Set Django IntegerField by choices=… name

Question

When you have a model field with a choices option you tend to have some magic values associated with human readable names. Is there in Django a convenient way to set these f

Answer 1

I'd probably set up the reverse-lookup dict once and for all, but if I hadn't I'd just use:

thing.priority = next(value for value, name in Thing.PRIORITIES
                      if name=='Normal')

which seems simpler than building the dict on the fly just to toss it away again;-).

Answer 2

My answer is very late and might seem obvious to nowadays-Django experts, but to whoever lands here, i recently discovered a very elegant solution brought by django-model-utils: https://django-model-utils.readthedocs.io/en/latest/utilities.html#choices

This package allows you to define Choices with three-tuples where:

• The first item is the database value
• The second item is a code-readable value
• The third item is a human-readable value

So here's what you can do:

from model_utils import Choices

class Thing(models.Model):
    PRIORITIES = Choices(
        (0, 'low', 'Low'),
        (1, 'normal', 'Normal'),
        (2, 'high', 'High'),
      )

    priority = models.IntegerField(default=PRIORITIES.normal, choices=PRIORITIES)

thing.priority = getattr(Thing.PRIORITIES.Normal)

This way:

• You can use your human-readable value to actually choose the value of your field (in my case, it's useful because i'm scraping wild content and storing it in a normalized way)
• A clean value is stored in your database
• You have nothing non-DRY to do ;)

Enjoy :)

Answer 3

Do as seen here. Then you can use a word that represents the proper integer.

Like so:

LOW = 0
NORMAL = 1
HIGH = 2
STATUS_CHOICES = (
    (LOW, 'Low'),
    (NORMAL, 'Normal'),
    (HIGH, 'High'),
)

Then they are still integers in the DB.

Usage would be thing.priority = Thing.NORMAL

Answer 4

As of Django 3.0, you can use:

class ThingPriority(models.IntegerChoices):
    LOW = 0, 'Low'
    NORMAL = 1, 'Normal'
    HIGH = 2, 'High'


class Thing(models.Model):
    priority = models.IntegerField(default=ThingPriority.LOW, choices=ThingPriority.choices)

# then in your code
thing = get_my_thing()
thing.priority = ThingPriority.HIGH

Answer 5

Originally I used a modified version of @Allan's answer:

from enum import IntEnum, EnumMeta

class IntegerChoiceField(models.IntegerField):
    def __init__(self, choices, **kwargs):
        if hasattr(choices, '__iter__') and isinstance(choices, EnumMeta):
            choices = list(zip(range(1, len(choices) + 1), [member.name for member in list(choices)]))

        kwargs['choices'] = choices
        super(models.IntegerField, self).__init__(**kwargs)

    def to_python(self, value):
        return self.choices(value)

    def get_db_prep_value(self, choice):
        return self.choices[choice]

models.IntegerChoiceField = IntegerChoiceField

GEAR = IntEnum('GEAR', 'HEAD BODY FEET HANDS SHIELD NECK UNKNOWN')

class Gear(Item, models.Model):
    # Safe to assume last element is largest value member of an enum?
    #type = models.IntegerChoiceField(GEAR, default=list(GEAR)[-1].name)
    largest_member = GEAR(max([member.value for member in list(GEAR)]))
    type = models.IntegerChoiceField(GEAR, default=largest_member)

    def __init__(self, *args, **kwargs):
        super(Gear, self).__init__(*args, **kwargs)

        for member in GEAR:
            setattr(self, member.name, member.value)

print(Gear().HEAD, (Gear().HEAD == GEAR.HEAD.value))

Simplified with the django-enumfields package package which I now use:

from enumfields import EnumIntegerField, IntEnum

GEAR = IntEnum('GEAR', 'HEAD BODY FEET HANDS SHIELD NECK UNKNOWN')

class Gear(Item, models.Model):
    # Safe to assume last element is largest value member of an enum?
    type = EnumIntegerField(GEAR, default=list(GEAR)[-1])
    #largest_member = GEAR(max([member.value for member in list(GEAR)]))
    #type = EnumIntegerField(GEAR, default=largest_member)

    def __init__(self, *args, **kwargs):
        super(Gear, self).__init__(*args, **kwargs)

        for member in GEAR:
            setattr(self, member.name, member.value)

Answer 6

Model's choices option accepts a sequence consisting itself of iterables of exactly two items (e.g. [(A, B), (A, B) ...]) to use as choices for this field.

In addition, Django provides enumeration types that you can subclass to define choices in a concise way:

class ThingPriority(models.IntegerChoices):
    LOW = 0, _('Low')
    NORMAL = 1, _('Normal')
    HIGH = 2, _('High')

class Thing(models.Model):
    priority = models.IntegerField(default=ThingPriority.NORMAL, choices=ThingPriority.choices)

Django supports adding an extra string value to the end of this tuple to be used as the human-readable name, or label. The label can be a lazy translatable string.

   # in your code 
   thing = get_thing() # instance of Thing
   thing.priority = ThingPriority.LOW

Note: you can use that using ThingPriority.HIGH, ThingPriority.['HIGH'], or ThingPriority(0) to access or lookup enum members.