I have a JsonField (col1) in a Model (model1) which contains values like below:
{"key_1":"val_1", "key_2":"{\"sub_key1\":sub_val1, \"sub_key2\":sub_val2}"}
I would like to aggregate sub_key1 and sub_key2. I tried below in Python but it does not work.
from django.db.models import FloatField, JSONField
from django.db.models import Sum, Value
from django.db.models.functions import Cast
## try 1
model1.objects.annotate(val=Cast('col1__key_2__sub_key1', FloatField())).aggregate(op=Sum('val'))
## try 2
model1.objects.annotate(
val=Cast('col1__key1', JSONField())).annotate(
val_2=Cast('val__sub_key1', FloatField())).aggregate(op=Sum('val_2'))
Is there a way I can aggregate over these sub keys?
Related
i'm trying to join together 2 QuerySets. Right now, I'm using the | operator, but doing it this way won't function as an "append".
My current code is:
df = RegForm((querysetA.all() | querysetB.all()).distinct())
I need the elements from querysetA to be before querysetB. Is it even possible to accomplish while keeping them just queries?
This can be solved by using annotate to add a custom field for ordering on the querysets, and use that in a union like this:
from django.db.models import Value
a = querysetA.annotate(custom_order=Value(1))
b = querysetB.annotate(custom_order=Value(2))
a.union(b).order_by('custom_order')
Prior to django-3.2, you need to specify the output_field for Value:
from django.db.models import IntegerField
a = querysetA.annotate(custom_order=Value(1, IntegerField()))
b = querysetB.annotate(custom_order=Value(2, IntegerField()))
I have a django model as follows:
class Order(models.Model):
cash=models.DecimalField(max_digits=11,decimal_places=2,default=0)
balance=models.DecimalField(max_digits=11,decimal_places=2,default=0)
current_ac=models.DecimalField(max_digits=11,decimal_places=2,default=0)
added_by = models.ForeignKey(User)
There can be multiple Orders and multiple users can create orders.
How can I get the sum of all orders for each columns for a particular user, something like
ord=Order.objects.filter(added_by.id=1).sum()
an SQL equivalent would be something like
Select sum(cash), sum (balance), sum(current_ac) from Orders where added_by = 1
You can aggregate, for example the sum of the current_ac with:
from decimal import Decimal
from django.db.models import Sum
ord=Order.objects.filter(added_by_id=1).aggregate(
total=Sum('current_ac')
)['total'] or Decimal()
or if you want to sum up the items for cash, balance and current_ac, you can work with:
from decimal import Decimal
from django.db.models import Sum
ord=Order.objects.filter(added_by_id=1).aggregate(
total_cash=Sum('current_ac'),
total_balance=Sum('balance'),
total_ac=Sum('current_ac')
)
here ord will be a dictionary that contains the corresponding values, for example:
{
'total_cash': Decimal('14.25'),
'total_balance': Decimal('13.02'),
'total_ac': Decimal('17.89')
}
or if you want to count the number of Orders, then we can work with:
from decimal import Decimal
ord=Order.objects.filter(added_by_id=1).count()
If you want to do that per User, it is more efficient to work with .annotate(…) [Django-doc].
If I get you correctly. You want to count the number of records, right? if that's the case. You can use filter and count. like in the example below:
numberOfRecords = Orders.filter(added_by=user_id).count
you can try this
from django.db.models import Sum
total=Order.objects.filter(added_by_id=1).aggregate(
total=Sum('current_ac')
)['total'] or 0
I have 2 tables
Class Billing(models.Model):
id=models.AutoField(primary_key=True)
.....
#Some more fields
....
Class BillInfo(models.Model):
id=models.AutoField(primary_key=True)
billing=models.ForeignKey(Billing)
testId=models.ForeignKey(AllTests)
costOfTest=models.IntegerField(default=0)
concession=models.IntegerField(default=0)
Here BillInfo is verticle table i.e one Billing has multiple BillInfo. Here I want to calculate the Sum(costOfTest - concession) for a single Billing.
Can I achieve this using single query?
Need help, Thanks in advance.
You can write this as:
from django.db.models import F, Sum
Billing.objects.annotate(
the_sum=Sum(F('billinfo__costOfTest') - F('billinfo__concession'))
)
Here every Billing object in this QuerySet will have an extra attribute .the_sum which is the sum of all costOfTests minus the concession of all related BillingInfo objects.
The SQL query that calculates this will look, approximately as:
SELECT billing.*
SUM(billinginfo.costOfTest - billinginfo.concession) AS the_sum
FROM billing
LEFT OUTER JOIN billinginfo ON billinginfo.billing_id = billing.id
GROUP BY billing.id
So when you "materialize" the query, the query will obtain the sum for all the Billing objects in a single call.
For Billing objects without any related BillingInfo, the the_sum attribute will be None, we can avoid that by using the Coalesce [Django-doc] function:
from django.db.models import F, Sum, Value
from django.db.models.functions import Coalesce
Billing.objects.annotate(
the_sum=Coalesce(
Sum(F('billinfo__costOfTest') - F('billinfo__concession')),
Value(0)
)
)
I'm using aggregate to get the count of a column of booleans. I want the number of True values.
DJANGO CODE:
count = Model.objects.filter(id=pk).aggregate(bool_col=Count('my_bool_col')
This returns the count of all rows.
SQL QUERY SHOULD BE:
SELECT count(CASE WHEN my_bool_col THEN 1 ELSE null END) FROM <table_name>
Here is my actual code:
stats = Team.objects.filter(id=team.id).aggregate(
goals=Sum('statistics__goals'),
assists=Sum('statistics__assists'),
min_penalty=Sum('statistics__minutes_of_penalty'),
balance=Sum('statistics__balance'),
gwg=Count('statistics__gwg'),
gk_goals_avg=Sum('statistics__gk_goals_avg'),
gk_shutout=Count('statistics__gk_shutout'),
points=Sum('statistics__points'),
)
Thanks to Peter DeGlopper suggestion to use django-aggregate-if
Here is the solution:
from django.db.models import Sum
from django.db.models import Q
from aggregate_if import Count
stats = Team.objects.filter(id=team.id).aggregate(
goals=Sum('statistics__goals'),
assists=Sum('statistics__assists'),
balance=Sum('statistics__balance'),
min_penalty=Sum('statistics__minutes_of_penalty'),
gwg=Count('statistics__gwg', only=Q(statistics__gwg=True)),
gk_goals_avg=Sum('statistics__gk_goals_avg'),
gk_shutout=Count('statistics__gk_shutout', only=Q(statistics__gk_shutout=True)),
points=Sum('statistics__points'),
)
Updated for Django 1.10. You can perform conditional aggregation now:
from django.db.models import Count, Case, When
query_set.aggregate(bool_col=Count(Case(When(my_bool_col=True, then=1))))
More information at:
https://docs.djangoproject.com/en/1.11/ref/models/conditional-expressions/#case
Update:
Since Django 1.10 you can:
from django.db.models import Count, Case, When
query_set.aggregate(
bool_col=Count(
Case(When(my_bool_col=True, then=Value(1)))
)
)
Read about the Conditional Expression classes
Old answer.
It seems what you want to do is some kind of "Conditional aggregation". Right now Aggregation functions do not support lookups like filter or exclude: fieldname__lt, fieldname__gt, ...
So you can try this:
django-aggregate-if
Description taken from the official page.
Conditional aggregates for Django queries, just like the famous SumIf and CountIf in Excel.
You can also first annotate the desired value for each team, I mean count for each team the ammount of True in the field you are interested. And then do all the aggregation you want to do.
Another Solution for count Bool is:
from django.db.models import Sum, IntegerField
from django.db.models.functions import Cast
Model.objects.filter(id=pk).annotate(bool_col=Sum(Cast('my_bool_col', IntegerField())))
Just convert False to 0 and True to 1, and then just Sum
I am using Django with MySQL. I have a model similar to the following:
class MM(models.Model):
a = models.IntegerField()
b = models.IntegerField()
c = models.DateTimeField(auto_now_add=True)
I have multiple rows that a is equal to b, and I want to perform the following SQL query:
SELECT a, b, MAX(c) AS max FROM MM GROUP BY b, a;
How can this be done with Django ORM? I have tried different approaches using annotations, but now luck so far.
Thanks a lot!
I think you can do something like:
MM.objects.all().values('b', 'a').annotate(max=Max('c'))
Note that you need to import something to use Max: from django.db.models import Max
values('b', 'a') will give GROUP BY b, a and annotate(...) will compute the MAX in your query.
You can try this also
from django.db.models import Max
mm_list=MM.objects.all().values('b','a').annotate(max=Max('c'))
for mm in mm_list:
a=mm['a']
b=mm['b']
max=mm['max']
Sum a field group by two fields.
from django.db.models import Sum
SQL
select caja_tipo_id, tipo_movimiento, sum(monto) from config_caja group by caja_tipo_id, tipo_movimiento
Django
objs = Caja.objects.values('caja_tipo__nombre','tipo_movimiento').order_by().annotate(total=Sum('monto'))