Remove polygon overlap based on id in Python

Question

I have a Shapefile consists of multiple polygons, where I want to erase the overlap based on the id. An example to visualize the problem:

import geopandas as gpd
from shapely.geometry import Polygon

polygon1 = Polygon([[7, 0], [8, 1], [8, 4], [6, 4], [6, 1]])
polygon2 = Polygon([[7, 3], [8, 4], [8, 7], [6, 7], [6, 4]])
polygon3 = Polygon([[7, 6], [8, 7], [8, 9], [6, 9], [6, 7]])

polygondata = gpd.GeoDataFrame()
polygondata['geometry'] = None

polygondata.loc[0, 'geometry'] = polygon1
polygondata.loc[1, 'geometry'] = polygon2
polygondata.loc[2, 'geometry'] = polygon3

This codes gives following output visible on the left side (Exported to QGIS)

Because the overlay is created systematically in one direction i want to erase every polygon from each other based on the id. The code would be:

erase i from i+1 --> erase 0 from 1, erase 1 from 2 ...

and should give result look like in picture 2 (right). In Python I have the data loaded into geopandas geodataframe, but I only know how to perform geometrical operations between geodataframes and not single features in it. I would split the data in single geometries put it in a list and perform the calculation based on a loop, but I'm not sure how it works. Is there maybe a solution in geopandas how to perform this process without splitting the data?

I'm using Windows 10 and Python 3.7

Answer 1

You don't need a loop. Use shift in difference operation. (I used df instead of polygondata for legibility)

df['geometry'] = df['geometry'].difference(df['geometry'].shift(1))
df.loc[0, 'geometry'] = polygon1