * distance_radians

mylats = []
mylons = []

# write the starting coordinates
mylats.append([])
mylons.append([])
mylats[0] = ptlat1
mylons[0] = ptlon1 

f = fractionalincrement
icounter = 1
while (icounter <  onelessthansegments):
        icountmin1 = icounter - 1
        mylats.append([])
        mylons.append([])
        # f is expressed as a fraction along the route from point 1 to point 2
        A=math.sin((1-f)*distance_radians)/math.sin(distance_radians)
        B=math.sin(f*distance_radians)/math.sin(distance_radians)
        x = A*math.cos(ptlat1_radians)*math.cos(ptlon1_radians) + B*math.cos(ptlat2_radians)*math.cos(ptlon2_radians)
        y = A*math.cos(ptlat1_radians)*math.sin(ptlon1_radians) +  B*math.cos(ptlat2_radians)*math.sin(ptlon2_radians)
        z = A*math.sin(ptlat1_radians) + B*math.sin(ptlat2_radians)
        newlat=math.atan2(z,math.sqrt(math.pow(x,2)+math.pow(y,2)))
        newlon=math.atan2(y,x)
        newlat_degrees = math.degrees(newlat)
        newlon_degrees = math.degrees(newlon)
        mylats[icounter] = newlat_degrees
        mylons[icounter] = newlon_degrees
        icounter += 1
        f = f + fractionalincrement

# write the ending coordinates
mylats.append([])
mylons.append([])
mylats[onelessthansegments] = ptlat2
mylons[onelessthansegments] = ptlon2

# Now, the array mylats[] and mylons[] have the coordinate pairs for intermediate points along the geodesic
# My mylat[0],mylat[0] and mylat[num_of_segments-1],mylat[num_of_segments-1] are the geodesic end points

# write a kml of the results
zipcounter = 0
kmlheader = "<?xml version=\"1.0\" encoding=\"UTF-8\"?><kml xmlns=\"http://www.opengis.net/kml/2.2\"><Document><name>LineString.kml</name><open>1</open><Placemark><name>unextruded</name><LineString><extrude>1</extrude><tessellate>1</tessellate><coordinates>"
print kmlheader
while (zipcounter < numberofsegments):
        outputstuff = repr(mylons[zipcounter]) + "," + repr(mylats[zipcounter]) + ",0 "
        print outputstuff
        zipcounter += 1
kmlfooter = "</coordinates></LineString></Placemark></Document></kml>"
print kmlfooter

GeographicLib有一个python接口：

这样可以在椭球上进行大地测量（将展平设置为零以获得大圆），并可以在大地测量上生成中间点（请参见示例中的“直线”命令）。

以下是在肯尼迪国际机场到樟宜机场（新加坡）的测地线上打印点的方法：

from geographiclib.geodesic import Geodesic
geod = Geodesic.WGS84

g = geod.Inverse(40.6, -73.8, 1.4, 104)
l = geod.Line(g['lat1'], g['lon1'], g['azi1'])
num = 15  # 15 intermediate steps

for i in range(num+1):
    pos = l.Position(i * g['s12'] / num)
    print(pos['lat2'], pos['lon2'])

->
(40.60, -73.8)
(49.78, -72.99)
(58.95, -71.81)
(68.09, -69.76)
(77.15, -65.01)
(85.76, -40.31)
(83.77, 80.76)
(74.92, 94.85)
...

pyproj具有Geod.npts函数，该函数将沿路径返回一个点数组。请注意，它在数组中不包含终端点，因此您需要考虑它们：

import pyproj
# calculate distance between points
g = pyproj.Geod(ellps='WGS84')
(az12, az21, dist) = g.inv(startlong, startlat, endlong, endlat)

# calculate line string along path with segments <= 1 km
lonlats = g.npts(startlong, startlat, endlong, endlat,
                 1 + int(dist / 1000))

# npts doesn't include start/end points, so prepend/append them
lonlats.insert(0, (startlong, startlat))
lonlats.append((endlong, endlat))

geopy Python的地理编码工具箱

http://code.google.com/p/geopy/wiki/GettingStarted#Calculating_distances

我没有使用过此软件包，但它似乎很有趣，并且是一种可能的解决方案：http : //trac.gispython.org/lab/wiki/Shapely