Chu*_*ung 8 python matplotlib point-in-polygon google-maps-api-3 shapely
我正在使用matplotlib和shapely测试多边形点函数.
这是一张包含百慕大三角形多边形的地图.
Google地图的多边形点函数清楚地表明,testsPoint和testingPoint2位于多边形内部,这是正确的结果.
如果我在matplotlib中测试两个点并且形状合理,则只有point2通过测试.
In [1]: from matplotlib.path import Path
In [2]: p = Path([[25.774252, -80.190262], [18.466465, -66.118292], [32.321384, -64.75737]])
In [3]: p1=[27.254629577800088, -76.728515625]
In [4]: p2=[27.254629577800088, -74.928515625]
In [5]: p.contains_point(p1)
Out[5]: 0
In [6]: p.contains_point(p2)
Out[6]: 1
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形状上显示与matplotlib相同的结果.
In [1]: from shapely.geometry import Polygon, Point
In [2]: poly = Polygon(([25.774252, -80.190262], [18.466465, -66.118292], [32.321384, -64.75737]))
In [3]: p1=Point(27.254629577800088, -76.728515625)
In [4]: p2=Point(27.254629577800088, -74.928515625)
In [5]: poly.contains(p1)
Out[5]: False
In [6]: poly.contains(p2)
Out[6]: True
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这里到底发生了什么?谷歌的算法比那两个好吗?
谢谢
Mik*_*e T 10
记住:世界并不平坦!如果Google Maps的投影是您想要的答案,则需要将地理坐标投影到球形墨卡托以获得不同的X和Y坐标集.Pyproj可以帮助解决这个问题,只需确保在之前反转坐标轴(即:X,Y或经度,纬度).
import pyproj
from shapely.geometry import Polygon, Point
from shapely.ops import transform
from functools import partial
project = partial(
pyproj.transform,
pyproj.Proj(init='epsg:4326'),
pyproj.Proj('+proj=merc +a=6378137 +b=6378137 +lat_ts=0.0 +lon_0=0.0 +x_0=0.0 +y_0=0 +k=1.0 +units=m +nadgrids=@null +no_defs'))
poly = Polygon(([-80.190262, 25.774252], [-66.118292, 18.466465], [-64.75737, 32.321384]))
p1 = Point(-76.728515625, 27.254629577800088)
# Old answer, using long/lat coordinates
poly.contains(p1) # False
poly.distance(p1) # 0.01085626429747994 degrees
# Translate to spherical Mercator or Google projection
poly_g = transform(project, poly)
p1_g = transform(project, p1)
poly_g.contains(p1_g) # True
poly_g.distance(p1_g) # 0.0 meters
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似乎得到了正确的答案.
虽然你已经接受了答案,但除了@ MikeT的答案,我会加入这对于谁可能想要做同样的将来游客matplotlib
和底图在mpl_toolkit
:
from mpl_toolkits.basemap import Basemap
from matplotlib.path import Path
# Mercator Projection
# http://matplotlib.org/basemap/users/merc.html
m = Basemap(projection='merc', llcrnrlat=-80, urcrnrlat=80,
llcrnrlon=-180, urcrnrlon=180, lat_ts=20, resolution='c')
# Poly vertices
p = [[25.774252, -80.190262], [18.466465, -66.118292], [32.321384, -64.75737]]
# Projected vertices
p_projected = [m(x[1], x[0]) for x in p]
# Create the Path
p_path = Path(p_projected)
# Test points
p1 = [27.254629577800088, -76.728515625]
p2 = [27.254629577800088, -74.928515625]
# Test point projection
p1_projected = m(p1[1], p1[0])
p2_projected = m(p2[1], p2[0])
if __name__ == '__main__':
print(p_path.contains_point(p1_projected)) # Prints 1
print(p_path.contains_point(p2_projected)) # Prints 1
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