Django按最小距离排序到点列表

Question

我想获取所有距指定的多个位置中的至少一个位置在给定距离内的所有作业的查询集，以最小距离对它们进行排序，并且不显示重复的作业。

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，我为过滤器建立了一个Q表达式，但是不确定是否有一种清晰的方法来标注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')

FYI使用带有PostGIS扩展名的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，通过从任意数量的表达式列表中选择最小值来起作用

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