使用GeoPandas计算包含在其他多边形内部的多边形面积

Question

我有两个 GeoSeries：

df1 = gpd.GeoSeries([Polygon([(0,0),     (2,0),   (2,2), (0,2)]),
                     Polygon([(1.5,1.5), (4,2),   (4,4), (2,4)]),
                     Polygon([(1,3.5),   (3,3.5), (1,2.5)]),
                     Polygon([(1,0),     (3,0),   (3,2.5)])])

df2 = gpd.GeoSeries([Polygon([(1,1), (3,1), (3,3), (1,3)]),
                     Polygon([(3,3), (5,3), (5,5), (3,5)]),
                     Polygon([(1,3), (1,5), (3,5), (2,3)]),
                     Polygon([(5,1), (3,1), (3,3), (3,5)])])

绘制此图形得到:

base = df2.plot()
df1.plot(ax=base, cmap='summer')

你如何计算df1和df2之间的重叠区域？

- Reda S

Answer 1

这是使用您的数据的可行代码。阅读提供的代码注释以获取更多信息。

import geopandas as gpd
from shapely.geometry import Polygon

# use GeoSeries to prepare data
gs1 = gpd.GeoSeries([Polygon([(0,0), (2,0), (2,2), (0,2)]),
                Polygon([(1.5,1.5), (4,2), (4,4), (2,4)]),
                Polygon([(1,3.5),(3,3.5),(1,2.5)]),
                Polygon([(1,0), (3,0), (3,2.5)])])

gs2 = gpd.GeoSeries([Polygon([(1,1), (3,1), (3,3), (1,3)]),
                Polygon([(3,3), (5,3), (5,5), (3,5)]),
                Polygon([(1,3), (1,5), (3,5),(2,3)]),
                Polygon([(5,1), (3,1), (3,3), (3,5)])])

#base = gs2.plot()
#gs1.plot(ax=base, cmap='summer')

# create geoDataFrame from GeoSeries obtained above
df1 = gpd.GeoDataFrame(gs1)
# assign geometry to the geoDataFrame
df1g = df1.rename(columns={0:'geometry'}).set_geometry('geometry')

# similarly, ...
df2 = gpd.GeoDataFrame(gs2)
df2g = df2.rename(columns={0:'geometry'}).set_geometry('geometry')

# perform polygon overlay betw the two geoDataFrames
ov_output = gpd.overlay(df1g, df2g, how="intersection")
ov_output.plot(figsize=(4,4), cmap="Set2")

# Calculating areas of all the resulting polygons
ov_output.geometry.area

输出文本，显示相交多边形的计算面积：

0    1.0000
1    1.8000
5    0.2500
9    0.9000
2    0.3500
6    0.5625
3    1.0000
4    1.8500
dtype: float64

并且，这是结果图。

- swatchai

Answer 2

你需要使用 geopandas.overlay：

gpd.overlay(df1, df2, how="intersection")

- iacob