在GeoDjango中获取最接近几何图形的最快方法 [英] Fastest way to get nearest geometry to a point in GeoDjango

问题描述

我在WGS84坐标系中有大量点(约150万个).这些点的面积很大，因此无法使用投影坐标系.我想找到最接近给定输入坐标对的点.我有一个有效的视图，但是执行时间太长(约2.5秒).

I've got a large number of points (~1.5 million) in the WGS84 coordinate system. The points span a large area, so I can't use a projected coordinate system. I want to find the closest point to a given pair of input coordinates. I have a working view, but it takes too long (~2.5 seconds) to execute.

这是我的模特

from django.contrib.gis.db import models

class Point(models.Model):

    id = models.IntegerField(primary_key=True)
    geom = models.PointField(srid=4326, spatial_index=True)

    objects = models.GeoManager()

这是视图中的查询(这是我从另一个SO问题获得的):

This is the query in the view (which I got from another SO question):

input_point = GEOSGeometry('POINT({0} {1})'.format(lon, lat))
point = Point.objects.distance(input_point).order_by('distance')[0]

有没有办法更快地做到这一点?我在SQLAlchemy/GeoAlchemy2中有一个等效的查询，执行该查询所需的时间不到0.5秒，所以我知道这是可能的.

Is there a way to do this faster? I have an equivalent query in SQLAlchemy/GeoAlchemy2, which takes less than 0.5 seconds to execute, so I know it's possible.

from geoalchemy2.elements import WKTElement

pt = WKTElement('POINT({0} {1})'.format(lon, lat), srid=4326)
q = session.query(Point).order_by(Point.geom.distance_box(pt)).first()

是否有更好的方法通过GeoDjango进行最近点"查询?

Is there a better way to do "nearest point" queries with GeoDjango?

推荐答案

不确定，这可能会更快.您可以使用Google投影( 900913 )来获取以米为单位的数据.

Not sure, this may be faster. You can use the google projection (900913) to get your data in meters.

from django.contrib.gis.measure import D

DISTANCE_LIMIT_METERS = 5000

input_point = Point(lon, lat, srid=4326)
input_point.transform(900913)
ModelContainingPoint.objects.filter(geom__dwithin=(input_point , D(m=DISTANCE_LIMIT_METERS)))

参考:

https://docs.djangoproject.com/en/dev/ref/contrib/gis/db-api/#distance-queries

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