Django 按点列表的最小距离排序

Question

我想获得一个查询集，其中包含在给定距离内的所有工作，这些工作至少到许多提供的位置之一，按最小距离排序，并且不显示重复的工作。

from django.db import models
from cities.models import City    

class Job(models.Model):
    title = models.CharField(max_length=255)
    cities = models.ManyToManyField(City)

如果只有一点我可以这样做：

from django.contrib.gis.db.models.functions import Distance
from django.contrib.gis.geos import Point
point = Point(x, y, srid=4326)  
Job.objects.filter(cities__location__dwithin=(point, dist)) \
           .annotate(distance=Distance("cities__location", point) \
           .order_by('distance')

但是当我有很多点时，我会为过滤器构建一个 Q 表达式，但不确定是否有一种干净的方法来注释作业到所有点的最小距离

query = Q()
for point in points:
    query |= Q(cities__location__dwithin=(point, dist))
Job.objects.filter(query).annotate(distance=Min(...)).order_by('distance')

仅供参考，使用带 PostGIS 扩展的 postgres 12.1

Answer 1

query = Q()
distances = []
for point in points:
    query |= Q(cities__location__dwithin=(point, dist))
    distances.append(Distance("cities__location", point))

# LEAST requires 2 or more expressions, MIN works for single expression
if len(distances) == 1:
    MIN_FUNC = Min
else:
    MIN_FUNC = Least

Job.objects.filter(query).annotate(distance=MIN_FUNC(*distances)).order_by('distance')

MIN 是一个聚合函数，它采用单个表达式，例如列名，并将多个输入减少为单个输出值
LEAST 是一个条件表达式，它通过从任意数量的表达式列表中选择最小值来发挥作用

https://docs.djangoproject.com/en/3.0/ref/models/querysets/#min https://docs.djangoproject.com/en/3.0/ref/models/database-functions/#least

Django 按点列表的最小距离排序

Django order by Min Distance to a list of points

python

django

postgresql

postgis

django-queryset