DRF：如何创建自定义 FilterSet 以按距离过滤最近的用户

Question

我正在尝试使用django-filter来创建用于通过距离过滤附近用户过滤附近用户的自定义 例如，如果我发送
想让所有附近的用户 等于

get/api/list/？dance = 300，我

latitude = models.DecimalField(  # [-90.000000, 90.000000]
    max_digits=8,
    decimal_places=6,
    null=True
)
longitude = models.DecimalField(  # [-180.000000, 180.000000]
    max_digits=9,
    decimal_places=6,
    null=True
)

objects = ClientManager()

300m > clientmanager 具有从模型中获取坐标的功能：

def get_geo_coordinates(self, pk):
    """
    :param pk: - client id
    :return: client's coords
    """

    instance = self.get(pk=pk)
    data = (instance.latitude, instance.longitude)
    return data

我的getListapiview

class GetClientListAPIView(ListAPIView):

    """
        Returns list with filtering capability
        Available filter fields:
            gender, first_name, last_name, distance
    """

    serializer_class = ClientSerializer
    queryset = Client.objects.all()
    permission_classes = [IsAuthenticated]
    filter_backends = [DjangoFilterBackend]
    filter_class = ClientFilter

我的clientfilter

class ClientFilter(FilterSet):

    distance = filters.NumberFilter(method='get_nearest_clients')

    def get_nearest_clients(self, queryset, name, value):
        sender_coords = Client.objects.get_geo_coordinates(pk=self.request.user.id)
        test_coords = Client.objects.get_geo_coordinates(pk=31)
        dist = get_great_circle_distance(sender_coords, test_coords)

    class Meta:
        model = Client
        fields = ['gender', 'first_name', 'last_name']

在这里，我正在使用我的功能来计算两个客户端之间的距离：

def get_great_circle_distance(first_coords, second_coords):
    """
        :param first_coords: (first_client_latitude, first_client_longitude) in degrees
        :param second_coords: (second_client_latitude, second_client_longitude) in degrees
        :return: distance
    """

    earth_radius = 6_400_000  # in metres

    la_1, lo_1 = map(radians, first_coords)
    la_2, lo_2 = map(radians, second_coords)

    coefficient = acos(
        cos(la_1) * cos(la_2) * cos(lo_1 - lo_2) +
        sin(la_1) * sin(la_2)
    )
    distance = earth_radius * coefficient

    return distance

我不知道如何过滤QuerySet，并从数据库访问侧面进行最佳操作。

Answer 1

I would recommend using the existing tools that solved this problem a long time ago. Doing accurate (and efficient) distance calculations on an irregularly shaped sphere is more complicated than this.

https://docs.djangoproject.com/en/4.0 /ref/conver/gis/install/postgis/

gis字段上的gis字段：

from django.contrib.gis.db.models import PointField

class Client(models.Model):
    location = PointField()

这为您提供了直接在querySet上进行距离计算的正确工具，并且在数据库侧AFAIK上进行了计算。（ https://djangoproject.com/44.0/ref /contrib/gis/tutorial/#spatial-queries)

It is a bit more overhead to setup GIS properly but it's worth the effort.

Sidenote：可以通过QuerySet Annotate（）和q和f表达方式手工做。过滤django-filter在您尝试的情况下，在客户端上过滤，几乎会失败的目的是首先使用django-filter。希望有助于获得更好的理解。

Answer 2

我解决了这个问题

class ClientFilter(FilterSet):
"""
     Custom ClientFilter
"""

distance = filters.NumberFilter(method='get_nearest_clients')

def get_nearest_clients(self, queryset: QuerySet, name: str, dist_value: int):
    sender_id = self.request.user.id
    sender_la, sender_lo = map(radians, Client.objects.get_geo_coordinates(pk=sender_id))
    earth_radius = 6_400_000

    queryset = (
        queryset.exclude(pk=sender_id)
        .alias(
            rad_lat=Radians('latitude'),
            rad_long=Radians('longitude'),
            distance=ExpressionWrapper(
                ACos(
                    Cos('rad_lat') * Cos(sender_la) * Cos(F('rad_long') - sender_lo)
                    + Sin('rad_lat') * Sin(sender_la)
                ) * earth_radius,
                output_field=FloatField()
            )
        )
        .exclude(distance__gte=dist_value)
        .order_by('distance')
    )

    return queryset