两条线串的交集 Geopandas

Intersection of Two LineStrings Geopandas

假设我有以下线串的 GeoDataFrames,其中之一代表道路,其中一个代表等高线。

>>> import geopandas as gpd
>>> import geopandas.tools
>>> import shapely
>>> from shapely.geometry import *
>>> 
>>> r1=LineString([(-1,2),(3,2.5)])
>>> r2=LineString([(-1,4),(3,0)])
>>> Roads=gpd.GeoDataFrame(['Main St','Spruce St'],geometry=[r1,r2], columns=['Name'])
>>> Roads
        Name                  geometry
0    Main St  LINESTRING (-1 2, 3 2.5)
1  Spruce St    LINESTRING (-1 4, 3 0)
>>> 

>>> c1=LineString(Point(1,2).buffer(.5).exterior)
>>> c2=LineString(Point(1,2).buffer(.75).exterior)
>>> c3=LineString(Point(1,2).buffer(.9).exterior)
>>> Contours=gpd.GeoDataFrame([100,90,80],geometry=[c1,c2,c3], columns=['Elevation'])
>>> Contours
   Elevation                                           geometry
0        100  LINESTRING (1.5 2, 1.497592363336099 1.9509914...
1         90  LINESTRING (1.75 2, 1.746388545004148 1.926487...
2         80  LINESTRING (1.9 2, 1.895666254004977 1.9117845...
>>> 

如果我绘制这些,它们看起来像这样:

有3条等高线和2条道路。我想找到每条道路上每个点的高程。基本上我想与道路和等高线相交(这应该给我 12 分)并保留两个地理数据框的属性(道路名称和高程)。

我可以通过使用两个地理数据框的并集的交集生成 12 个点:

>>> Intersection=gpd.GeoDataFrame(geometry=list(Roads.unary_union.intersection(Contours.unary_union)))
>>> Intersection
                                        geometry
0    POINT (0.1118644118110415 2.13898305147638)
1   POINT (0.2674451642029509 2.158430645525369)
2   POINT (0.3636038969321072 2.636396103067893)
3   POINT (0.4696699141100895 2.530330085889911)
4   POINT (0.5385205980649126 2.192315074758114)
5   POINT (0.6464466094067262 2.353553390593274)
6    POINT (1.353553390593274 1.646446609406726)
7    POINT (1.399321982208571 2.299915247776072)
8     POINT (1.530330085889911 1.46966991411009)
9    POINT (1.636396103067893 1.363603896932107)
10   POINT (1.670759586114587 2.333844948264324)
11   POINT (1.827239686607525 2.353404960825941)
>>> 

但是,我现在如何获取这 12 个点中每个点的道路名称和海拔高度?空间连接的行为并不像我预期的那样,只有 returns 4 个点(所有 12 个点都应该与线文件相交,因为它们是按照定义的方式创建的)。

>>> gpd.tools.sjoin(Intersection, Roads)
                                       geometry  index_right       Name
2  POINT (0.3636038969321072 2.636396103067893)            1  Spruce St
3  POINT (0.4696699141100895 2.530330085889911)            1  Spruce St
5  POINT (0.6464466094067262 2.353553390593274)            1  Spruce St
6   POINT (1.353553390593274 1.646446609406726)            1  Spruce St
>>> 

关于如何执行此操作有什么建议吗?

编辑: 问题似乎与交点的创建方式有关。如果我对道路和等高线进行少量缓冲,十字路口会按预期工作。见下文:

>>> RoadsBuff=gpd.GeoDataFrame(Roads, geometry=Roads.buffer(.000005))
>>> ContoursBuff=gpd.GeoDataFrame(Contours, geometry=Contours.buffer(.000005))
>>> 
>>> Join1=gpd.tools.sjoin(Intersection, RoadsBuff).drop('index_right',1).sort_index()
>>> Join2=gpd.tools.sjoin(Join1, ContoursBuff).drop('index_right',1).sort_index()
>>> 
>>> Join2
                                             geometry       Name  Elevation
0   POLYGON ((1.636395933642091 1.363596995290097,...  Spruce St         80
1   POLYGON ((1.530329916464109 1.469663012468079,...  Spruce St         90
2   POLYGON ((1.353553221167472 1.646439707764716,...  Spruce St        100
3   POLYGON ((0.5385239436706243 2.192310454047735...    Main St        100
4   POLYGON ((0.2674491823047923 2.158426108877007...    Main St         90
5   POLYGON ((0.1118688004427904 2.138978561144256...    Main St         80
6   POLYGON ((0.6464467873602107 2.353546141571978...  Spruce St        100
7   POLYGON ((0.4696700920635739 2.530322836868614...  Spruce St         90
8   POLYGON ((0.3636040748855915 2.636388854046597...  Spruce St         80
9   POLYGON ((1.399312865255344 2.299919147068011,...    Main St        100
10  POLYGON ((1.670752113626148 2.333849053114361,...    Main St         90
11  POLYGON ((1.827232214119086 2.353409065675979,...    Main St         80
>>> 

以上是所需的输出,虽然我不确定为什么我必须缓冲线以使它们与从线的交点创建的点相交。

请注意,操作 unary_unionintersection 是在 GeoDataFrame 内的几何图形上进行的,因此您丢失了存储在其余列中的数据。我认为在这种情况下,您必须通过访问数据框中的每个几何体来手动完成。以下代码:

import geopandas as gpd
from shapely.geometry import LineString, Point

r1=LineString([(-1,2),(3,2.5)])
r2=LineString([(-1,4),(3,0)])
roads=gpd.GeoDataFrame(['Main St','Spruce St'],geometry=[r1,r2], columns=['Name'])

c1=LineString(Point(1,2).buffer(.5).exterior)
c2=LineString(Point(1,2).buffer(.75).exterior)
c3=LineString(Point(1,2).buffer(.9).exterior)
contours=gpd.GeoDataFrame([100,90,80],geometry=[c1,c2,c3], columns=['Elevation'])

columns_data = []
geoms = []
for _, n, r in roads.itertuples():
    for _, el, c in contours.itertuples():
        intersect = r.intersection(c)
        columns_data.append( (n,el) )
        geoms.append(intersect)

all_intersection = gpd.GeoDataFrame(columns_data, geometry=geoms, 
                    columns=['Name', 'Elevation'])

print all_intersection 

产生:

        Name  Elevation                                           geometry
0    Main St        100  (POINT (0.5385205980649125 2.192315074758114),...
1    Main St         90  (POINT (0.2674451642029509 2.158430645525369),...
2    Main St         80  (POINT (0.1118644118110415 2.13898305147638), ...
3  Spruce St        100  (POINT (0.6464466094067262 2.353553390593274),...
4  Spruce St         90  (POINT (0.4696699141100893 2.53033008588991), ...
5  Spruce St         80  (POINT (0.363603896932107 2.636396103067893), ...

请注意,每个几何图形都有两个点,如果您需要逐点信息,您可以稍后访问,或者您可以为每个点创建一行,引入一个迭代这些点的 for 循环,例如:

for p in intersect:
    columns_data.append( (n,el) )
    geoms.append(p)

但在这种情况下,您需要知道每个交叉点都会产生一个多几何图形。

关于您使用 sjoin 函数的其他方法,我无法测试它,因为我使用的 geopandas 版本没有提供 tools 模块。尝试输入 buffer(0.0) 看看会发生什么。