Goal is to group polygon (1,2,4) and polygon (3) based on overlap. All polygons are part of the same shapefile layer.
In ArcMap I can simply do dissolve and uncheck Create multipart features.
However dissolve in geopandas requires you to set an attribute to dissolve on. What would be the easiest alternative in geopandas?
Using this example GeoSeries:
s = geopandas.GeoSeries([Polygon([(0, 0), (0, 2), (2, 2), (2, 0)]), Polygon([(0, 1), (0, 3), (2, 3), (2, 1)]),Polygon([(1, 0), (1, 2), (3, 2), (3, 0)]), Polygon([(4, 4), (4, 6), (6, 6), (6, 4)])])
s.plot(alpha=0.5, cmap='Set1')
We could create a matrix indicating which geometries are overlapping:
In [55]: overlap_matrix = s.apply(lambda x: s.overlaps(x)).values.astype(int)
In [56]: overlap_matrix
Out[56]:
array([[0, 1, 1, 0],
[1, 0, 1, 0],
[1, 1, 0, 0],
[0, 0, 0, 0]])
And from that get the groups using scipy's connected components:
In [57]: from scipy.sparse.csgraph import connected_components
In [58]: connected_components(overlap_matrix)
Out[58]: (2, array([0, 0, 0, 1], dtype=int32))
In [59]: n, ids = connected_components(overlap_matrix)
And use dissolve based on those groups:
In [60]: df = geopandas.GeoDataFrame({'geometry': s, 'group': ids})
In [61]: res = df.dissolve(by='group')
In [62]: res
Out[62]:
geometry
group
0 POLYGON ((0 0, 0 1, 0 2, 0 3, 2 3, 2 2, 3 2, 3...
1 POLYGON ((4 4, 4 6, 6 6, 6 4, 4 4))
In [63]: res.plot(cmap='Set1')
But note: I am certainly not sure if creating such a matrix is the most efficient way.
Further note, this would actually be a nice enhancement to geopandas to enable this somehow (or at least to have a good solution in the examples)
intersects instead of overlaps.
I found a workaround:
def explode(gdf):
"""
Will explode the geodataframe's muti-part geometries into single
geometries. Each row containing a multi-part geometry will be split into
multiple rows with single geometries, thereby increasing the vertical size
of the geodataframe. The index of the input geodataframe is no longer
unique and is replaced with a multi-index.
The output geodataframe has an index based on two columns (multi-index)
i.e. 'level_0' (index of input geodataframe) and 'level_1' which is a new
zero-based index for each single part geometry per multi-part geometry
Args:
gdf (gpd.GeoDataFrame) : input geodataframe with multi-geometries
Returns:
gdf (gpd.GeoDataFrame) : exploded geodataframe with each single
geometry as a separate entry in the
geodataframe. The GeoDataFrame has a multi-
index set to columns level_0 and level_1
"""
gs = gdf.explode()
gdf2 = gs.reset_index().rename(columns={0: 'geometry'})
gdf_out = gdf2.merge(gdf.drop('geometry', axis=1), left_on='level_0', right_index=True)
gdf_out = gdf_out.set_index(['level_0', 'level_1']).set_geometry('geometry')
gdf_out.crs = gdf.crs
return gdf_out
df_all = df1.append(df2,ignore_index=True)
df_all["group"] = 1
dissolved = df_all.dissolve(by="group")
gdf_out = explode(dissolved)
gdf_out2 = gdf_out.reset_index()
Notebook with the workaround (different polygons):
df1 and df2 here? As you mention you have a single shapefile with those polygons?
I've tried to enhance @joris's answer
from shapely.geometry import Polygon
import geopandas as gpd
s = gpd.GeoDataFrame(data=[1, 2, 3, 4],
geometry=[Polygon([(0, 0), (0, 2), (2, 2), (2, 0)]),
Polygon([(0, 1), (0, 3), (2, 3), (2, 1)]),
Polygon([(1, 0), (1, 2), (3, 2), (3, 0)]),
Polygon([(4, 4), (4, 6), (6, 6), (6, 4)])],
columns=['column'])
column geometry
0 1 POLYGON ((0.00000 0.00000, 0.00000 2.00000, 2....
1 2 POLYGON ((0.00000 1.00000, 0.00000 3.00000, 2....
2 3 POLYGON ((1.00000 0.00000, 1.00000 2.00000, 3....
3 4 POLYGON ((4.00000 4.00000, 4.00000 6.00000, 6....
s.plot(alpha=0.5, cmap='Set1')
s_ = gpd.GeoDataFrame(
geometry=[s.unary_union]).explode(
index_parts=False).reset_index(
drop=True)
s_.plot(alpha=0.5, cmap='Set1')
If you need to work with data you can use sjoin and dissolve functions.
s_ = gpd.sjoin(s_, s, how='left').drop(columns=['index_right'])
s_.dissolve(s_.index, aggfunc='first')
geometry column
0 POLYGON ((3.00000 0.00000, 2.00000 0.00000, 1.... 1
1 POLYGON ((6.00000 6.00000, 6.00000 4.00000, 4.... 4