Django Sort ManyToMany Field in A Nonsignificant Order

Question

I have two models as below:

class Stop(models.Model):
    """
    Showing bus stops in İzmir.
    """

    code = models.PositiveIntegerField(
        unique=True,
        primary_key=True,
        verbose_name="Code"
    )

    label = models.CharField(
        null=False,
        blank=False,
        max_length=64,
        verbose_name="Label"
    )

    coor = ArrayField(
        models.FloatField(),
        size=2,
        verbose_name="Coordination"
    )

    class Meta:
        verbose_name = "Stop"
        verbose_name_plural = "Stops"
        ordering = ["label"]

    def __str__(self):
        return self.label

class Route(models.Model):
    """
    Bus routes of İzmir.
    """

    code = models.PositiveSmallIntegerField(
        unique=True,
        primary_key=True,
        verbose_name="Code"
    )

    stops = models.ManyToManyField(
        Stop,
        null=True,
        blank=True,
        related_name="routes",
        verbose_name="Stops"
    )

    terminals = ArrayField(
        models.CharField(
            null=False,
            blank=False,
            max_length=32,
        ),
        size=2,
        default=[],
        verbose_name="Terminals"
    )

    departure_times = ArrayField(
        ArrayField(
            models.TimeField(
                null=False,
                blank=False
            ),
            null=True,
            default=[]
        ),
        default=[],
        size=6,
        verbose_name="Departure Times"
    )

    class Meta:
        verbose_name = "Route"
        verbose_name_plural = "Routes"
        ordering = ["code"]

    def __str__(self):
        return "{}: {} - {}".format(str(self.code), self.terminals[0], self.terminals[1])

As you can see, Route has a ManyToManyFields which takes Stop instances.

I put the instances with a script which it scraps a couple of web page, it seems I will use crontab to keep them updated. In the data I am scraping, Stop objects are ordered. The thing is, there are no significant filter to sort e.g. a Stop instance comes after another.

Django (or Django Rest Framework) returns Stop instances of Route instance in alphabetic order, e.g.

{
    "code": 285,
    "terminals": [
        "EVKA 1",
        "KONAK"
    ],
    "stops": [
        40586,
        40633,
        12066,
        40645,
        40627,
        40647,
        40588,
        40592,
        40623,
        40016,
        40506,
        40508,
        40528,
        40462,
        40631,
        40014,
        40619,
        40530,
        12060,
        40661,
        40504,
        40488,
        40653,
        40590,
        40512,
        40464,
        10240,
        10036,
        12068,
        40514,
        40510,
        40658,
        40002,
        40649,
        12070,
        40004,
        40010,
        40656,
        12064,
        40614,
        40012
    ],
    ...
}

In which stops[0] returns a Stop instance beginning with A and sorts like that.

So, is there a way to order like a list in Python? Like, there is no significant point, you just append to the end and return so.

Environment

python 3.5.1
django 1.9.7
djangorestframework 3.3.3
psycopg2 2.6.2
postgresql 9.5

If you put a time stamp on the `Stop` model, then you could order based on that time stamp. — Aaron Lelevier, Jul 20 '16 at 21:05
The thing is, a `Stop` instance might belong to more than one Route objects. So even if I put timestamp to it, the `timestamp` field of a `Stop` instance would change when it comes to build another `Route` instance. — Eray Erdin, Jul 21 '16 at 05:06

Todor · Answer 1 · 2016-07-21T05:46:59.343

1

The position of a stop is relative to a Route, e.g. one stop can be first for route 1, 2nd for route 2 and etc. So this is a perfect example that you need more metadata about the Route-Stop relation. Djagno solves this by letting you provide a Intermediate Table with two ForeignKey and the metadata you need for the relation.

class Stop(models.Model):
    #...

class Route(models.Model):

    #...

    stops = models.ManyToManyField(Stop, through='RouteStop', blank=True, related_name="routes", verbose_name="Stops")


class RouteStop(models.Model):
    stop = models.ForeignKey(Stop)
    route = models.ForeignKey(Route)
    position = models.PositiveSmallIntegerField()

    class Meta:
        unique_together = (("stop", "route"),)

Now when you get Routes you can order route.stops by RouteStop.position, something like:

Route.objects.all().prefetch_related(
    Prefetch('stops', queryset=Stop.objects.all().order_by('routestop__position'))
)

edited Jul 21 '16 at 05:46

answered Jul 20 '16 at 21:44

Todor

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Should your `route = models.ForeignKey(Stop)` line be `route = models.ForeignKey(Route)`? Or maybe I am wrong and code is valid? – Eray Erdin Jul 21 '16 at 05:12
It also raises when I try to migrate: `ValueError: Cannot alter field izmir.Route.stops into izmir.Route.stops - they are not compatible types (you cannot alter to or from M2M fields, or add or remove through= on M2M fields)` – Eray Erdin Jul 21 '16 at 05:21
Yes, it should be `Route` (a typo clearly). About the migration, it won't happen automatically, you will need to make a [data migration](http://stackoverflow.com/questions/33257530/how-to-add-through-option-to-existing-manytomanyfield-with-migrations-and-data-i) – Todor Jul 21 '16 at 05:49

Django Sort ManyToMany Field in A Nonsignificant Order

Environment

1 Answers1