我有一个geopandas数据框,其中包含一系列POINT几何形状。另一列带有ID的列表,用于指定每个点所属的唯一多边形。简化的输入代码为:
import pandas as pd
from shapely.geometry import Point, LineString, Polygon
from geopandas import GeoDataFrame
data = [[1,10,10],[1,15,20],[1,20,10],[2,30,30],[2,35,40],[2,40,30]]
df_poly = pd.DataFrame(data, columns = ['poly_ID','lon', 'lat'])
geometry = [Point(xy) for xy in zip(df_poly.lon, df_poly.lat)]
geodf_poly = GeoDataFrame(df_poly, geometry=geometry)
geodf_poly.head()
我想对poly_ID进行分组,以便将几何图形从POINT转换为POLYGON。该输出基本上看起来像:
poly_ID geometry
1 POLYGON ((10 10, 15 20, 20 10))
2 POLYGON ((30 30, 35 40, 40 30))
我想这很简单,但是我很难使其正常工作。我发现以下代码使我可以将其转换为开放式折线,但无法为多边形弄清楚。谁能建议如何适应这一点?
geodf_poly = geodf_poly.groupby(['poly_ID'])['geometry'].apply(lambda x: LineString(x.tolist()))
简单地用Polygon替换LineString会导致TypeError:类型为'Point'的对象没有len()
您的请求在Pandas中很难完成,因为在您的输出中,您需要文本'POLYGON',但括号内的数字。
请参阅以下选项为您工作
from itertools import chain
df_poly.groupby('poly_ID').agg(list).apply(lambda x: tuple(chain.from_iterable(zip(x['lon'], x['lat']))), axis=1).reset_index(name='geometry')
输出
poly_ID geometry
0 1 (10, 10, 15, 20, 20, 10)
1 2 (30, 30, 35, 40, 40, 30)
或
from itertools import chain
df_new =df_poly.groupby('poly_ID').agg(list).apply(lambda x: tuple(chain.from_iterable(zip(x['lon'], x['lat']))), axis=1).reset_index(name='geometry')
df_new['geometry']=df_new.apply(lambda x: 'POLYGON ('+str(x['geometry'])+')',axis=1 )
df_new
输出
poly_ID geometry
0 1 POLYGON ((10, 10, 15, 20, 20, 10))
1 2 POLYGON ((30, 30, 35, 40, 40, 30))
注意:列geometry
是字符串,我不确定您可以直接将其输入Shapely