Develop Reference

How to compare what values two models/querysets share in a view for later display in the template?

I'm trying to compare all the things in ModelOne with ModelTwo, to check which things are or are not in one or the other model, then put this in the view context for display in the template.
class Things(model.Model):
name = models.CharField()
class ModelOne(models.Model):
things = models.ManyToManyField(Things)
class ModelTwo(models.Model):
things = models.ManyToManyField(Things)
How would you do this?

one_instance = ModelOne.objects.get(id=one_id)
two_instance = ModelTwo.objects.get(id=two_id)
one_thing_ids = set(one_instance.things.values_list("id", flat=True))
two_thing_ids = set(two_instance.things.values_list("id", flat=True))
shared_thing_ids = one_thing_ids & two_thing_ids
thing_ids_in_one_not_in_two = one_thing_ids - two_thing_ids
thing_ids_in_two_not_in_one = two_thing_ids - one_thing_ids
shared_things = Thing.objects.filter(id__in=shared_thing_ids)
You can then pass shared_things queryset into the template for display.
If your Thing model only has a name field and the names are unique we can simplify a little by altering the model:
class Things(model.Model):
name = models.CharField(unique=True)
or even:
class Things(model.Model):
name = models.CharField(primary_key=True, unique=True)
(in this case the db table will not have an id column, it's not needed)
Either way we can then eliminate the extra Thing query at the end:
one_instance = ModelOne.objects.get(id=one_id)
two_instance = ModelTwo.objects.get(id=two_id)
one_thing_names = set(one_instance.things.values_list("name", flat=True))
two_thing_names = set(two_instance.things.values_list("name", flat=True))
shared_thing_names = one_thing_names & two_thing_names
thing_names_in_one_not_in_two = one_thing_names - two_thing_names
thing_names_in_two_not_in_one = two_thing_names - one_thing_names
...and just pass sets of string names into the template.

How to add extra key-value to QuerySet lists?

I am working with Django,i need to retrieve data from multiple database, which has different database name but with same table column structure.
So I use model.using(database).all()to get queryset and merge them into one.
I want to add extra databasename to indicate the data's database name, this is my code.
model:
class Sections(models.Model):
apply_id = models.PositiveIntegerField()
pathology_id = models.CharField(max_length=128)
user_id = models.PositiveIntegerField()
updated_at = models.DateTimeField(blank=True, null=True)
get_queryset:
def get_queryset(self):
slideset = []
database_config = ['database1', 'database2', 'database3']
for i, x in database_config:
slides = Sections.objects.using(x).all()
#### I want to add extra databasename column in every query object.
for x1 in slides:
x1.databasename = x
######
slideset.append(slides)
# merge QuerySet
query = functools.reduce(lambda a, b: a|b, slideset)
return query.order_by("updated_at").reverse()
the one return will be :
{
"apply_id": 1123,
"pathology_id": 1235,
"user_id": 1,
"updated_at": "202106011430",
# add extra databasename.
"databasename": "database1".
}
Because the column can't be modify, so I had to leave Sections model unchange, just add extra key-value to query, can someone help me on that?

thanks to #Abdul Aziz Barkat
use annotate
from django.db.models import CharField, Value
slides = Sections.objects.using(x).annotate(databasename=Value(databasename, output_field=CharField())

MongoEngine Query Optimization

I have two collections ScenarioDrivers and ModelDrivers which has One to Many relationship with each other.
class ScenarioDrivers(Document):
meta = {
'collection': 'ScenarioDrivers'
}
ScenarioId = ReferenceField('ModelScenarios')
DriverId = ReferenceField('ModelDrivers')
DriverCalibrationMethod = StringField()
SegmentName = StringField()
DriverValue = ListField()
CalibrationStatus = StringField()
AdjustedValues = ListField(default=[])
CreateDate = DateTimeField(default=ObjectId().generation_time)
LastUpdateDate = DateTimeField(default=datetime.utcnow())
class ModelDrivers(Document):
meta = {
'collection': 'ModelDrivers'
}
PortfolioModelId = ReferenceField('PortfolioModels')
DriverName = StringField()
CreateDate = DateTimeField(default=ObjectId().generation_time)
LastUpdateDate = DateTimeField(default=datetime.utcnow())
FieldFormat = StringField()
DriverData = ListField()
My query is like this.
class GetCalibratedDrivers(Resource):
def get(self, scenario_id):
scenario_drivers_list = []
scenario_drivers = ScenarioDrivers.objects(ScenarioId=scenario_id).exclude('ScenarioId').select_related(1)
for scenario_driver in scenario_drivers:
scenario_driver_dict = {
'id': str(scenario_driver.id),
'DriverId': str(scenario_driver.DriverId.id),
'SegmentName': scenario_driver.SegmentName,
'CalibrationMethod': scenario_driver.DriverCalibrationMethod,
'CalibratedValues': exchange(scenario_driver.DriverValue),
'AdjustedValues': scenario_driver.AdjustedValues,
'LastUpdateDate': formatted_date(scenario_driver.LastUpdateDate),
'FieldFormat': scenario_driver.DriverId.FieldFormat
}
scenario_drivers_list.append(scenario_driver_dict)
return {
'DriverCalibrations': scenario_drivers_list
}
The Query matches 1140 records and then I construct a dictionary and make it a list.
But this API call takes 30s to process just 1140 records. Where I am missing? Please help. I am using latest version of Pymongo and MongoEngine.

I think the problem is not with your query, it is with you looping over 1140 records. I do not see any use of referenced objects so you should consider removing select_related(1). Once you do that, if you want to convert reference object ids to string, you can use as_pymongo() which will do that by default for you. And finally if you must read some data in specific format like formatted_date or exchange, it is better to save them as part of your document. i.e. save FormattedLastUpdateDate with LastUpdateDate. In MongoDB, you have to think about your read specific logic when you save the document.

django - can't assign a foreign key

For unknown reasons, I cannot assign a foreign key instance of Item_rarity table into Detailed_item table. Django throws an error:
Cannot assign "u'Basic'": "Detailed_item.rarity" must be a "Item_rarity" instance.
... But in Item_rarity dictionary "Basic" record exists - I can choose it from admin panel and create Detailed_item record manually.
I have defined models:
class Detailed_item(models.Model):
item_id = models.IntegerField(primary_key=True)
name = models.CharField(max_length=50)
level = models.IntegerField()
icon = models.CharField(max_length=150)
rarity = models.ForeignKey('Item_rarity')
general_type = models.ForeignKey('Item_type')
detailed_type = models.ForeignKey('Item_detailed_type')
class Item_rarity(models.Model):
name = models.CharField(max_length=15, primary_key=True)
class Item_type(models.Model):
name = models.CharField(max_length=15, primary_key=True)
class Item_detailed_type(models.Model):
name = models.CharField(max_length=20, primary_key=True)
In views, I try to populate it in this manner (inserting multiple items):
...
items = get_all_items() #get dict of items
for element in items:
tmp_det_type = ''
for key, val in element.iteritems():
#get 'detailed type' from inner dict
if key == "type":
tmp_det_type = val
item = Detailed_item(
item_id=element['id'],
name=element['name'],
level=element['level'],
icon=element['icon'],
rarity=element['rarity'], #error
general_type=element['type'],
detailed_type=tmp_det_type,
)
item.save()
...
I even tried to hard code "Basic" string, but it doesn't work either.
* Solved *
Next two entries, that is Item_type and Item_detailed_type were also invalid.
Correct code:
from app.models import Detailed_item, Item_rarity, Item_type, Item_detailed_type
...
items = get_all_items() #get dict of items
for element in items:
tmp_det_type = ''
for key, val in element.iteritems():
#get 'detailed type' from inner dict
if key == "type":
tmp_det_type = val
#create objects with string values
obj_rarity = Item_rarity(name=element['rarity'])
obj_item_type = Item_type(name=element['type'])
obj_item_detailed_type = Item_detailed_type(name=tmp_det_type)
item = Detailed_item(
item_id=element['id'],
name=element['name'],
level=element['level'],
icon=element['icon'],
rarity=obj_rarity,
general_type=obj_item_type,
detailed_type=obj_item_detailed_type,
)
item.save()
...

Item_rarity instance should be passed while storing Detailed_item object since Item_rarity is a foreign key related object in Detailed_item.
Its that you might have passed the Basic string instead of the <Basic Object> itself.
While creating an object in django using its ORM, any foreign_key related object should be provided with the instance itself instead of the id(pk) of the object, where as while fetching the data from the database you can use either of instance or the id(pk) of the instance.
class ParentModel(models.Model):
model_field = models.CharField(max_length=16)
class MyModel(models.Model):
some_field = models.ForeignKey('ParentModel')
parent_model = ParentModel.objects.create(model_field='some_data')
my_model = MyModel.objects.create(some_field=parent_model)
^^^^^^^^^^^^
Note here that the parent_model object itself is passed instead of the id
While fetching the data back,
parent_model = ParentModel.objects.get(model_field='some_data')
my_model = MyModel.objects.get(some_field=parent_model)
or
my_model = MyModel.objects.get(some_field=parent_model.id)
Both would work in case of data fetch.

You do not have to provide the related object on creation if you change the kwarg in to rarity_name:
item = Detailed_item(
item_id=element['id'],
name=element['name'],
level=element['level'],
icon=element['icon'],
rarity_name=element['rarity'], # no error
general_type=element['type'],
detailed_type=tmp_det_type,
)
I have only tested this with the regular id field (the auto pk) but it
should work with your primary key just fine.
E.g.
class SimpleModel(Model):
value = TextField(blank=True)
class ComplexModel(Model):
simple = ForeingKey(SimpleModel)
title = TextField(unique=True)
ComplexModel.objects.create(title='test', simple_id=1)

I need to query for a set of objects whose primary keys are contained inside of a list

As the title says, I need a way to perform this query. I have tried the following:
user_list_ids = []
user_lists = []
user_entries = OwnerEntry.objects.filter(name=request.user)
for user in user_entries:
user_list_ids.append(user.list_id)
user_lists = ListEntry.objects.filter(id__in=user_list_ids)
for user in user_entries:
user_list_ids.append(user.list_id)
user_lists = ListEntry.objects.filter(id__in=user_list_ids)
However, I get an error on the last line: int() argument must be a string or a number, not 'ListEntry'
Here are the relevant models:
class OwnerEntry(models.Model):
name = models.CharField(max_length=32)
list_id = models.ForeignKey(ListEntry)
class Meta:
ordering = ('name',)
class ListEntry(models.Model):
name = models.CharField(max_length=64)
# active_date = models.DateTimeField('date of last list activity')
expire_date = models.DateField('date of expiration')
create_date = models.DateField('date created')

to answer your question directly, please note that you have a list_id rather than list as a ForeignKey name (OwnerEntry model). In order to actually extract the fk value, you should use list_id_id instead (or rename list_id to list ;))
Please also note that django supports object references, like so:
someowner = OwnerEntry.objects.get( ... )
ownerslist = someowner.listentry_set.all()
cheers!

You can define OwnerEntry's foreign key to ListEntry as :
list_id = models.ForeignKey(ListEntry, related_query_name='owner_entry')
and then do this one-liner in your code:
user_lists = ListEntry.objects.filter(owner_entry__name=request.user)
What this does is exactly filter every ListEntry which has at least one owner_entry whose name is equal to request.user's.
The redefinition of the foreign key is just for the sake of giving a nice name to the query attribute.
For more details on queries that work with backward relationships: https://docs.djangoproject.com/en/dev/topics/db/queries/#lookups-that-span-relationships