Efficiently update multiple fields in Django

Question

Django beginner here. I have a many-to-many model and want to update multiple values at once.

This is the model (simplified):

class Inventory(models.Model):
    item = models.ForeignKey(Items, models.CASCADE)
    player = models.ForeignKey(Players, models.CASCADE)
    count = models.PositiveIntegerField()

What I do right now is this:

for key in dict:
    change_value = Inventory.objects.get(player=player, item=key)
    change_value.count += dict[key]
    change_value.save()

At least on my HDD test environment this is very slow. Is there a more efficient way of doing this? In other posts it was suggested to use update instead, so this would be:

for key in dict:
    Inventory.objects.get(player=player, item=key).update(count=dict[key])

Would this be faster? Could this be improved even further by updating multiple entries at once? Also, is there an efficient way of adding multiple new entries?

willeM_ Van Onsem · Accepted Answer · 2020-04-13 15:24:28Z

1

It is better to use .filter(..) over .get(..) since then the number of queries is half, so:

for k, v in data.items():
    Inventory.objects.filter(player=player, item=k).update(count=v)

Alternatively you can even perform the mapping in a single query with:

from django.db.models import Case, PositiveIntegerField, Value, When

Inventory.objects.filter(player=player, item__in=data).update(
    count=Case(
        *[When(item=k, then=Value(v)) for k, v in data.items()]
        output_field=PositiveIntegerField()
    )
)

edited Apr 13, 2020 at 15:24

answered Apr 13, 2020 at 15:17

willeM_ Van Onsem

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