返回ZipCode,如果指向多边形(形状)

时间:2015-11-22 22:27:07

标签: python polygon shapely point-in-polygon spatial-data-frame

我有一个带有一列点的DataFrame A:

   points                      Code
0  Point(1.23, 1.34)            ?
1  Point(1.32, 3.56)            ?
2  Point(-1.09, 2.11)           ?
.
.

我还有另一个DataFrame B,其中包含Polygon列:

   Code   Polygon
0  123    Polygon((-2,3),(1,4),(3,3),(-1,-2))
1  203    Polygon((-1,2),(0,2),(4,1),(-2,-1))
.
.

当点在多边形内时,如何将B中的代码传递给A?

1 个答案:

答案 0 :(得分:0)

假设您有两个数据帧:

df1 = GeoDataFrame(
    [[Point(1.23, 1.34)],
    [Point(1.32, 3.56)],
    [Point(-1.09, 2.11)]],
    columns=['geometry'],  
    geometry='geometry')

df2 = GeoDataFrame(
    [[Polygon([(-2,3),(1,4),(3,3),(-1,-2)]), 123],
    [Polygon([(-1,2),(0,2),(4,1),(-2,-1)]), 203]],
    columns=['geometry', 'Code'],  
    geometry='geometry')

您可以手动执行此操作:

# add the Code column:
df1.loc[:,'Code'] = [None]*len(df1.geometry)
# compute the intersections
for i, p in enumerate(df1.geometry):
    for j, pol in enumerate(df2.geometry):
        if pol.contains(p):
            df1['Code'][i] = df2['Code'][j]
            break

或者你可以通过空间连接来实现:

df1 = gpd.sjoin(df1, df2, how="inner", op='intersects')
# remove the index_right extra column that is generated:
df1.drop('index_right', axis=1, inplace=True)

注意,如果一个点与多个列相交,第二个方法将重复点行,第一个不会。