如何获得Django .filter（something__in = some_set）查询的* UN *匹配条件？

Question

Not infrequently, I find myself with a set of identifiers and want to retrieve all the objects in a table that match any of the identifiers, but also want to see all the identifiers that didn't have any matched objects in the database.

The way I do this now is:

some_ids = ("a", "b", "c")
matched_objects = MyModel.objects.filter(my_key__in=some_ids)
caught_ids = set()
for obj in matched_objects:
    caught_ids.add(obj.my_key)
unmatched_ids = set(some_ids) - caught_ids

This feels very verbose. Is there a better way to do this?

Answer 1

For the second part of your question get distict values for your my_key field

caught_ids = matched_objects.values_list('my_key').distinct()

then get unmatched_ids just like you did before.

unmatched_ids = set(some_ids) - set(caught_ids)

Your code would look like this:

some_ids = ("a", "b", "c")
matched_objects = MyModel.objects.filter(my_key__in=some_ids)
caught_ids = matched_objects.values_list('my_key').distinct()
unmatched_ids = set(some_ids) - set(caught_ids)

Answer 2

# Get a list of keys that match your IDs

some_ids = set(some_ids)
keys = set(MyModel.objects.filter(my_key__in = some_ids).values_list('my_key',flat=True))

Presumably keys will contain all those IDs that existed in some_ids . Now we can use typical set operations to identify those elements that match or don't match. There are a number of examples here ( http://docs.python.org/2/library/sets.html#set-objects ) but from your comment, it sounds like you want:

unmatched = some_ids not in keys