Question

现在我有一个GeoJson文件和以下函数使用匀称：

它接受一个坐标并返回邻域名称

    def get_neighb(lat, lon):
    """Input Latitude and Longitude, Returns Neighborhood Name"""
    point = Point(lon, lat)
    found = False
    for feature in geo_data['features']:
        polygon = shape(feature['geometry'])
        if polygon.contains(point):
            return(feature['properties']['neighborhood'])
            found = True
    if found is False:
        return('NA')

# Initialize list
tn = ['']*data.shape[0]
for i in range(len(tn)):
    tn[i] = get_neighb(data.latitude[i], data.longitude[i])

这很有效，但它真的很慢，对我如何加速它的任何想法，目前在4,000,000行上运行。

Answer 1

你必须找到一个不检查每一行的策略。最简单的方法是将所有形状转储到地理感知数据库中并进行查询。像后期gis或弹性搜索这样的东西。

另一种策略可能是找到所有邻域的质心，然后使用KD树仅过滤附近质心的邻域。

Answer 2

如果你想避免使用PostGIS数据库的重型机器，可能会感兴趣的是使用rtree包作为（如文档所述）＆＃34; cheapo空间数据库＆＃34 ;。这个想法主要如下：

#!/usr/bin/env python
from itertools import product
from random import uniform, sample, seed
from rtree import index
from shapely.geometry import Point, Polygon, box, shape
from shapely.affinity import translate

seed(666)

#generate random polygons, in your case, the polygons are stored
#in geo_data['features']
P = Polygon([(0, 0), (0.5, 0), (0.5, 0.5), (0, 0.5), (0, 0)])
polygons = []
for dx, dy in product(range(0, 100), range(0, 100)):
    polygons.append(translate(P, dx, dy))

#construct the spatial index and insert bounding boxes of all polygons
idx = index.Index()
for pid, P in enumerate(polygons):
    idx.insert(pid, P.bounds)

delta = 0.5
for i in range(0, 1000):
    #generate random points
    x, y = uniform(0, 10), uniform(0, 10)
    pnt = Point(x, y)

    #create a region around the point of interest
    bounds = (x-delta, y-delta, x+delta, y+delta)

    #also possible, but much slower
    #bounds = pnt.buffer(delta).bounds

    #the index tells us which polygons are worth checking, i.e.,
    #the bounding box of which intersects with the region constructed in previous step
    for candidate in idx.intersection(bounds):
        P = polygons[candidate]

        #test only these candidates
        if P.contains(pnt):
            print(pnt, P)

如何在python中反转地理编码大量的点？

2 个答案: