What is the most Pythonic way to organise constants? - python

I use marshmallow and have a series of object serialization. Because of technical reasons, the names of the field and class can be the same.
For example,
class Sample(Schema):
SampleField = fields.Nested(SampleField)
class SampleField(Schema):
# Some other fields
When I need to reference SampleField as a string in the code. I create constant and name it SAMPLE_FIELD = SampleField. Often I need to have the same constant in both places in the definition class and where it is used as a field.
How to organize it the way, it will not become messy?

The marshmallow field does not have to be named after the field name in serialiazed data.
Would this solve your problem?
class SampleSchema(Schema):
sample_field = fields.Nested(SampleFieldSchema, data_key='SampleField')
class SampleFieldSchema(Schema):
# Some other fields

Related

Django, treat OneToOne related field as my own field

I'm essentially trying to come up with my own inheritance scheme because Django's inheritance doesn't fit my needs.
I'd like parent table(class) hold common data fields.
sub classess would have its own additional data in a separate table.
class ProductBase(models.Model):
common = models.IntegerField()
def get_price(self):
return some_price
class FooProduct(ProductBase):
# no field because I'm proxy
class Meta:
proxy = True
def get_price(self):
return price_using_different_logic
class FooExtra(models.Model):
base = models.OneToOneField(ProductBase, primary_key=True)
phone = models.CharField(max_length=10)
My question is, would it be able to treat as if Foo has FooExtra's fields?
I'd like to do things like following..
foo = FooProduct.objects.create()
foo.phone = "3333" # as django does with its multiple inheritance
foo.save()
FooProduct.objects.filter(phone="3333")
I'd like to list Products of different kind(data)
I need to list them together, so abstract Base inheritance is out
from the list, I'd like to treat each model as polymorphic model, when iterating over ProductBase.objects.all(), product.get_price() will use appropriate classe's method. (without incurring join if don't have to)
When and only when I want, I retrieve the addtional table data (by something like .select_related('fooextra')
Django-polymorphic is close to what I want, but it is rather obscure what it does so I'm afraid to use it, and I think it fails #3.
If I understand well, you want inheritance and you want the fields that are specific to the child class to be on a separate table.
As far as I know, you don't need a proxy class to achieve that, you could just implement multi-table inheritance as specified in the manual at https://docs.djangoproject.com/en/1.9/topics/db/models/#multi-table-inheritance e.g.:
class Base(models.Model):
common = models.IntegerField()
class Foo(Base):
phone = models.CharField(max_length=10)
This, as explained at the link above, will automatically create a one-to-one relationship. And of course you can do foo.phone = "3333" (where foo is of type Foo) as in your example above. And the neat thing is that you can also access foo.common whereas in your example it would have been foo.base.common.
It doesn't seem like you want anything different to Django's standard inheritance.
class ProductBase(models.Model):
common1 = models.IntegerField()
common2 = models.IntegerField()
class FooProduct(ProductBase):
fooextra = models.IntegerField()
class BarProduct(ProductBase):
barextra = models.IntegerField()
If you create instances of each:
foo1 = FooProduct(common1=1, common2=1, fooextra=1)
foo2 = FooProduct(common1=1, common2=1, fooextra=2)
bar1 = BarProduct(common1=1, common2=1, barextra=1)
bar2 = BarProduct(common1=1, common2=1, barextra=2)
You can loop over all products:
for product in ProductBase.objects.all():
print product.common1, product.common2
From a ProductBase object that is actually a FooProduct, you can get the custom field with:
product.foo.fooextra
From a ProductBase object that is actually a BarProduct, you can get the custom field with:
product.bar.barextra
You can still do querying:
foo = FooProduct.objects.get(fooextra=1)
bar = BarProduct.objects.get(barextra=2)
And you can access the common fields directly on those objects:
foo.common1
bar.common2
You can use the InheritanceManager from django-model-utils if you need more control over querying etc - and this should address point 3, too: ProductBase.objects.filter(...).select_subclasses() would give you the FooProduct and BarProduct objects instead of ProductBase objects.

How to define a django model field with the same name as a Python keyword

I need to define a Django model field with the name in, which is a Python language keyword. This is a syntax error:
class MyModel(models.Model):
in = jsonfield.JSONField()
How can I make this work?
The reason I need this name is when I use django-rest-framework's ModelSerializer class, field name is used as the key for serialization output, and I thought it might be easier to manipulate django's Model class instead of ModelSerializer class to get the output I want.
Generally speaking, you don't. Avoid the use of keywords in your identifiers. The general Python convention is to add an underscore to such names; here that'd be in_:
class MyModel(models.Model):
in_ = jsonfield.JSONField()
However, Django prohibits names ending in an underscore because the underscore clashes with their filter naming conventions, so you have to come up with a different name still; pick one that still describes your case; I picked contained in rather than in, as a guess to what you want to do here:
class MyModel(models.Model):
contained_in = jsonfield.JSONField()
If you are trying to match an existing database schema, use the db_column attribute:
class MyModel(models.Model):
contained_in = jsonfield.JSONField(db_column='in')
If you want to be stubborn, in normal classes you could use setattr() after creating the class to use a string instead of an identifier:
class Foo:
pass
setattr(Foo, 'in', 'some value')
but you'll have to use setattr(), getattr(), delattr() and/or vars() everywhere in your code to be able to access this.
In Django you'll have the added complication that a models.Model subclass uses a metaclass to parse out your class members into others structures, and adding an extra field with setattr() doesn't work without (a lot of) extra work to re-do what the metaclass does. You could instead use the field.contribute_to() method, calling it after the class has been prepared by Django (technique taken from this blog post):
from django.db.models.signals import class_prepared
def add_field(sender, **kwargs):
if sender.__name__ == "MyModel":
field = jsonfield.JSONField('in')
field.contribute_to_class(sender, 'in')
class_prepared.connect(add_field)
but you have to make sure this hook is registered before you create your model class.
There is no way to make it work, and it's a bad idea anyway. Choose a different name.
If, for some reason, you want to have column name that matches some reserved keyword, use db_column argument for that field.
in_something = models.CharField(db_column='in', max_length=100)
You mentioned the use of django rest framework. Here's how to make it work on the serializer layer. The keyword used is from. to is just an example of a non-keyword if you want it mapped to a different name.
from django.db import models
from rest_framework import serializers
SP_FIELD_MAP = {
'from': 'sender'
}
# would be in models.py
class Transaction(models.Model):
recipient = models.CharField(max_length=16)
sender = models.CharField(max_length=64)
# would be in serializers.py
class TransactionSerializer(serializers.ModelSerializer):
to = serializers.CharField(source='recipient')
class Meta:
model = Transaction
fields = ('id', 'to', 'from')
# `from` is a python keyword hence this
extra_kwargs = {'from': {'source': 'sender'}}
def build_field(self, field_name, info, model_class, nested_depth):
# Catches python keywords like `from` and maps to its proper field
field_name = SP_FIELD_MAP.get(field_name, field_name)
return super(TransactionSerializer, self).build_field(
field_name, info, model_class, nested_depth)
Tested on CharField using POST and GET methods only but I don't see how it won't work on other methods. You might need special stuff for other field types though. I suggest going into the source. There's tons of fun stuff going on in DRF's source.
You should be giving all your variables descriptive names that clearly state what they are to be used for, and where possible it should be easy to assertain what type of variable it is.
in, to me, would appear at first glance to be a boolean so in order to use this variable in my own extension to the code I'd need to find other usages of it before I knew how I could use it.
Therefore, simply don't try to hack something together just so you can get this terrible variable name into your model, it offers no value to you to do so, its not really any quicker to type since intellisense is available in most places. Figure out what "in" relates to and then formulate a proper name that is descriptive.

Django creating multiple tables/model classes from same base class with factory function

I have been trying to figure out the best way to automate creating multiple SQL tables based on separate but identical models, all based on the same base class. I'm basically creating pseudo message boards or walls with different Groups, and I wanted each Group to have its own db_table of Posts, each Post containing the user id, timestamp, etc.
My first thought was to have one base class of Posts and just include a field for Group name, but I thought this would be bad practice. My rationale was that one table containing every Post for all Groups would get really big (in theory anyway) and slow down filtering, and also that the extra field for group name would in the long run be a waste of memory when I could have separate tables per group and skip this field.
I've also considered using a ForeignKey with a Many-to-One relationship, but as far as I can tell this has the same drawbacks. Am I wrong to think that? Or are these size concerns not really an issue?
So my next idea was to make Posts an abstract class, and then create subclasses based on each Group. This is ultimately what I did. However, I found myself having to copy and paste the code over and over and change the class name each time. This felt very unPythonic to me. It was something like:
class Posts(models.Model):
timestamp = models.DateTimeField(auto_now_add=True, unique=False)
user_id = ...
#etc.
#
class Meta:
abstract = True
class GroupA(Posts):
class Meta(Posts.Meta):
db_table = 'groupa_board'
class GroupB(Posts):
class Meta(Posts.Meta):
db_table = 'groupb_board'
class GroupC...etc.
What I really was looking for was a factory function to do this for me. I tried this sort of thing:
def makeBoard(group):
class Board(Posts):
class Meta(Posts.Meta):
db_table = group
return board #note I tried with and without this line
And then I ran a simple for loop using a list of groups.
for group in groups:
makeBoard(group)
I found myself hitting a RuntimeError: conflicting models in application, and I probably deserved it. So then I figured what I need is something like:
def makeBoard(group):
class group(Posts): #***group here being a variable, not the class name
class Meta(Posts.Meta):
db_table = '%s' % group #maybe issues here too, but the table
return group #name is not that important if the class
#name works
But I couldn't figure out how to make this work! Is there a way to pass a variable from a list to a class name?
Anyway if you're still with me I appreciate it. I've been on stackoverflow all day and while I've found guides for creating abstract base classes and subclasses to solve similar issues, I didn't see a way to create a function to do this for me. I ultimately punted here and just make a subclass for each group by hand. If there is a way to automate this process, I'd love to hear it.
Also, if I'm being stupid for not just going with one db table containing every post, I'd like to know that too, and why! Or if there's a better way to implement this kind of system altogether. I apologize if this has been answered before, I really couldn't find it.
Thank you!
Using a single table would not be bad practice. The extra memory is minimal, on modern systems that shouldn't be a problem. You shouldn't worry about performance either, premature optimization (not including the actual system design) is considered bad practice, but if you run into performance problems you can always specify an index on the group column:
group = models.CharField(max_length=100, db_index=True)
That's not to say that it is the best option, or that your method isn't good. Also, it is entirely possible to dynamically create models, using the type() built-in function. The only difference with dynamically creating models and creating other classes is that you must specifically pass the __module__ attribute. You can create subclasses for Posts in the following way:
def fabric(names, baseclass=Posts):
for name in names:
class Meta:
db_table = '%s_table' % name.lower()
attrs = {'__module__': baseclass.__module__, 'Meta': Meta}
# specify any other class attributes here. E.g. you can specify extra fields:
attrs.update({'my_field': models.CharField(max_length=100)})
newclass = type(str(name), (baseclass,), attrs)
globals()[name] = newclass
fabric(['GroupA', 'GroupB', 'GroupC', etc...])
Put that code in your models.py after your Posts class, and all classes will be created for you. They can be used in any way normal classes can be used: Django doesn't even know you dynamically created this class. Though your Meta class doesn't inherit from Posts.Meta, your meta settings should still be preserved.
Tested with Django 1.4.
Try smth like this
import app.models as group_models
from django.db.models.base import ModelBase
def fabric(group):
for item in dir(group_models):
c = getattr(group_models, item)
if type(c) is ModelBase:
if c._meta.db_table == '%s_table' % group:
return c
return None

Problem with model inheritance and polymorphism

i came with new django problem. The situtaion: i have a model class UploadItemModel, i subcallss it to create uploadable items, like videos, audio files ...
class UploadItem(UserEntryModel):
category = 'abstract item'
file = models.FileField(upload_to=get_upload_directory)
i subclass it like this:
class Video(UploadItem):
category = 'video'
I need to access category attributes from a custom tag. The problem si that i am getting category='abstract item' even if the class is actually Video.
Any clue?
EDIT: I need to use hierarchy because i have several types of item that user can uplaod(Video, Audio files, PDF text). I need to create a class for each type, but there are lot of things in common between those classes(eg forms).
Any clue?
Yes. AFAIK it doesn't work the way you're hoping. Django Models aren't trivially Python classes. They're more like metaclasses which create instances of a kind of "hidden" class definition. Yes, the expected model class exists, but it isn't quite what you think it is. For one thing, the class you use was built for you from your class definition. That's why some static features of Python classes don't work as you'd expect in Django models.
You can't really make use of class-level items like this.
You might want to create an actual field with a default value or something similar.
class UploadItem(UserEntryModel):
category = models.CharFIeld( default='abstract item' )
file = models.FileField(upload_to=get_upload_directory)
Even after the comments being added to the question, I'm still unclear on why this is being done. There do not seem to be any structural or behavioral differences. These all seem like a single class of objects. Subclasses don't seem to define anything new.
Options.
Simply use the class name instead of this "category" item at the class level. Make the class names good enough that you don't need this "category" item.
Use a property
class UploadItem(UserEntryModel):
file = models.FileField(upload_to=get_upload_directory)
#property
def category( self ):
return self.__class__.__name__
You will need to create an additional field that will be a descriptor for that type.
There is a good tutorial here explaining how to use inheritance in Django models
Can you try overriding the __init__ method of the class to assign a category to each instance? For e.g.
class Video(UploadItem):
def __init__(self, *args, **kwargs):
super(Video, self).__init__(*args, **kwargs)
self.category = 'video'

Can't use an inheriting Django model's Meta class to configure a field defined in an inherited abstract model

I would like to use properties from an inheriting model's Meta class to configure a field defined in an abstract model higher up the inheritance tree:
class NamedModel(models.Model):
class Meta:
abstract = True
verbose_name = 'object'
name = models.CharField("Name",
max_length=200,
db_index=True,
help_text="A meaningful name for this %s." % Meta.verbose_name)
# see what I'm trying to do here?
)
...
class OwnedModel(NamedModel):
class Meta(NamedModel.Meta):
verbose_name = 'owned object'
I would like the help text on the name field of OwnedModel forms to say 'A meaningful name for this owned object'. But it does not: the word 'owned' is missing, which would suggest that the verbose_name from the NamedModel.Meta is used when the model is set up, not OwnedModel.Meta.
This isn't quite what I expect from an inheritance point of view: is there some way to get the field to be created whereby Meta.verbose_name refers to the value on the non-abstract model class, not the abstract one on which the field was defined?
Or am I being daft?
(This may seem like a trivial example, and it is: but it's just to illustrate the point of something more important and complex I am trying to do)
Many thanks in advance.
Why don't you try to make a class.
class BaseNamedModelMeta:
abstract = True
verbose_name = "your text"
And then inherit and override whatever you want like this:
class OwnedModel(NamedModel):
class Meta(BaseNamedModelMeta):
verbose_name = 'owned object'
I think this happens because Meta.verbose_name is used and NamedModel.name is created when class NamedModel is parsed. So later, when class OwnedModel gets parsed, there is no chance to change anything.
Maybe you can set the help_text property on OwnedModel.name later on, but this may change NamedModel.name also.
In similar situations I have put the variable parts in class attribute of the model (not Meta) and then used the by run time methods/properties to generate the texts I need.
In fact I ended up doing the following. The base model gets given a dynamic_field_definition() class method, which can be used to patch up the fields, with the cls argument being the correct (inheriting) class. That means that that cls' Meta attributes are of that correct child, not the original base.
I then wire up that method to get called on the class_prepared signal, so that you know everything's otherwise ready.
class NamedModel(models.Model):
...
#classmethod
def dynamic_field_definition(cls):
pass
def dynamic_field_definition(sender, **kwargs):
if issubclass(sender, NamedModel):
sender.dynamic_field_definition()
class_prepared.connect(dynamic_field_definition)
Then the field properties that vary with model class are simply reconfigured by that class method (or more likely the method as overridden in derived classes).
It's a slightly hacky way to bring a last little bit of OO-ness to Django models, but it works fine for my purpose.

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