use*_*696 5 python point shapely
我试图使用shapely的'within'函数来做一个Linestring和一个点文件的'空间连接'(fyi - 点文件是使用线串上的插值函数生成的).问题是 - 什么都没有归还.
我错过了什么?
# this condition is never satisfied
if point.within(line):
# here I write stuff to a file
Run Code Online (Sandbox Code Playgroud)
和
point = POINT (-9763788.9782693591000000 5488878.3678984242000000)
line = LINESTRING (-9765787.998118492 5488940.974948905, -9748582.801636808 5488402.127570709)
Run Code Online (Sandbox Code Playgroud)
Mik*_*e T 14
在线上找到点时存在浮点精度误差.请使用具有适当阈值的距离.
from shapely.geometry import Point, LineString
line = LineString([(-9765787.9981184918, 5488940.9749489054), (-9748582.8016368076, 5488402.1275707092)])
point = Point(-9763788.9782693591, 5488878.3678984242)
line.within(point) # False
line.distance(point) # 7.765244949417793e-11
line.distance(point) < 1e-8 # True
Run Code Online (Sandbox Code Playgroud)