Django: How to get parameters from QuerySet

Question

Consider the following Django model:

class SomeModel(models.Model):
    some_var = models.IntergerField()

Whats the best way to get an array of a models parameters from a search results?

Example (not working of course):

a = SomeModel.objects.all().some_var

would give something like [3, 4, 1, 9, 1, 2]

Sardorbek Imomaliev · Accepted Answer · 2016-11-07 10:57:18Z

5

There is existing queryset method called values_list https://docs.djangoproject.com/en/1.10/ref/models/querysets/#values-list

a = SomeModel.objects.all().values_list('some_var', flat=True)

We are using flat=True so that it will give flat list

Entry.objects.values_list('id').order_by('id')
[(1,), (2,), (3,), ...]
Entry.objects.values_list('id', flat=True).order_by('id')
[1, 2, 3, ...]

edited Nov 7, 2016 at 10:57

answered Nov 7, 2016 at 10:43

Sardorbek Imomaliev

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6 Comments

Prakhar Trivedi Over a year ago

That's correct,So the upvote.Wondering, What will be the structure of result?? Comma separated list maybe..??

Moinuddin Quadri Over a year ago

@PrakharTrivedi: Yes, a list (which are comma separated ;) ) of values in some_var column

Назар Топольський Over a year ago

@PrakharTrivedi strictly speaking, the result won't be a list, it would be a queryset. If all you want to do is to iterate over it or pass as a value to some other queryset.filter(), it will behave just the same. But if you want to concatenate it to some other list, use negative indexing (like qs[-2]), you would have to explicitly convert it to a list.

Sardorbek Imomaliev Over a year ago

@PrakharTrivedi Actually it will be QueryList not QuerySet.

Назар Топольський Over a year ago

@SardorbekImomaliev I can't find anything named QueryList in Django (1.8.13) code. And calling type(qs.values_list()) yields ValuesListQuerySet. So, while it is not a simple QuerySet, it is a subclass.

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Fontinalis · Accepted Answer · 2016-11-07 12:51:57Z

0

You can yield the list, this would like the following:

foo = [i.some_var for i in SomeModel.objects.all()]

But I've been notified that this would waste the memory.

edited Nov 7, 2016 at 12:51

answered Nov 7, 2016 at 10:50

Fontinalis

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1 Comment

Назар Топольський Over a year ago

List comprehension yields a list, not a generator. And if SomeModel has many fields, this would be wasteful in terms of RAM, since Django would have to load everything into memory.

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