在Shapely

Question

我试图用两行来分割两个部分的shapely.geometry.Polygon实例。例如，在下面的代码中，polygon是一个环，如果我们用line1和line2切割它，我们应该得到两个部分环，一个w / 270度，一个90度。会有一个干净的方法吗？

from shapely.geometry import Point, LineString, Polygon

polygon = Point(0, 0).buffer(2).difference(Point(0, 0).buffer(1))
line1 = LineString([(0, 0), (3, 3)])
line2 = LineString([(0, 0), (3, -3)])

Answer 1

肯·沃特福德回答here关于使用buffer和difference来解决问题的方法，其缺点是失去了一点区域。下面的示例代码：

from shapely.geometry import Point, LineString, Polygon

polygon = Point(0, 0).buffer(2).difference(Point(0, 0).buffer(1))
line1 = LineString([(0, 0), (3, 3)])
line2 = LineString([(0, 0), (3, -3)])

line1_pol = line1.buffer(1e-3)
line2_pol = line2.buffer(1e-3)

new_polygon = polygon.difference(line1_pol).difference(line2_pol)

现在可以使用，我有兴趣看看是否还有另一种（可能没有丢失的区域）方法！

Answer 2

从1.6.0版（2017年8月）开始，Shapely中有一个功能可以将一个几何图形拆分为另一个几何图形，因此不再需要滚动自己的几何图形。请参阅以下文档：shapely.ops.split(geom, splitter)

请注意，此线程上可接受的答案写在 之前，拆分功能在Shapely中-现在已过时了。

Answer 3

from shapely.ops import linemerge, unary_union, polygonize
from shapely.geometry import LineString, Polygon

# Define the Polygon and the cutting line
line = LineString([(-5, -5), (5, 5)])
polygon = Polygon([(-1, -1), (1, -1), (1, 1), (-1, 1)])


def cut_polygon_by_line(polygon, line):
    merged = linemerge([polygon.boundary, line])
    borders = unary_union(merged)
    polygons = polygonize(borders)
    return list(polygons)

def plot(shapely_objects, figure_path='fig.png'):
    from matplotlib import pyplot as plt
    import geopandas as gpd
    boundary = gpd.GeoSeries(shapely_objects)
    boundary.plot(color=['red', 'green', 'blue', 'yellow', 'yellow'])
    plt.savefig(figure_path)

result = cut_polygon_by_line(polygon, line)
print(result)
plot(result)
print(result[0].intersection(result[1]))

结果是

[<shapely.geometry.polygon.Polygon object at 0x7f50dcf46d68>, 
 <shapely.geometry.polygon.Polygon object at 0x7f50dcf46da0>]
LINESTRING (-1 -1, 1 1)