Django的按距離排序

Question

我有以下型號：

class Vacancy(models.Model): 
    lat = models.FloatField('Latitude', blank = True) 
    lng = models.FloatField('Longitude', blank = True) 

我應該如何進行查詢按距離排序（距離爲無窮大）？
使用PosgreSQL，GeoDjango（如果需要）。

謝謝。

Answer 1

首先，這是更好地做，而不是把經緯度和LNT分離點場：

from django.contrib.gis.db import models 

    location = models.PointField(null=False, blank=False, srid=4326, verbose_name="Location") 

然後，您可以像其過濾：

from django.contrib.gis.geos import * 
    from django.contrib.gis.measure import D 

    distance = 2000 
    ref_location = Point(1.232433, 1.2323232) 

    res = yourmodel.objects.filter(location__distance_lte=(ref_location, D(m=distance))).distance(ref_location).order_by('distance') 

已經有像stackoverlow中的幾個問題，如果你檢查它們會更好。

Answer 2

這是一個不需要GeoDjango使用自定義管理器的解決方案。

class LocationManager(models.Manager): 
    def nearby(self, latitude, longitude, proximity): 
     """ 
     Return all object which distance to specified coordinates 
     is less than proximity given in kilometers 
     """ 
     # Great circle distance formula 
     gcd = """ 
       6371 * acos(
       cos(radians(%s)) * cos(radians(latitude)) 
       * cos(radians(longitude) - radians(%s)) + 
       sin(radians(%s)) * sin(radians(latitude)) 
      ) 
       """ 
     return self.get_queryset()\ 
        .exclude(latitude=None)\ 
        .exclude(longitude=None)\ 
        .annotate(distance=RawSQL(gcd, (latitude, 
                longitude, 
                latitude)))\ 
        .filter(distance__lt=proximity)\ 
        .order_by('distance') 


class Location(models.Model): 
    objects = LocationManager() 
    latitude = models.FloatField() 
    longitude = models.FloatField() 
    ... 

用途如下：

eiffel_tower_5k = Location.objects.nearby(48.8582, 2.2945, 5) 

如果你使用SQLite，你需要的地方添加

from django.db.backends.signals import connection_created 
from django.dispatch import receiver 


@receiver(connection_created) 
def extend_sqlite(connection=None, **kwargs): 
    if connection.vendor == "sqlite": 
     # sqlite doesn't natively support math functions, so add them 
     cf = connection.connection.create_function 
     cf('acos', 1, math.acos) 
     cf('cos', 1, math.cos) 
     cf('radians', 1, math.radians) 
     cf('sin', 1, math.sin) 

Answer 3

的.distance(ref_location)在Django被刪除> = 1.9，你應該用註解來替代。

from django.contrib.gis.db.models.functions import Distance 
from django.contrib.gis.measure import D 

ref_location = Point(1.232433, 1.2323232, srid=4326) 
yourmodel.objects.filter(location__distance_lte=(ref_location, D(m=2000))).annotate(distance=Distance('location', ref_location)).order_by('distance') 

也應該縮小與使用空間索引的dwithin運營商的搜索，距離不使用減慢查詢了指數：

yourmodel.objects.filter(location__dwithin=(geom, 0.02)) 
    .filter(location__distance_lte=(ref_location, D(m=2000))) 
    .annotate(distance=Distance('location', ref_location)) 
    .order_by('distance') 

看到this post爲location__dwithin=(geom, 0.02)解釋

Answer 4

如果你不想/沒有機會使用gis，這裏是溶劑（django orm sql中的半徑距離函數寫入器）：

lat = 52.100 
lng = 21.021 

earth_radius=Value(6371.0, output_field=FloatField()) 

f1=Func(F('latitude'), function='RADIANS') 
latitude2=Value(lat, output_field=FloatField()) 
f2=Func(latitude2, function='RADIANS') 

l1=Func(F('longitude'), function='RADIANS') 
longitude2=Value(lng, output_field=FloatField()) 
l2=Func(longitude2, function='RADIANS') 

d_lat=Func(F('latitude'), function='RADIANS') - f2 
d_lng=Func(F('longitude'), function='RADIANS') - l2 

sin_lat = Func(d_lat/2, function='SIN') 
cos_lat1 = Func(f1, function='COS') 
cos_lat2 = Func(f2, function='COS') 
sin_lng = Func(d_lng/2, function='SIN') 

a = Func(sin_lat, 2, function='POW') + cos_lat1 * cos_lat2 * Func(sin_lng, 2, function='POW') 
c = 2 * Func(Func(a, function='SQRT'), Func(1 - a, function='SQRT'), function='ATAN2') 
d = earth_radius * c 

Shop.objects.annotate(d=d).filter(d__lte=10.0) 

PS 改款車型，變化濾波器ORDER_BY，更改關鍵字和參數化

PS2 爲sqlite3的，你應當確保有可用功能SIN，COS，弧度，ATAN2，SQRT