python: how to access class.__dict__ from class variable?

Question

I need to define a class variable, named "class". I want to do this directly in class namespace, not in a class method. Obviously, I cannot directly say:

class MyClass(object):
    a = 1
    b = 2
    class = 3

So, I want to do something like:

class MyClass(object):
    a = 1
    b = 2
    self.__dict__["class"] = 3

Where " self " should be replaced with a reference to the class. So, how do I refer to a class from class namespace?

---

NOTE: This question might seem contrived, but it stems from a practical goal.

In fact, MyClass is a Django REST Framework serializer and I need a "class" field to be defined on it, because this REST endpoint has to follow a certain protocol.

There's a metaclass defined for Serializers, which calls __new__() upon class creation and that __new__() aggregates all the fields, defined on class and populates a registry of fields with them. So, I have to define my variable class before the class is created. Also see: Django REST Framework: how to make verbose name of field differ from its field_name?

Answer 1

You could do:

class MyClass(object):
    a = 1
    b = 2
    vars()['class'] = 3

But since class is a reserved keyword, then you have to access the variable using getattr and setattr , so that class remains a string.

>>> m = MyClass()
>>> getattr(m, 'class')
3

Answer 2

You can create your class from type and add the attribute class to the class dictionary:

>>> MyClass = type('MyClass', (), {'class': 3, 'a':1, 'b':2})
>>> getattr(MyClass, 'class')
3

You can't directly access the name class with a dot reference, you'll need to use getattr :

>>> MyClass.class
  File "<stdin>", line 1
    MyClass.class
                ^
SyntaxError: invalid syntax

FWIW, you can define the class methods like you would do conventionally and then bind them to the class later on.

Caveat : While this works, I wouldn't use this hack myself as the keyword class is too much of a keyword to tamper with.

Answer 3

You don't need to name the attribute class , which can lead to all kinds of problems. You can name the attribute class_ , but still have it pull from a source attribute named class and render out to JSON as class .

You can do this by overriding the metaclass for Serializers. Here is an example of a serializers.py file (the models and classes are largely pulled straight from the tutorial ).

The main magic is this section of the metaclass

# Remap fields (to use class instead of class_)
fields_ = []
for name, field in fields:
    if name.endswith('_'):
        name = name.rstrip('_')
    fields_.append((name, field))

This takes any field you define in the serializer that ends in an underscore (ie. field_ ) and removes the underscore from the name when it binds the Fields and sets the _declared_fields attribute on the serializer.

from collections import OrderedDict

from rest_framework import serializers
from rest_framework.fields import Field
from snippets.models import Snippet, LANGUAGE_CHOICES, STYLE_CHOICES

class MyMeta(serializers.SerializerMetaclass):

    @classmethod
    def _get_declared_fields(cls, bases, attrs):
        fields = [(field_name, attrs.pop(field_name))
                  for field_name, obj in list(attrs.items())
                  if isinstance(obj, Field)]
        fields.sort(key=lambda x: x[1]._creation_counter)

        # If this class is subclassing another Serializer, add that Serializer's
        # fields.  Note that we loop over the bases in *reverse*. This is necessary
        # in order to maintain the correct order of fields.
        for base in reversed(bases):
            if hasattr(base, '_declared_fields'):
                fields = list(base._declared_fields.items()) + fields

        # Remap fields (to use class instead of class_)
        fields_ = []
        for name, field in fields:
            if name.endswith('_'):
                name = name.rstrip('_')
            fields_.append((name, field))

        return OrderedDict(fields_)


class SnippetSerializer(serializers.Serializer):

    __metaclass__ = MyMeta

    pk = serializers.IntegerField(read_only=True)
    title = serializers.CharField(required=False, allow_blank=True, max_length=100)
    class_ = serializers.CharField(source='klass', label='class', default='blah')

    def create(self, validated_data):
        """
        Create and return a new `Snippet` instance, given the validated data.
        """
        return Snippet.objects.create(**validated_data)

    def update(self, instance, validated_data):
        """
        Update and return an existing `Snippet` instance, given the validated data.
        """
        instance.title = validated_data.get('title', instance.title)
        instance.class_ = validated_data.get('class', instance.class_)
        instance.save()
        return instance

Here is the models.py file for reference (django doesn't allow field names to end in an underscore)

from django.db import models

class Snippet(models.Model):
    title = models.CharField(max_length=100, blank=True, default='')
    klass = models.CharField(max_length=100, default='yo')

This is how it looks from the django shell

$ python manage.py shell

>>> from snippets.models import Snippet
>>> from snippets.serializers import SnippetSerializer
>>> from rest_framework.renderers import JSONRenderer
>>> from rest_framework.parsers import JSONParser
>>> snippet = Snippet(title='test')
>>> snippet.save()
>>> serializer = SnippetSerializer(snippet)
>>> serializer.data
{'title': u'test', 'pk': 6, 'class': u'yo'}

Answer 4

You cannot it while creating class - technically that object does not exist yet.

You could consider:

class MyClass(object):
    a = 1
    b = 2

# class is already created
MyClass.__dict__["class"] = 3

But MyClass.__dict__ is not a dict, but a dictproxy , and 'dictproxy' object does not support item assignment , so there would be TypeError raised.

Answer 5

Use '''setattr''' to set a class attribute immediately after you finish the class definition. Outside the definition, of course. Pass '''MyClass''' for parameter, and it will create an attribute of your class.

Dict of members should not be used, especially for modifying an object. In fact, it is rarely needed. Most of things (though not all) people usually intend it to do are better done by '''setattr''' and '''getattr'''.

Finally, as one of those who answered noticed, you do not really need a field named '''class''', but that's another story, different from your original question.