group by two values and get the third values - django

Question

I have this Django model with three CharFields, which I want to run a query on to get the existing values for the two of them, and for each combination get the existing values of the third field.
a = models.CharField(null=False, max_length=8000)
b = models.CharField(null=False, max_length=8000)
c = models.CharField(null=False, max_length=8000)
if assume that these values are in the database:
a | b | c |
---------------
a1 | b2 | c3 |
a1 | b2 | c1 |
a2 | b2 | c3 |
a1 | b3 | c3 |
a1 | b2 | c2 |
I want some result in this form :
{"a1-b2" : [c3, c1, c2], "a2-b2" : [c3], "a1-b3" : [c3]}
or
{"a1" : {"b2":[c3, c1, c2], "b3": [c3]}, "a2": {"b2" : [c3]}}

Answer 1

TLDR:
items = MyModel.objects.annotate(custom_field=Concat('a', Values('-'), 'b').values('custom_field', 'c')
Explanation
With the part .annotate(custom_field=Concat('a', Values('-'), 'b'), you are basically doing a group_by operation in SQL and creating a temporary new column with name custom_field in your queryset which will have the value of a-b.
This gives you the following structure:
a | b | c | custom_field
a1 b1 c1 a1-b1
a2 b2 c2 a2-b2
a1 b1 c3 a1-b1
The .values('custom_field', 'c') portion fetches only the custom_field and c columns from this queryset. Now all you have to do is serialize your data.
EDIT
If you want your data in that specific format, you can concatenate column c. Please read the SO accepted answer in this post. Django making a list of a field grouping by another field in model. You can then create a new field during serialization which will split() the concatenated c field into a list.

Answer 2

Couldn't think of good pure SQL solution, but here's pythonic one using groupby:
from itertools import groupby
# Order by key fields so it will be easier to group later
items = YOUR_MODEL.objects.order_by('a', 'b')
# Group items by 'a' and 'b' fields as key
groups = groupby(items, lambda item: (item.a, item.b))
# Create dictionary with values as 'c' field from each item
res = {
'-'.join(key): list(map(lambda item: item.c, group))
for key, group in groups
}
# {'a1-b2': ['c3', 'c1', 'c2'], 'a1-b3': ['c3'], 'a2-b2': ['c3']}