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Given a dataframe, is it possible to find polygons that lie within another polygon in the same dataframe and remove them based on a condition?

For example, consider the following dataframe

                                            geometry          category            
0  POLYGON ((332294.432 2794730.319, 332294.032 2...            A                
1  POLYGON ((332193.232 2794730.719, 332192.832 2...            B                
2  POLYGON ((332271.632 2794731.519, 332271.232 2...            C                
3  POLYGON ((332271.632 2794731.519, 332271.232 2...            C

Suppose polygon 3 lies within polygon 2. Since they are both of category C, I would like to remove polygon 3. If they were of different categories, I would not do a removal.

Is there a way to do this within geopandas?

ashnair1
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1 Answers1

2

enter image description here

import geopandas as gpd
from itertools import permutations

df = gpd.read_file(r'/home/bera/Desktop/within.shp')
category_field = 'cat'


indices = df.index.to_list()


to_delete = []
for p1, p2 in permutations(indices,2): #https://stackoverflow.com/questions/104420/how-to-generate-all-permutations-of-a-list
    #print(p1, p2)
    if df.iloc[p1][category_field] == df.iloc[p2][category_field] and df.iloc[p1].geometry.within(df.iloc[p2].geometry):
        to_delete.append(p1)


#>>to_delete
#>>Out[18]: [1]


if len(to_delete)>0:
    df = df.drop(df.index[to_delete])
BERA
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  • Can you please tell how did you access GeoPandas in QGIS? – Taras Aug 23 '20 at 14:12
  • I didnt. I did the coding in Anaconda and just displayed the polygons in QGIS – BERA Aug 23 '20 at 14:14
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    Thank you, I know that it is possible, but still can not achieve that ... – Taras Aug 23 '20 at 14:15
  • This seems like it could be extremely slow with a layer containing many features as the formula for permutations is P(n,r) = n! / (n-r)!. Maybe an intermediate step could be used to reduce n by comparing the distances from centroids to nearest neighboring centroids and filtering them with some threshold value. – Kartograaf Sep 15 '22 at 20:54