Convert Multipolygon to Polygon in Python [closed]

Answer 1

An approximate approach could be to:

1. extract the outer boundary of individual components of the MultiPolygon of interest
2. expand-shrink each of the outer boundaries in order to fill "holes" which these boundaries approximately encompass, e.g., to handle a "doughnut with a cut"
3. merge all geometries obtained in previous step

For example:

#!/usr/bin/env python
from shapely.geometry import MultiPolygon, Polygon
from shapely.ops import cascaded_union

a = 0.25
delta = 0.49

P = MultiPolygon([
    (
        ((0,0),(0,3),(3,3),(3,2-delta),(2,2-delta),(2,2),(1,2),(1,1),(2,1),(2,1+delta),(3,1+delta),(3,0),(0,0)),
        [((a, a), (1-a,a), (1-a,1-a), (a,1-a), (a,a))]
    )
])


eps = 0.01

omega = cascaded_union([
    Polygon(component.exterior).buffer(eps).buffer(-eps) for component in P
])
for x,y in zip(*omega.exterior.coords.xy):
    print(x, y)

The MultiPolygon P looks like:

while the script listed above produces as expected an approximate square with side of length 3, i.e., it fills the hole in the lower-left corner as well as the "empty space" in the center of the MultiPolygon which is rendered equivalent to a hole in the expand-shrink procedure with sufficient high value of parameter eps.

Answer 2

If it's enough to fill the holes of some MultiPolygon m, you could do this:

no_holes = MultiPolygon(Polygon(p.exterior) for p in m)

If you also need to fill the holes that arise from touching Polygons inside your MultiPolygon, the following should work:

# Create a polygon `b` that contains `m`
xmin, ymin, xmax, ymax = m.bounds
b = Polygon([(xmin, ymin), (xmin, ymax), (xmax, ymax), (xmax, ymin)]).buffer(1, resolution=1)

# Take the complement of `m` in `b`, which is a MultiPolygon.
# Pick the outer polygon and take the complement in `b`.
no_holes = b - min(b - m, key=lambda p: p.bounds)