QuerySet Optimisations in Django - django

Question

I was just wondering, I have the following two pseudo-related queries:
organisation = Organisation.objects.get(pk=org_id)
employees = Employee.objects.filter(organisation=organisation).filter(is_active=True)
Each Employee has a ForeignKey relationship with Organisation.
I was wondering if there is anything I can leverage to do the above in one Query in the native Django ORM?
Also, would:
employees = Employee.objects.filter(organisation__id=organisation.id).filter(is_active=True)
Be a quicker way to fetch employees?
For Willem's reference, employees is then used as:
# Before constructing **parameters, it is neccessary to filter out any supurfluous key, value pair that do not correspond to model attributes:
if len(request.GET.getlist('gender[]')) > 0:
parameters['gender__in'] = request.GET.getlist('gender[]')
employees = employees.filter(**parameters)
if len(request.GET.getlist('age_group[]')) > 0:
parameters['age_group__in'] = request.GET.getlist('age_group[]')
employees = employees.filter(**parameters)
results = SurveyResult.objects.filter(
user__in=employees,
created_date__range=date_range,
).annotate(
date=TruncDate('created_date'),
).values(
'survey',
'date',
).annotate(
score=Sum('normalized_score'),
participants=Count('user'),
).order_by(
'survey',
'date',
)
I omitted this as it seemed like unnecessary complications to my original goal.

Answer 1

Also, would:
employees = Employee.objects.filter(organisation__id=organisation.id).filter(is_active=True)
Be a quicker way to fetch employees?
No, or perhaps marginally, since that is in essence what the Django ORM will do itself: it will simply obtain the primary key of the organisation and then make a query like the one you describe.
If you do not need the organisation itself, you can query with:
employees = Employee.objects.filter(organisation_id=org_pk, is_active=True)
Furthermore you can for example perform a .select_related(..) [Django-doc] on the organisation, to load the data of the organisation in the same query as the one of the employee, although reducing one extra query, usually does not make that much of a difference. Performance is more an issue if iut results in N+1 queries.
We can for example "piggyback" fetching the Organisation details with fetching the employees, like:
employees = list(
Employee.objects.select_related('organization').filter(
organisation_id=org_pk, is_active=True
)
)
if employees: # at least one employee
organization = employees[0].organization
But anyway, as said before the difference between one or two queries is not that much. It is usually more of a problem if you have N+1 queries. It is a bit of a pitty that Django/Python does not seem to have a Haxl [GitHub] equivalent, to enable fast retrieval of (remote) resources through algebraic analysis.
In case you are interested in the Employee servey results, you can query with:
results = SurveyResult.objects.filter(
user__organization_id=org_pk,
created_date__range=date_range,
).annotate(
date=TruncDate('created_date'),
).values(
'survey',
'date',
).annotate(
score=Sum('normalized_score'),
participants=Count('user'),
).order_by(
'survey',
'date',
)
You can thus omit a separate querying of Employees if you do not need these anyway.
You can furthermore add the filters to your query, like:
emp_filter = {}
genders = request.GET.getlist('gender[]')
if genders:
emp_filter['user__gender__in'] = genders
age_groups = request.GET.getlist('age_group[]')
if age_groups:
emp_filter['user__age_group__in'] = age_groups
results = SurveyResult.objects.filter(
user__organization_id=org_pk,
created_date__range=date_range,
**emp_filter
).annotate(
date=TruncDate('created_date'),
).values(
'survey',
'date',
).annotate(
score=Sum('normalized_score'),
participants=Count('user'),
).order_by(
'survey',
'date',
)

Answer 2

if you have a foreign key relation between organisation and employees then you can get the employees using the select_related like this:
employees = Employee.objects.selected_related('organisation').filter(is_active=True)
OR
organisation = Organisation.objects.get(pk=org_id)
employees =organisation.employee_set.all() #your_employee_model_name_set.all