How to filter a django model with latitude and longitude coordinates that fall within a certain radius

坚强是说给别人听的谎言 提交于 2019-12-20 14:21:32

问题


I have the following model.

class Location(models.Model):
    name = models.CharField(max_length = 128, blank = True)
    address =models.CharField(max_length = 200, blank= True)
    latitude = models.DecimalField(max_digits=6, decimal_places=3)
    longitude = models.DecimalField(max_digits=6, decimal_places=3)

    def __unicode__(self):
        return self.name

If my current latitude & longitude is:

current_lat = 43.648
current_long = 79.404

I did some research and came across the Haversine Equation which calculates the distance between two location coordinates. Below is the equation I found:

import math

def distance(origin, destination):
    lat1, lon1 = origin
    lat2, lon2 = destination
    radius = 6371 # km

    dlat = math.radians(lat2-lat1)
    dlon = math.radians(lon2-lon1)
    a = math.sin(dlat/2) * math.sin(dlat/2) + math.cos(math.radians(lat1)) \
        * math.cos(math.radians(lat2)) * math.sin(dlon/2) * math.sin(dlon/2)
    c = 2 * math.atan2(math.sqrt(a), math.sqrt(1-a))
    d = radius * c

    return d

I would like to return all the Location objects that fall within a 10 km radius, how can I filter it in such a way that it will only return all the Location objects that fall within this 10 km radius?

LocationsNearMe = Location.objects.filter(#This is where I am stuck)

Is there anyway I can implement the Haversine equation into the filtering so that it only returns the location objects that fall within a 10 km radius?

I'm looking for a well detailed answer. Appreciate the help.


回答1:


But you can always make proposed by Brian approach better by filtering the results from previous step (which hoepfully should be smaller subset) and for each you check either they are within the radius.

Your user is in black point. Square approximation given by Brian return green but also orange points. The divernce in distance can be significant in worst case user have to go sqrt(2) times further than expected (extra 40% of distance). So for all orange and green points it is worth to check if their distance from black point (e.g euclidian one if this are really short distances e.g navigation in city) is not greater than assumed radius.

UPDATE:

If you would like to use Haversine distance or (better) mentioned GeoDjango hava a look on this snippet comparing two django views dealing with nearby search:

https://gist.github.com/andilabs/4232b463e5ad2f19c155




回答2:


You can do range queries with filter.

LocationsNearMe = Location.objects.filter(latitude__gte=(the minimal lat from distance()),
                                          latitude__lte=(the minimal lat from distance()),
                                          (repeat for longitude))

Unfortunately, this returns results in the form of a geometric square (instead of a circle)




回答3:


If you don't want to use GeoDjango, then you could consider writing it out with Django's Database functions. In contrast to raw SQL, this also gives you the advantage of being able to easily append/prepend other ORM filters.

from django.db.models.functions import Radians, Power, Sin, Cos, ATan2, Sqrt, Radians
from django.db.models import F

dlat = Radians(F('latitude') - current_lat)
dlong = Radians(F('longitude') - current_long)

a = (Power(Sin(dlat/2), 2) + Cos(Radians(current_lat)) 
    * Cos(Radians(F('latitude'))) * Power(Sin(dlong/2), 2)
)

c = 2 * ATan2(Sqrt(a), Sqrt(1-a))
d = 6371 * c

LocationsNearMe = Location.objects.annotate(distance=d).order_by('distance').filter(distance__lt=10)


来源:https://stackoverflow.com/questions/17682201/how-to-filter-a-django-model-with-latitude-and-longitude-coordinates-that-fall-w

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