问题
I have a linestring which spans various polygons, stored as GeoJsons. I want to split the line into individual parts within each polygon area. I have not been able to achieve this yet, however. This is a reproducible example for what I have so far:
from shapely.geometry import Point, LineString, Polygon
from shapely.ops import split
from matplotlib import pyplot as plt
poly = {
'type': "Feature",
'geometry': {
"type": "Polygon",
"coordinates": ([(2,2),(2,4),(4,4),(4,2)]),
},
'properties': {
'id': 'A'
}
}
line = {
'type': "Feature",
'geometry': {
"type": "Linestring",
"coordinates": ([(3,3),(5,1)]),
},
'properties': {
'id': 'A'
}
}
poly = Polygon(poly['geometry']['coordinates'])
line = LineString(line['geometry']['coordinates'])
x,y = poly.exterior.xy
x2,y2 = line.coords
plt.plot(x, y, x2, y2)
plt.show()
This code produces the following square polygon with a linestring crossing it:
I've then tried to split the line via the polygon, as so:
new_lines = split(line, poly)
But I get the following output which doesn't seem correct:
GEOMETRYCOLLECTION (LINESTRING (3 3, 4 2), LINESTRING (4 2, 5 1))
I was expecting three lines, one existing inside the square polygon, and then two existing separately outside the polygon.
回答1:
It looks like split is working as you expect, but the line's coordinates are plotted in a misleading way - should use .xy
to get the line xs
and ys
as with the polygon:
from shapely.geometry import Point, LineString, Polygon
from shapely.ops import split
from matplotlib import pyplot as plt
poly = Polygon([(2,2),(2,4),(4,4),(4,2)])
original_line = LineString([(3,3),(5,1)])
line = LineString([(3,1), (3,5)])
x, y = poly.exterior.xy
x2, y2 = line.xy
x3, y3 = original_line.xy
plt.plot(x, y, x2, y2)
plt.show()
plt.plot(x, y, x3, y3)
plt.show()
image with original coords
image with expected coords
>>> print(split(line, poly))
GEOMETRYCOLLECTION (LINESTRING (3 1, 3 2), LINESTRING (3 2, 3 4), LINESTRING (3 4, 3 5))
来源:https://stackoverflow.com/questions/54498791/split-a-linestring-at-each-point-it-crosses-a-polygon-in-shapely