用Python/Shapely聚合地理点的最佳方法

Question

我想将一长串纬度/经度坐标转换为它们所属的美国州(或县).考虑到我具有状态几何,一种可能的解决方案是针对所有状态检查每个点.

for point in points:
    for state in states:
        if point.within(state['shape']):
            print state.name

有没有更优化的方法来做到这一点,可能在O(1)？

Answer 1

使用Rtree作为空间索引,可以非常快速地识别零个或多个多边形的边界框中的点,然后使用Shapely确定该点所在的多边形.

与此示例类似,/sf/answers/1036305651/

from shapely.geometry import Polygon, Point
from rtree import index

# List of non-overlapping polygons
polygons = [
    Polygon([(0, 0), (0, 1), (1, 1), (0, 0)]),
    Polygon([(0, 0), (1, 0), (1, 1), (0, 0)]),
]

# Populate R-tree index with bounds of polygons
idx = index.Index()
for pos, poly in enumerate(polygons):
    idx.insert(pos, poly.bounds)

# Query a point to see which polygon it is in
# using first Rtree index, then Shapely geometry's within
point = Point(0.5, 0.2)
poly_idx = [i for i in idx.intersection((point.coords[0]))
            if point.within(polygons[i])]
for num, idx in enumerate(poly_idx, 1):
    print("%d:%d:%s" % (num, idx, polygons[idx]))

如果您剖析列表推导,您将看到list(idx.intersection((point.coords[0])))实际匹配两个多边形的边界框.另外,请注意边界上的点Point(0.5, 0.5)与任何内容都不匹配within,但两者都匹配intersects.所以要准备好匹配0,1个或更多的多边形.