问题
Is it possible to convert a Multipolygon
into a Polygon
which fills all holes or missing inner areas using Shapely? I have been trying since a while but I can't find it in the documentation. The following image I show an example of a multipolygon with those holes I want to fill, and those squares I want to remove.
回答1:
An approximate approach could be to:
- extract the outer boundary of individual components of the
MultiPolygon
of interest - 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"
- 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
.
回答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)
来源:https://stackoverflow.com/questions/48082553/convert-multipolygon-to-polygon-in-python