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Django get total count and count by unique value in queryset

I have models Software and Domain described loosely as:
class Software(models.Model)
id = models.BigInteger(primary_key=True, db_index=True, null=False)
company = models.ForeignKey('Company')
domain = models.ForeignKey('Domain')
type = models.CharField(null=False)
vendor = models.CharField(null=False)
name = models.CharField(null=False)
class Domain(models.Model):
id = models.BigInteger(primary_key=True, db_index=True, null=False)
type = models.CharField()
importance = models.DecimalField(max_digits=11, decimal_places=10, null=False)
And I get a Software queryset with:
qs = Software.objects.filter(company=c).order_by('vendor')
The desired output should have an aggregated Domain importance with total count for each unique Software, i.e.
[
{
'type': 'type_1', \
'vendor': 'ajwr', | - unique together
'name': 'nginx', /
'domains': {
'total_count': 4,
'importance_counts': [0.1: 1, 0.5: 2, 0.9: 1] # sum of counts = total_count
},
},
{
...
},
]
I feel like the first step here should be to just group the type, vendor, name by Domain so each Software object has a list of Domains instead of just one but I'm not sure how to do that. Doing this in memory would make it a lot easier but it seems like it would be a lot slower than using querysets / SQL.

So I would do it like this:
from django.db.models import Sum
qs = Software.objects.filter(company=c).prefetch_related(
'domain'
).annotate(
total_count=Sum('domain__importance')
).order_by('vendor')
output = []
for obj in qs:
domains = obj.domain.all() # using prefetched domains, no db query
output.append({
# ...
'domains': {
'total_count': obj.total_count,
'importance_counts': [d.importance for d in domains]
}
})
And I belive it should be fast enough. Only if finding that it isn't I would try to improve. Remember "Premature optimization is the root of all evil"

Python SQLAlchemy TypeError: unhashable type: 'dict' when trying to instantiate model with one-to-many relationship

I'm trying to use SQLAlchemy to create a model which contains a one-to-many relationship. One recipe may have many directions associated with it. However, when I try to instantiate a recipe I get TypeError: unhashable type: 'dict'. If I remove the directions argument everything works fine and it creates the recipe without any directions. Is there something I'm missing that won't allow the directions parameter to be a list?
app.py
data = {
'cook_time': 15,
'description': 'Recipe description',
'directions': [{'order': 1, 'text': 'First direction'},
{'order': 2, 'text': 'Second direction'}],
'image_url': 'https://via.placeholder.com/800x300?text=Recipe+Image',
'name': 'Test recipe 2',
'prep_time': 15,
'servings': 6
}
recipe = models.Recipe(
name=data['name'],
description=data['description'],
image_url=data['image_url'],
prep_time=data['prep_time'],
cook_time=data['cook_time'],
servings=data['servings'],
directions=data['directions']
)
models.py
class Recipe(db.Model):
__tablename__ = 'recipes'
id = db.Column(db.Integer, primary_key=True)
name = db.Column(db.String(200), index=True)
description = db.Column(db.String(2000))
image_url = db.Column(db.String(200))
prep_time = db.Column(db.Integer)
cook_time = db.Column(db.Integer)
servings = db.Column(db.Integer)
directions = db.relationship('RecipeDirection', backref='recipes', lazy='dynamic')
class RecipeDirection(db.Model):
__tablename__ = 'recipe_directions'
id = db.Column(db.Integer, primary_key=True)
recipe_id = db.Column(db.Integer, db.ForeignKey('recipes.id'))
order = db.Column(db.Integer)
text = db.Column(db.String(1000))

You are getting the error because SQLAlchemy is expecting directions to be a list of RecipeDirection. To fix, create a list of RecipeDirection first.
data = {
'cook_time': 15,
'description': 'Recipe description',
'directions': [{'order': 1, 'text': 'First direction'},
{'order': 2, 'text': 'Second direction'}],
'image_url': 'https://via.placeholder.com/800x300?text=Recipe+Image',
'name': 'Test recipe 2',
'prep_time': 15,
'servings': 6
}
# Create a list of `RecipeDirection`
directions = []
for direction in data.get("directions", []):
directions.append(models.RecipeDirection(**direction))
recipe = models.Recipe(
name=data['name'],
description=data['description'],
image_url=data['image_url'],
prep_time=data['prep_time'],
cook_time=data['cook_time'],
servings=data['servings'],
directions=directions # Now list of RecipieDirection not list of dicts
)
I would also suggest looking into a serilizer that will take care of some of the details of marshalling and serilizing nested data structures for you, such as marshmallow-sqlalchemy

python django sort with lambda with if statement

I have date and some dollar gross model:
class FirstDate(models.Model):
gross = models.DecimalField(max_digits=12, decimal_places=2, default=0)
updated = models.DateTimeField(auto_now=True)
class SecondDate(models.Model):
gross = models.DecimalField(max_digits=12, decimal_places=2, default=0)
updated = models.DateTimeField(auto_now=True)
And want to sort it by gross, and if gross is the same, then sort it with updated field
For example,
qs1 = SoloDate.objects.all()[:2]
qs2 = GroupDate.objects.all()[:2]
result_list = sorted(
chain(qs1, qs2),
key=lambda x: x.gross # and if gross is the same, for the gross same objects, x.updated and then update was also the same, x.pk,
reverse=True
)
I mean, let me say that there is two objects individually from qs1 and qs2.
# objects from qs1
qs1_obj1 = {
'pk': 1,
'gross': 5,
'updated': 2018-11-24 10:53:23.360707+00:00
}
qs1_obj2 = {
'pk': 2,
'gross': 5,
'updated': 2018-11-25 10:53:23.360707+00:00
}
# objects from qs2
qs2_obj1 = {
'pk': 3,
'gross': 5,
'updated': 2018-11-24 10:53:23.360707+00:00
}
qs2_obj2 = {
'pk': 4,
'gross': 1,
'updated': 2018-11-23 10:53:23.360707+00:00
}
It's result_list order will be qs1_obj1, qs2_obj1, qs1_obj2, qs_2_obj_2.
Reasons is it:
qs1_obj1: 1.by gross, 2.by updated, 3.by pk,
qs2_obj1: 1.by gross, 2.by updated, 3. but pk was not good,
qs1_obj2: 1.by gross, 2.but by dpdated was late,
qs2_obj2: 1.gross was small.
Maybe it is not good question or bothersome question, I need help.
Question line is that :
key=lambda x: x.gross # and if gross is the same, for the same gross objects, x.updated and then update was also the same, x.pk,
How can I do this?

Try sorting by multiple fields like so:
result_list = sorted(
chain(qs1, qs2),
key=lambda x: (x.gross, -x.updated.timestamp(), x.pk) # and if gross is the same, for the gross same objects, x.updated and then update was also the same, x.pk,
reverse=True
)

How can I use parent_id in Django?

In my model:
class HomePageFirstModule(models.Model):
name = models.CharField(max_length=8, unique=True)
is_active = models.BooleanField(default=True) # 是否启用
class HomePageSecondModule(models.Model):
name = models.CharField(max_length=16, unique=True)
is_active = models.BooleanField(default=True) # 是否启用
home_page_first_module = models.ForeignKey(to=HomePageFirstModule) # 所属的第一级模块
class HomePageThridModule(models.Model):
name = models.CharField(max_length=16, unique=True)
url = models.CharField(max_length=128)
is_active = models.BooleanField(default=True) # 是否启用
home_page_second_module = models.ForeignKey(to=HomePageSecondModule) # 所属的第二级模块
Then I use filter method to query out the data:
def get_homepage_module_list():
"""
获取到可以使用的模块信息
:return:
"""
data_query_list = models.HomePageThridModule.objects.filter(
home_page_second_module__home_page_first_module="1"
).values('id', 'name', 'is_active', 'home_page_second_module__name',
'home_page_second_module__home_page_first_module__name',
'home_page_second_module__home_page_first_module__is_active',
'home_page_second_module__is_active'
)
data_list_del = []
data_list = list(data_query_list)
for item in data_list:
if (item['is_active'] == False) or (
item['home_page_second_module__is_active'] == False
) or (
item['home_page_second_module__home_page_first_module__is_active'] == False
):
data_list_del.append(item)
for item_del in data_list_del:
data_list.remove(item_del)
return data_list
========================
How can I convert this list data :
[
{
"home_page_second_module__name": "云主机",
"home_page_second_module__home_page_first_module__name": "产品",
"id": 1,
"name": "云主机子1"
},
{
"home_page_second_module__name": "云主机",
"home_page_second_module__home_page_first_module__name": "产品",
"id": 4,
"name": "云主机子4"
},
{
"home_page_second_module__name": "云硬盘",
"home_page_second_module__home_page_first_module__name": "产品",
"id": 2,
"name": "云硬盘子2"
},
{
"home_page_second_module__name": "云硬盘",
"home_page_second_module__home_page_first_module__name": "产品",
"id": 3,
"name": "云硬盘子3"
}
]
to this:
[
{"name":"产品",
"data":[
{"name":"云主机",
"data":[{"name":"云主机子1",
"data":{"id":1}},
{"name":"云主机子2",
"data":{"id":2}}]},
{"name":"云硬盘",
"data":[{"name":"云硬盘子1",
"data":{"id":3}},
{"name":"云硬盘子2",
"data":{"id":4}}]}
]
}
]
There should has a arithmetic method to do this, but I tried, do not get that.
I only think of this below little things:
home_page_second_module__name_list = []
home_page_second_module__home_page_first_module__name_list = []
id_list = []
name_list = []
for home_page_second_module__name,home_page_second_module__home_page_first_module__name,id,name in ori_list:
if not (home_page_second_module__name_list.__contains__(home_page_second_module__name)):
home_page_second_module__name_list.append(home_page_second_module__name)
if not (home_page_second_module__home_page_first_module__name_list.__contains__(home_page_second_module__home_page_first_module__name_list)):
home_page_second_module__home_page_first_module__name_list.append(home_page_second_module__home_page_first_module__name)
But now I think this is very difficult to do that, and I think mine is wrong way to do that.
Is there a convenient way to realize it?
EDIT
The 产品, 云主机, 云硬盘 may can be deed as parent id.

You can use django-rest-framework, and use related serializers. relations serializers

Outdated code, as the original question just asked how to transform a list of dicts into another list of dicts with different outcome
I bet this can optimized a lot but... assuming your dict is named old, I think this might do it:
new = {'name': i['home_page_second_module__home_page_first_module__name'] for i in old if not i['home_page_second_module__home_page_first_module__name'] in old}
new['data'] = [['name': i['home_page_second_module__name'], 'data':[{'name': i['home_page_second_module__name'], 'data': {'id': i['id']}}]] for i in old]

Django REST serialize output - group by foreign keys

I have models like below.
Restaurant Model
class Restaurant(models.Model):
name = models.CharField(max_length=40, verbose_name='Name')
Menu Model
class Menu(models.Model):
name = models.CharField(max_length=40, unique=True, verbose_name='menu name')
Item Model
class Item(models.Model):
restaurant = models.ForeignKey(Restaurant)
menu = models.ForeignKey(Menu)
name = models.CharField(max_length=500)
price = models.IntegerField(default=0)
I want to get the menus for the shop id.
How can I group my results by menu for the restaurant id ?
call GET /menus/restaurant_id
Sample.
{
name: menu name 1
items: [ {item1}, {item2}]
},
{
name: menu name 2
items: [ {item1}, {item2}]
}
Thanks..

The only thing i can find it's postgres specific aggregation function ArrayAgg
You can use it like this:
from django.contrib.postgres.aggregates import ArrayAgg
Item.objects.filter(restaurant_id=1).values('menu__name').annotate(items=ArrayAgg('name'))
# example output:
# [
# {
# 'menu__name': 'menu1',
# 'items': ['item1', 'item2']
# },
# {
# 'menu__name': 'menu2',
# 'items': ['item3', 'item4']
# },
# ]
Such qs performs next raw sql query:
SELECT
"appname_menu"."name",
ARRAY_AGG("appname_item"."name") AS "items"
FROM "appname_item"
INNER JOIN "appname_menu" ON ("appname_item"."menu_id" = "appname_menu"."id")
WHERE "appname_item"."restaurant_id" = 1
GROUP BY "appname_menu"."name"
Probably it can help you.