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假设我对线串的 GeoDataFrames 有以下内容，其中一个代表道路，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：

在此处输入图像描述

有3条等高线和2条道路。我想找到每条道路上每个点的海拔高度。基本上我想使道路和轮廓相交（这应该给我 12 分）并保留两个地理数据框的属性（道路名称和海拔）。

我可以通过使用两个地理数据框的并集的交集来生成 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 个点中的每一个的道路名称和标高？空间连接的行为不像我预期的那样，只返回 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以这种方式创建的）。

>>> 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
>>>

关于我如何做到这一点的任何建议？

编辑：看来问题与如何创建交点有关。如果我对道路和轮廓进行少量缓冲，则交叉口将按预期工作。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，尽管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。

4

1 回答 1

9

Notice that operations unary_union and intersection are made over the geometries inside the GeoDataFrame, so you lose the data stored in the rest of the columns. I think in this case you have to do it by hand by accessing each geometry in the data frames. The following code:

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

produces:

        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), ...

Notice each geometry has two points, that you can access later if you want point by point information, or you can create a row for each point introducing a for loop that iterates over the points, something like: 

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

But in this case you depend on knowing that each intersection produces a multi-geometry.

About your other approach using the sjoin function, I couldn't test it because the version of geopandas I'm using does not provide the tools module. Try to put buffer(0.0) to see what happens.

于 2015-12-14T16:25:08.507 回答