从多边形列表中减去内环

时间:2019-11-29 14:38:18

标签: python gis polygon shapely

我有大量的多边形列表(> 10 ^ 6),其中大多数是不相交的,但是其中一些多边形是另一个多边形的孔(〜10 ^ 3种情况)。这里是一个解释问题的图像,较小的多边形是较大的多边形中的孔,但在多边形列表中都是独立的多边形。 A large polygon with smaller polygon on the inside.

现在,我想有效地确定哪些多边形是孔并减去孔,即减去完全位于另一个多边形内部的较小多边形,并返回“已清理”多边形的列表。一对孔和父多边形应按以下方式进行转换(因此基本上是从父级减去孔): enter image description here

在Stackoverflow和gis.stackexchange.com上有很多类似的问题,但是我还没有找到能真正解决这个问题的问题。以下是一些相关的问题: 1. https://gis.stackexchange.com/questions/5405/using-shapely-translating-between-polygons-and-multipolygons 2. https://gis.stackexchange.com/questions/319546/converting-list-of-polygons-to-multipolygon-using-shapely

这是示例代码。

from shapely.geometry import Point
from shapely.geometry import MultiPolygon
from shapely.ops import unary_union
import numpy as np

#Generate a list of polygons, where some are holes in others; 
def generateRandomPolygons(polygonCount = 100, areaDimension = 1000, holeProbability = 0.5):
    pl = []
    radiusLarge = 2 #In the real dataset the size of polygons can vary
    radiusSmall = 1 #Size of holes can also vary

    for i in range(polygonCount):
        x, y = np.random.randint(0,areaDimension,(2))
        rn1 = np.random.random(1)
        pl.append(Point(x, y).buffer(radiusLarge))
        if rn1 < holeProbability: #With a holeProbability add a hole in the large polygon that was just added to the list
            pl.append(Point(x, y).buffer(radiusSmall))
    return pl

polygons = generateRandomPolygons()
print(len(pl))

输出看起来像这样:enter image description here

现在如何在删除孔的情况下创建多边形的新列表。 Shapely提供了从一个多边形中减去另一个多边形(difference)的功能,但是多边形列表是否有类似的功能(也许像unary_union之类的东西,但是重叠部分被删除了)?或者,如何有效地确定哪些是孔,然后从较大的多边形中减去它们?

1 个答案:

答案 0 :(得分:6)

您的问题是您不知道哪些是“漏洞”,对吗?要“有效地确定哪些多边形是孔”,可以使用rtree来加速相交检查:

from rtree.index import Index

# create an rtree for efficient spatial queries
rtree = Index((i, p.bounds, None) for i, p in enumerate(polygons))
donuts = []

for i, this_poly in enumerate(polygons):
    # loop over indices of approximately intersecting polygons
    for j in rtree.intersection(this_poly.bounds):
        # ignore the intersection of this polygon with itself
        if i == j:
            continue
        other_poly = polygons[j]
        # ensure the polygon fully contains our match
        if this_poly.contains(other_poly):
            donut = this_poly.difference(other_poly)
            donuts.append(donut)
            break  # quit searching

print(len(donuts))