jtam
jtam

Reputation: 902

Shapely point geometry in geopandas df to lat/lon columns

I have a geopandas df with a column of shapely point objects. I want to extract the coordinate (lat/lon) from the shapely point objects to generate latitude and longitude columns. There must be an easy way to do this, but I cannot figure it out.

I know you can extract the individual coordinates like this:

lon = df.point_object[0].x
lat = df.point_object[0].y

And I could create a function that does this for the entire df, but I figured there was a more efficient/elegant way.

Upvotes: 31

Views: 70360

Answers (3)

Kum_R
Kum_R

Reputation: 378

The solution to extract the center point (latitude and longitude) from the polygon and multi-polygon.

import geopandas as gpd
df = gpd.read_file(path + 'df.geojson')
#Find the center point
df['Center_point'] = df['geometry'].centroid
#Extract lat and lon from the centerpoint
df["long"] = df.Center_point.map(lambda p: p.x)
df["lat"] = df.Center_point.map(lambda p: p.y)

Upvotes: 2

Mandi
Mandi

Reputation: 331

Without having to iterate over the Dataframe, you can do the following:

df['lon'] = df['geometry'].x
df['lat'] = df['geometry'].y

Upvotes: 23

joris
joris

Reputation: 139162

If you have the latest version of geopandas (0.3.0 as of writing), and the if df is a GeoDataFrame, you can use the x and y attributes on the geometry column:

df['lon'] = df.point_object.x
df['lat'] = df.point_object.y

In general, if you have a column of shapely objects, you can also use apply to do what you can do on individual coordinates for the full column:

df['lon'] = df.point_object.apply(lambda p: p.x)
df['lat'] = df.point_object.apply(lambda p: p.y)

Upvotes: 43

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