32

Given a GeoPandas's GeoDataFrame containing a series of polygons, I would like to get the area in km sq of each feature in my list.

This is a pretty common problem, and the usual suggested solution in the past has been to use shapely and pyproj directly (e.g. Calculating area in km² for Polygon in WKT using Python and robinkraft/projected_area.py).

Is there a way to do this in pure GeoPandas?

4 Answers 4

33

If the crs of the GeoDataFrame is known (EPSG:4326 unit=degree, here), you don't need Shapely, nor pyproj in your script because GeoPandas uses them).

import geopandas as gpd
test = gpd.read_file("test_wgs84.shp")
print test.crs
test.head(2)

enter image description here

Now copy your GeoDataFrame and change the projection to a Cartesian system (EPSG:3857, unit= m as in the answer of ResMar)

tost = test.copy()
tost= tost.to_crs({'init': 'epsg:3857'})
print tost.crs
tost.head(2)

enter image description here

Now the area in square kilometers

tost["area"] = tost['geometry'].area/ 10**6
tost.head(2)

enter image description here

But the surfaces in the Mercator projection are not correct, so with other projection in meters.

tost= tost.to_crs({'init': 'epsg:32633'})
tost["area"] = tost['geometry'].area/ 10**6
tost.head(2)

enter image description here

0
9

Yes, simply be sure to reproject your shape in Cylindrical equal-area format with {'proj':'cea'} that preserve area measure.

Then you can use .area method of your GeoDataFrame.

Your also need to divide by 1000000 because .area method give area in square meters.

import geopandas as gpd

gdf = gpd.read_file("YOUR_SHAPE_FILE.shp")
gdf = gdf['geometry'].to_crs({'proj':'cea'}) 

gdf.area / 10**6
8

I believe yes. The following ought to work:

gdf['geometry'].to_crs({'init': 'epsg:3395'})\
               .map(lambda p: p.area / 10**6)

This converts the geometry to an equal-area projection, fetches the shapely area (returned in m^2), and maps that to a km^2 (this last step is optional).

0
0

Just a quick thought on the appropriate EPSG code for an equal-area estimation - 6933 may be a better "generic" solution (see https://epsg.io/6933 / https://www.mdpi.com/2220-9964/1/1/32)

No perfect solution for obvious reasons, but 6933 does seem to nicely merge the benefits of a cea and Lambert equal area.

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