Using aggregation on subquery in Django ORM - sql

Question

I have a table like this
Category
Subcategory
Sub_subcategory
Cat_1
Subcat_1
Sub_subcat_1
Cat_1
Subcat_1
Sub_subcat_2
Cat_1
Subcat_2
Sub_subcat_3
Cat_2
Subcat_2
Sub_subcat_4
Cat_3
Subcat_3
Sub_subcat_5
And I need to find in how many categories each subcategory appears.
So my expected output based on above table would be:
Subcategory
Total
Subcat_2
2
Subcat_1
1
Subcat_3
1
So I can get that by running following SQL query:
SELECT subcategory, count(*) total FROM (
SELECT DISTINCT subcategory, category FROM table_1
) as temp_table GROUP BY subcategory ORDER BY total DESC
I spent a lot of time trying to get the same result with Django ORM but wasn't able to get it done.
I expected this code to work:
subquery = Table1.objects.values('subcategory', 'category').distinct()
results = subquery.annotate(total=Count('*')).values('subcategory', 'total').order_by('-total')
But it works exactly the same as without 'distinct()' in subquery, so it counts all categories for every subcategory.
I also tried to find similar case in other questions, but those with subqueries usually relate to JOINing tables and using OuterRef, here it is more like getting results based on temporary table that is created by subquery.
Does anyone know how can I achieve that (or if it's even possible)?

Answer 1

this should do the job
dictionary = {}
queryset = YOUR_queryset_with_table_HERE.objects.all()
for i in queryset:
subcategory = i.subcategory
#check if subcategory already in dictionary
#if not create it with 1, else +=1
if dictionary["subcategory"]:
dictionary["subcategory"] += 1
else:
dictionary["subcategory"] = 1
print(dictionary)

Answer 2

Say you have two models in a many-to-many relationship similar to this:
class Category(models.Model):
category_name = models.CharField(max_length=30)
class SubCategory(models.Model):
subcategory_name = models.CharField(max_length=30)
category = models.ManyToManyField(Category)
Then this should give you the total number of categories in which each subcategory appears:
results = SubCategory.objects.all().annotate(total=Count('category')).values('subcategory_name', 'total')