我有一组由地理（WGS84）坐标指定的多边形：它们生活在一个球体上。
我有一个纬度 - 经度对指定的点。
我想（有效地）找到点和多边形之间的最小圆圈距离。
我目前的筹码包括fiona，shapely，gdal和proj。
关于StackOverflow的类似问题似乎主要是将特征投射到一个平面上并在那里找到距离，或者（令人不安地）省略了对投影的提及或完全没有提及。
答案 0 :(得分：2)
这不是特别有效，因为大部分操作都是在Python中进行的，而不是在编译库中进行，但它确实完成了工作：
import shapely
import numpy as np
import math
def Pairwise(iterable):
"""
Iterate through an itertable returning adjacent pairs
:param iterable An iterable
:returns: Pairs of sequential, adjacent entries from the iterable
"""
it = iter(iterable)
a = next(it, None)
for b in it:
yield (a, b)
a = b
def LatLonToXYZ(lat,lon,radius):
"""
Convert a latitude-longitude pair to 3D-Cartesian coordinates
:param lat Latitude in degrees
:param lon Longitude in degrees
:param radius Radius in arbitrary units
:returns: x,y,z coordinates in arbitrary units
"""
lat = np.radians(lat)
lon = np.radians(lon)
x = radius * np.cos(lon) * np.cos(lat)
y = radius * np.sin(lon) * np.cos(lat)
z = radius * np.sin(lat)
return x,y,z
def XYZtoLatLon(x,y,z):
"""
Convert 3D-Cartesian coordinates to a latitude-longitude pair
:param x x-coordinate in arbitrary units
:param y y-coordinate in arbitrary units
:param z z-coordinate in arbitrary units
:returns A (lat,lon) pair in degrees
"""
radius = np.sqrt(x*x+y*y+z*z)
lat = np.degrees(np.arcsin(z/radius))
lon = np.degrees(np.arctan2(y, x))
return lat,lon
def Haversine(lat1, lon1, lat2, lon2):
"""
Calculate the Great Circle distance on Earth between two latitude-longitude
points
:param lat1 Latitude of Point 1 in degrees
:param lon1 Longtiude of Point 1 in degrees
:param lat2 Latitude of Point 2 in degrees
:param lon2 Longtiude of Point 2 in degrees
:returns Distance between the two points in kilometres
"""
Rearth = 6371
lat1 = np.radians(lat1)
lon1 = np.radians(lon1)
lat2 = np.radians(lat2)
lon2 = np.radians(lon2)
#Haversine formula
dlon = lon2 - lon1
dlat = lat2 - lat1
a = np.sin(dlat/2)**2 + np.cos(lat1) * np.cos(lat2) * np.sin(dlon/2)**2
c = 2 * np.arcsin(np.sqrt(a))
return Rearth*c
#http://stackoverflow.com/a/1302268/752843
def NearestPointOnGC(alat1,alon1,alat2,alon2,plat,plon):
"""
Calculate the location of the nearest point on a Great Circle to a query point
:param lat1 Latitude of start of arc in degrees
:param lon1 Longtiude of start of arc in degrees
:param lat2 Latitude of end of arc in degrees
:param lon2 Longtiude of end of arc in degrees
:param plat Latitude of query point in degrees
:param plon Longitude of query point in degrees
:returns: A (lat,lon) pair in degrees of the closest point
"""
Rearth = 6371 #km
#Convert everything to Cartesian coordinates
a1 = np.array(LatLonToXYZ(alat1,alon1,Rearth))
a2 = np.array(LatLonToXYZ(alat2,alon2,Rearth))
p = np.array(LatLonToXYZ(plat, plon, Rearth))
G = np.cross(a1,a2) #Plane of the Great Circle containing A and B
F = np.cross(p,G) #Plane perpendicular to G that passes through query pt
T = np.cross(G,F) #Vector marking the intersection of these planes
T = Rearth*T/LA.norm(T) #Normalize to lie on the Great Circle
tlat,tlon = XYZtoLatLon(*T)
return tlat,tlon
def DistanceToGCArc(alat,alon,blat,blon,plat,plon):
"""
Calculate the distance from a query point to the nearest point on a
Great Circle Arc
:param lat1 Latitude of start of arc in degrees
:param lon1 Longtiude of start of arc in degrees
:param lat2 Latitude of end of arc in degrees
:param lon2 Longtiude of end of arc in degrees
:param plat Latitude of query point in degrees
:param plon Longitude of query point in degrees
:returns: The distance in kilometres from the query point to the great circle
arc
"""
tlat,tlon = NearestPointOnGC(alat,alon,blat,blon,plat,plon) #Nearest pt on GC
abdist = Haversine(alat,alon,blat,blon) #Length of arc
atdist = Haversine(alat,alon,tlat,tlon) #Distance arc start to nearest pt
tbdist = Haversine(tlat,tlon,blat,blon) #Distance arc end to nearest pt
#If the nearest point T on the Great Circle lies within the arc, then the
#length of the arc is approximately equal to the distance from T to each end
#of the arc, accounting for floating-point errors
PRECISION = 1e-3 #km
#We set the precision to a relatively high value because super-accuracy is not
#to needed here and a failure to catch this can lead to vast under-estimates
#of distance
if np.abs(abdist-atdist-tbdist)<PRECISION: #Nearest point was on the arc
return Haversine(tlat,tlon,plat,plon)
#Okay, the nearest point wasn't on the arc, so the nearest point is one of the
#ends points of the arc
apdist = Haversine(alat,alon,plat,plon)
bpdist = Haversine(blat,blon,plat,plon)
return min(apdist,bpdist)
def Distance3dPointTo3dPolygon(lat,lon,geom):
"""
Calculate the closest distance between a latitude-longitude query point
and a `shapely` polygon defined by latitude-longitude points using only
spherical mathematics
:param lat Latitude of query point in degrees
:param lon Longitude of query point in degrees
:param geom A `shapely` geometry whose points are in latitude-longitude space
:returns: The minimum distance in kilometres between the polygon and the
query point
"""
if geom.type == 'Polygon':
dist = math.inf
xy = features[0]['geometry'][0].exterior.xy
#Polygons are closed rings, so the first-last pair is automagically delt with
for p1, p2 in Pairwise(zip(*xy)):
dist = min(dist,DistanceToGCArc(p1[1],p1[0],p2[1],p2[0],lat,lon))
elif geom.type == 'MultiPolygon':
dist = min(*[Distance3dPointTo3dPolygon(lat,lon,part) for part in geom])
return dist
当然，您可以通过仅考虑点并忽略大圆弧来加快速度，这对于具有合适密集边界规范的多边形是合适的：
def Distance3dPointTo3dPolygonQuick(lat,lon,geom):
"""
Calculate the closest distance between a polygon and a latitude-longitude
point, using only spherical considerations. Ignore edges.
:param lat Latitude of query point in degrees
:param lon Longitude of query point in degrees
:param geom A `shapely` geometry whose points are in latitude-longitude space
:returns: The minimum distance in kilometres between the polygon and the
query point
"""
if geom.type == 'Polygon':
dist = math.inf
x,y = geom.exterior.xy
#Polygons are closed rings, so the first-last pair is automagically delt with
dist = np.min(Haversine(x,y,lat,lon))
elif geom.type == 'MultiPolygon':
dist = min(*[Distance3dPointTo3dPolygonQuick(lat,lon,part) for part in geom])
return dist