Spatial join in geopandas when overlapping polygons?

Question

I have two datasets, one with points (shops) and one with polygons (districts).

The districts dataset sometimes has overlapping polygons (as I have buffered them).

I want to know if each polygon has any matching points?

joined = geopandas.sjoin(districts,shops, op='contains', how='inner')
joined

The above code probably give me only one of the matching polygons. How do I check each polygon?

Answer 1

TL/DR

gpd.sjoin(districts, shops, how="left", op="contains") \
.reset_index()\
.rename(columns={"index": "districts"})\
.groupby(["districts"])\
.agg(nshops=("index_right", "nunique"), lshops=("index_right", "unique"))\
.astype(str)\
.replace("[nan]", "")

Explanations

Problem setup

Let's said that we have 3 districts ( d1 , d2 and d3 ). Two district overlap ( d1 and d2 ). We have 4 shops. s1 is inside district d1 , s2 is inside district d2 . s12 inside district d1 and d2 . s3 is not in any district.

We generate this geometry in python using shapely :

from shapely.geometry import Point
from shapely.geometry import Polygon
import matplotlib.pyplot as plt
# Create Polygons for the districts
d1 = Polygon([(0, 0), (3, 0), (3, 3), (0, 3)])
d2 = Polygon([(1, 1), (4, 1), (4, 4), (1, 4)])
d3 = Polygon([(5, 2), (6, 2), (6, 3), (5, 3)])
# Create Points for the shops
s1 = Point(0.5, 0.5)
s2 = Point(3.5, 3.5)
s3 = Point(4.5, 2)
# This shop is in distric 1 and distric 2.
s12 = Point(2, 2)

Saving the geometry into GeoPandas DataFrame and using matplotlib we can have a look at the configuration:

import geopandas as gpd
import matplotlib.pyplot as plt
districts = gpd.GeoDataFrame(index=['d1', 'd2', 'd3'], geometry=[d1, d2, d3])
shops = gpd.GeoDataFrame(index=['s1', 's12', 's2', 's3'], geometry=[s1, s12, s2, s3])
ax = districts.boundary.plot()
shops.plot(ax=ax, color='red')
plt.show()

Now let's have a look at how spatial join are working in GeoPandas. We have to be careful at the order of the dataframe because the operation is not commutative. Meaning gpd.sjoin(shops, districts, how="inner", op="contains") is not equal to gpd.sjoin(districts, shops, how="inner", op="contains") .

Now let's have a look to six arrangement:

1 gpd.sjoin(shops, districts, how="left", op="contains")

	geometry	index_right
s1	POINT (0.5 0.5)	nan
s12	POINT (2 2)	nan
s2	POINT (3.5 3.5)	nan
s3	POINT (4.5 2)	nan

Keep the shops as index and filling districts columns with NaN.

2 gpd.sjoin(shops, districts, how="right", op="contains")

	index_left	geometry
d1	nan	POLYGON ((0 0, 3 0, 3 3, 0 3, 0 0))
d2	nan	POLYGON ((1 1, 4 1, 4 4, 1 4, 1 1))
d3	nan	POLYGON ((5 2, 6 2, 6 3, 5 3, 5 2))

Keep the districts as index and filling shops columns with NaN.

3 gpd.sjoin(shops, districts, how="inner", op="contains")

geometry	index_right

This return an empty dataframe because points can't contains polygons.

4 gpd.sjoin(districts, shops, how="left", op="contains")

	geometry	index_right
d1	POLYGON ((0 0, 3 0, 3 3, 0 3, 0 0))	s1
d1	POLYGON ((0 0, 3 0, 3 3, 0 3, 0 0))	s12
d2	POLYGON ((1 1, 4 1, 4 4, 1 4, 1 1))	s12
d2	POLYGON ((1 1, 4 1, 4 4, 1 4, 1 1))	s2
d3	POLYGON ((5 2, 6 2, 6 3, 5 3, 5 2))	nan

Keep the districts as index.

5 gpd.sjoin(districts, shops, how="right", op="contains")

	index_left	geometry
s1	d1	POINT (0.5 0.5)
s12	d1	POINT (2 2)
s12	d2	POINT (2 2)
s2	d2	POINT (3.5 3.5)
s3	nan	POINT (4.5 2)

Keep shop as index .

6 gpd.sjoin(districts, shops, how="inner", op="contains")

	geometry	index_right
d1	POLYGON ((0 0, 3 0, 3 3, 0 3, 0 0))	s1
d1	POLYGON ((0 0, 3 0, 3 3, 0 3, 0 0))	s12
d2	POLYGON ((1 1, 4 1, 4 4, 1 4, 1 1))	s12
d2	POLYGON ((1 1, 4 1, 4 4, 1 4, 1 1))	s2

Close to what we obtain with left, keep districts as index, but drop NaN.

Answer to the question

    gpd.sjoin(districts, shops, how="left", op="contains") \
    .reset_index()\
    .rename(columns={"index": "districts"})\
    .groupby(["districts"])\
    .agg(nshops=("index_right", "nunique"), lshops=("index_right", "unique"))\
    .astype(str)\
    .replace("[nan]", "")

districts	nshops	lshops
d1	2	['s1' 's12']
d2	2	['s12' 's2']
d3	0

Like this we can known if each polygon has any matching points