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实现这一点，可能是在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来确定该点所在的多边形。

与此示例类似https://stackoverflow.com/a/14804366/327026

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])))实际上匹配两个多边形的边界框。另请注意，Point(0.5, 0.5)等边界上的点与within的任何内容都不匹配，但会与intersects匹配。所以要准备匹配0,1个或更多的多边形。

用Python / Shapely聚合地理点的最佳方法

1 个答案: