如何找到哪些点与 geopandas 中的多边形相交?
How to find which points intersect with a polygon in geopandas?
我一直在尝试在地理数据框上使用 "intersects" 功能,以查看哪些点位于多边形内。但是,只有框架中的第一个特征 return 为真。我做错了什么?
from geopandas.geoseries import *
p1 = Point(.5,.5)
p2 = Point(.5,1)
p3 = Point(1,1)
g1 = GeoSeries([p1,p2,p3])
g2 = GeoSeries([p2,p3])
g = GeoSeries([Polygon([(0,0), (0,2), (2,2), (2,0)])])
g1.intersects(g) # Flags the first point as inside, even though all are.
g2.intersects(g) # The second point gets picked up as inside (but not 3rd)
根据 documentation:
Binary operations can be applied between two GeoSeries, in which case
the operation is carried out elementwise. The two series will be
aligned by matching indices.
您的示例不应该起作用。所以如果你想测试每个点都在一个多边形中,你必须这样做:
poly = GeoSeries(Polygon([(0,0), (0,2), (2,2), (2,0)]))
g1.intersects(poly.ix[0])
输出:
0 True
1 True
2 True
dtype: bool
或者,如果您想测试特定 GeoSeries 中的所有几何图形:
points.intersects(poly.unary_union)
Geopandas 依靠 Shapely 进行几何工作。直接使用它有时很有用(也更容易阅读)。以下代码也像宣传的那样工作:
from shapely.geometry import *
p1 = Point(.5,.5)
p2 = Point(.5,1)
p3 = Point(1,1)
poly = Polygon([(0,0), (0,2), (2,2), (2,0)])
for p in [p1, p2, p3]:
print(poly.intersects(p))
你也可以看看
边界上的点可能出现的问题。
解决此问题的一种方法似乎是获取特定条目(这不适用于我的应用程序,但可能适用于其他人的应用程序:
from geopandas.geoseries import *
p1 = Point(.5,.5)
p2 = Point(.5,1)
p3 = Point(1,1)
points = GeoSeries([p1,p2,p3])
poly = GeoSeries([Polygon([(0,0), (0,2), (2,2), (2,0)])])
points.intersects(poly.ix[0])
另一种方法(对我的应用程序更有用)是与第二层的特征的一元并集相交:
points.intersects(poly.unary_union)
您可以使用下面这个简单的函数轻松检查哪些点位于多边形内:
import geopandas
from shapely.geometry import *
p1 = Point(.5,.5)
p2 = Point(.5,1)
p3 = Point(1,1)
g = Polygon([(0,0), (0,2), (2,2), (2,0)])
def point_inside_shape(point, shape):
#point of type Point
#shape of type Polygon
pnt = geopandas.GeoDataFrame(geometry=[point], index=['A'])
return(pnt.within(shape).iloc[0])
for p in [p1, p2, p3]:
print(point_inside_shape(p, g))
由于 geopandas 最近进行了许多性能增强更改,因此此处的答案已过时。 Geopandas 0.8 引入了许多变化,使处理大型数据集的速度更快。
import geopandas
from shapely.geometry import Polygon
p1 = Point(.5,.5)
p2 = Point(.5,1)
p3 = Point(1,1)
points = GeoSeries([p1,p2,p3])
poly = GeoSeries([Polygon([(0,0), (0,2), (2,2), (2,0)])])
geopandas.overlay(points, poly, how='intersection')
我认为最快的方法是使用 geopandas.sjoin
。
import geopandas as gpd
gpd.sjoin(pts, poly, how='left', op='intersects')
检查示例:link
我一直在尝试在地理数据框上使用 "intersects" 功能,以查看哪些点位于多边形内。但是,只有框架中的第一个特征 return 为真。我做错了什么?
from geopandas.geoseries import *
p1 = Point(.5,.5)
p2 = Point(.5,1)
p3 = Point(1,1)
g1 = GeoSeries([p1,p2,p3])
g2 = GeoSeries([p2,p3])
g = GeoSeries([Polygon([(0,0), (0,2), (2,2), (2,0)])])
g1.intersects(g) # Flags the first point as inside, even though all are.
g2.intersects(g) # The second point gets picked up as inside (but not 3rd)
根据 documentation:
Binary operations can be applied between two GeoSeries, in which case the operation is carried out elementwise. The two series will be aligned by matching indices.
您的示例不应该起作用。所以如果你想测试每个点都在一个多边形中,你必须这样做:
poly = GeoSeries(Polygon([(0,0), (0,2), (2,2), (2,0)]))
g1.intersects(poly.ix[0])
输出:
0 True
1 True
2 True
dtype: bool
或者,如果您想测试特定 GeoSeries 中的所有几何图形:
points.intersects(poly.unary_union)
Geopandas 依靠 Shapely 进行几何工作。直接使用它有时很有用(也更容易阅读)。以下代码也像宣传的那样工作:
from shapely.geometry import *
p1 = Point(.5,.5)
p2 = Point(.5,1)
p3 = Point(1,1)
poly = Polygon([(0,0), (0,2), (2,2), (2,0)])
for p in [p1, p2, p3]:
print(poly.intersects(p))
你也可以看看
解决此问题的一种方法似乎是获取特定条目(这不适用于我的应用程序,但可能适用于其他人的应用程序:
from geopandas.geoseries import *
p1 = Point(.5,.5)
p2 = Point(.5,1)
p3 = Point(1,1)
points = GeoSeries([p1,p2,p3])
poly = GeoSeries([Polygon([(0,0), (0,2), (2,2), (2,0)])])
points.intersects(poly.ix[0])
另一种方法(对我的应用程序更有用)是与第二层的特征的一元并集相交:
points.intersects(poly.unary_union)
您可以使用下面这个简单的函数轻松检查哪些点位于多边形内:
import geopandas
from shapely.geometry import *
p1 = Point(.5,.5)
p2 = Point(.5,1)
p3 = Point(1,1)
g = Polygon([(0,0), (0,2), (2,2), (2,0)])
def point_inside_shape(point, shape):
#point of type Point
#shape of type Polygon
pnt = geopandas.GeoDataFrame(geometry=[point], index=['A'])
return(pnt.within(shape).iloc[0])
for p in [p1, p2, p3]:
print(point_inside_shape(p, g))
由于 geopandas 最近进行了许多性能增强更改,因此此处的答案已过时。 Geopandas 0.8 引入了许多变化,使处理大型数据集的速度更快。
import geopandas
from shapely.geometry import Polygon
p1 = Point(.5,.5)
p2 = Point(.5,1)
p3 = Point(1,1)
points = GeoSeries([p1,p2,p3])
poly = GeoSeries([Polygon([(0,0), (0,2), (2,2), (2,0)])])
geopandas.overlay(points, poly, how='intersection')
我认为最快的方法是使用 geopandas.sjoin
。
import geopandas as gpd
gpd.sjoin(pts, poly, how='left', op='intersects')
检查示例:link