Suppose that I have the following python base class:
class BaseClass(object):
def a():
"""This method uses method b(), defined in the inheriting class"""
And also a class that inherites BaseClass:
class UsedByUser(BaseClass):
def b():
"""b() is defined here, yet is used by the base class"""
My user would only create instances of class UsedByUser. Typical use would be:
if __name__ == '__main__':
# initialize the class used by the user
usedByUser = UsedByUser()
# invoke method a()
usedByUser.a()
My questions is, is the above use problematic? is this a valid approach, or must I also define method b() in BaseClass and then override it in UsedByUser?
I would define the b method in the BaseClass too:
class BaseClass(object):
def b(self):
raise NotImplementedError('b must be implemented by a subclass')
Remember: explicit is better than implicit and given that the method a needs the method b anyways, better raise a meaningful exception rather than a general AttributeError.
It is worth to point out that this is absolutely NOT needed from a syntactic point of view, but it adds clarity to the code and enforces the subclass to provide an implementation.
The use is correct. Classes can define methods that can be overriden
from subclasses, but those can also define new methods. Defining
every method needed for subclasses in the superclass seems a bit
senseless. (Since object then would also need to have every
function defined ?)
A subclass often has a different bahviour to another subclass.
class Vehicle(object):
def Refuel(self):
# ...
class Plane(Vehicle):
def Fly(self):
# ...
class Car(Vehicle):
def Drive(self):
# ...
Edit: I misread the code.
If only you create a subclass of it and make sure subclasses have B(), then it's theoratically ok, but bad style. It makes more sense and is safer to give the superclass attributes and methods that are used by the superclass. -> Define B()
Sounds like you want A to call a protected member function of UsedByUser that can't go in the abstract BaseClass). Try prefixing the protected function with an underscore (although note this is just a convention used by Python and not strictly checked, as mentioned here).
class BaseClass(object):
def A(self):
print "Grettings from A"
self._B()
def _B(self):
raise NotImplementedError('b must be implemented by a subclass')
class UsedByUser(BaseClass):
def _B(self):
""" prefix with underscore is a python convention for a protected member function """
print "B rocks!"
if ( __name__=='__main__' ):
usedByUser = UsedByUser()
usedByUser.A()
Find more on this convention in the PEP guidelines.
Edit:
As GaretJax suggested, I added a BaseClass _B method for clarity. Nice tip!
BaseClass can't assume that a UsedByUser object exists, so it can't use a B() method from it. You probably want to define a B() in BaseClass, that either does nothing (if it's sensible to attempt B() with something that doesn't support it) or raises an exception (if it's not sensible to attempt B()).
If you tell us what A and B are in your use case, we may be able to advise you better.
Related
class X():
#decorator
def method(self):
return
class Y(X):
def method(self):
return
Is there any way where the applied decorators still applies on child class method without explicitly decorating them?
Nope, since overriding methods creates a completely new object. It is the same logic that super(...).__init__ does not get called automatically, and the general Python guideline of be explicit - for example, if that happened by default, but you did not want the decorator, how would it work?
Does not seem like too much work to be explicit here, and decorate.
In Java, for example, you can make a class MyClass with certain methods that are specified but not implemented in MyClass, but must be implemented in any class MySubClass that inherits from MyClass. So basically there is some common functionality among all subclasses you want, so you put it in MyClass, and there is some functionality unique (but required) for each subclass, so you want it in each subclass. How can this behavior be achieved in Python?
(I know there are concise terms to describe what I'm asking, so feel free to let me know what these are and how I can better describe my question.)
A very basic example but the abc docs provide a few more
import abc
class Foo():
__metaclass__ = abc.ABCMeta
#abc.abstractmethod
def bar(self):
raise NotImplemented
class FooBar(Foo):
pass
f = FooBar()
TypeError: Can't instantiate abstract class FooBar with abstract methods bar
You can't require the implementation of a method in a subclass in a way that will break at compile-time, but the convention on writing a method on the base class that must be implemented in the subclasses is to raise NotImplementedError.
Something like this:
class MyBase(object):
def my_method(self, *args, **kwargs):
raise NotImplementedError("You should implement this method on a subclass of MyBase")
Then your subclasses can implement my_method, but this will break only when the method is called. If you have comprehensive unit tests, as you should, this won't be a problem.
When you decorate a method, it is not bound yet to the class, and therefor doesn't have the im_class attribute yet. I looking for a way to get the information about the class inside the decorator. I tried this:
import types
def decorator(method):
def set_signal(self, name, value):
print name
if name == 'im_class':
print "I got the class"
method.__setattr__ = types.MethodType(set_signal, method)
return method
class Test(object):
#decorator
def bar(self, foo):
print foo
But it doesn't print anything.
I can imagine doing this:
class Test(object):
#decorator(klass=Test)
def bar(self, foo):
print foo
But if I can avoid it, it would make my day.
__setattr__ is only called on explicit object.attribute = assignments; building a class does not use attribute assignment but builds a dictionary (Test.__dict__) instead.
To access the class you have a few different options though:
Use a class decorator instead; it'll be passed the completed class after building it, you could decorate individual methods on that class by replacing them (decorated) in the class. You could use a combination of a function decorator and a class decorator to mark which methods are to be decorated:
def methoddecoratormarker(func):
func._decorate_me = True
return func
def realmethoddecorator(func):
# do something with func.
# Note: it is still an unbound function here, not a method!
return func
def classdecorator(klass):
for name, item in klass.__dict__.iteritems():
if getattr(item, '_decorate_me', False):
klass.__dict__[name] = realmethoddecorator(item)
You could use a metaclass instead of a class decorator to achieve the same, of course.
Cheat, and use sys._getframe() to retrieve the class from the calling frame:
import sys
def methoddecorator(func):
callingframe = sys._getframe(1)
classname = callingframe.f_code.co_name
Note that all you can retrieve is the name of the class; the class itself is still being built at this time. You can add items to callingframe.f_locals (a mapping) and they'll be made part of the new class object.
Access self whenever the method is called. self is a reference to the instance after all, and self.__class__ is going to be, at the very least, a sub-class of the original class the function was defined in.
My strict answer would be: It's not possible, because the class does not yet exist when the decorator is executed.
The longer answer would depend on your very exact requirements. As I wrote, you cannot access the class if it does not yet exists. One solution would be, to mark the decorated method to be "transformed" later. Then use a metaclass or class decorator to apply your modifications after the class has been created.
Another option involves some magic. Look for the implementation of the implements method in zope.interfaces. It has some access to the information about the class which is just been parsed. Don't know if it will be enough for your use case.
You might want to take a look at descriptors. They let you implement a __get__ that is used when an attribute is accessed, and can return different things depending on the object and its type.
Use method decorators to add some marker attributes to the interesting methods, and use a metaclass which iterates over the methods, finds the marker attributes, and does the logic. The metaclass code is run when the class is created, so it has a reference to the newly created class.
class MyMeta(object):
def __new__(...):
...
cls = ...
... iterate over dir(cls), find methods having .is_decorated, act on them
return cls
def decorator(f):
f.is_decorated = True
return f
class MyBase(object):
__metaclass__ = MyMeta
class MyClass(MyBase):
#decorator
def bar(self, foo):
print foo
If you worry about that the programmer of MyClass forgets to use MyBase, you can forcibly set the metaclass in decorator, by exampining the globals dicitionary of the caller stack frame (sys._getframe()).
I'm trying to do something I consider pretty ordinary in Object Oriented programming
but can't find any documentation on it for Python 2.7
I want to create a subclass of a superclass where, when I'm done
superclass
is bound to the superclass
superclass.subclass is bound to the subclass and
subclass is bound to nothing.
Here's the best solution I can come up with:
class superclass:
pass
class subclass(superclass):
pass
superclass.subclass = subclass
del subclass
Sometimes I want subclass to be in its own file, other times not.
Is there a more elgant solution where I don't have to manually perform the
last two lines?
Although
class superclass:
class subclass:
pass
almost does the right thing, subclass doesn't actually inherit from superclass.
And if I try to do:
class superclass:
class subclass(superclass):
pass
I get an error message that superclass is unbound.
I agree with everyone else that this is a silly thing to do and I don't think you should do it, but in the interest of knowledge:
class Superclass(object):
pass
Superclass.Subclass = type('Subclass', (Superclass,), {
'foo': lambda self: self,
})
where the dictionary argument contains any methods. I don't think there's a nice way to do this with the class syntax, but that's really just syntactic sugar for calling type anyway.
You don't have to define the methods as lambdas; they can be normal functions with def as long as they have the self argument. You'll probably want to hide their definitions if you do that, though....
Here's a simple class decorator to do the referencing:
def refer(cls):
cls.__bases__[0].subclass = cls
return cls
Here's how you use it:
>>> class Superclass:
pass
>>> #refer
class Subclass(SuperClass):
pass
You will still need to delete Subclass if you don't want the name in the module namespace.
Look at this code:
class MyClass():
# Why does this give me "NameError: name 'self' is not defined":
mySelf = self
# But this does not?
def myFunction(self):
mySelf2 = self
Basically I want a way for a class to refer to itself without needing to name itself specifically, hence I want self to work for the class, not just methods/functions. How can I achieve this?
EDIT: The point of this is that I'm trying to refer to the class name from inside the class itself with something like self.class._name_ so that the class name isn't hardcoded anywhere in the class's code, and thus it's easier to re-use the code.
EDIT 2: From what I've learned from the answers below, what I'm trying to do is impossible. I'll have to find a different way. Mission abandoned.
EDIT 3: Here is specifically what I'm trying to do:
class simpleObject(object):
def __init__(self, request):
self.request = request
#view_defaults(renderer='string')
class Test(simpleObject):
# this line throws an error because of self
myClassName = self.__class__.__name__
#view_config(route_name=myClassName)
def activateTheView(self):
db = self.request.db
foo = 'bar'
return foo
Note that self is not defined at the time when you want the class to refer to itself for the assignment to work. This is because (in addition to being named arbitrarily), self refers to instances and not classes. At the time that the suspect line of code attempts to run, there is as of yet no class for it to refer to. Not that it would refer to the class if there was.
In a method, you can always use type(self). That will get the subclass of MyClass that created the current instance. If you want to hard-code to MyClass, that name will be available in the global scope of the methods. This will allow you to do everything that your example would allow if it actually worked. E.g, you can just do MyClass.some_attribute inside your methods.
You probably want to modify the class attributes after class creation. This can be done with decorators or on an ad-hoc basis. Metaclasses may be a better fit. Without knowing what you actually want to do though, it's impossible to say.
UPDATE:
Here's some code to do what you want. It uses a metaclass AutoViewConfigMeta and a new decorator to mark the methods that you want view_config applied to. I spoofed the view_config decorator. It prints out the class name when it's called though to prove that it has access to it. The metaclass __new__ just loops through the class dictionary and looks for methods that were marked by the auto_view_config decorator. It cleans off the mark and applies the view_config decorator with the appropriate class name.
Here's the code.
# This just spoofs the view_config decorator.
def view_config(route=''):
def dec(f):
def wrapper(*args, **kwargs):
print "route={0}".format(route)
return f(*args, **kwargs)
return wrapper
return dec
# Apply this decorator to methods for which you want to call view_config with
# the class name. It will tag them. The metaclass will apply view_config once it
# has the class name.
def auto_view_config(f):
f.auto_view_config = True
return f
class AutoViewConfigMeta(type):
def __new__(mcls, name, bases, dict_):
#This is called during class creation. _dict is the namespace of the class and
# name is it's name. So the idea is to pull out the methods that need
# view_config applied to them and manually apply them with the class name.
# We'll recognize them because they will have the auto_view_config attribute
# set on them by the `auto_view_config` decorator. Then use type to create
# the class and return it.
for item in dict_:
if hasattr(dict_[item], 'auto_view_config'):
method = dict_[item]
del method.auto_view_config # Clean up after ourselves.
# The next line is the manual form of applying a decorator.
dict_[item] = view_config(route=name)(method)
# Call out to type to actually create the class with the modified dict.
return type.__new__(mcls, name, bases, dict_)
class simpleObject(object):
__metaclass__ = AutoViewConfigMeta
class Test(simpleObject):
#auto_view_config
def activateTheView(self):
foo = 'bar'
print foo
if __name__=='__main__':
t = Test()
t.activateTheView()
Let me know if you have any questions.
Python has an "explict is better than implicit" design philosophy.
Many languages have an implicit pointer or variable in the scope of a method that (e.g. this in C++) that refers to the object through which the method was invoked. Python does not have this. Here, all bound methods will have an extra first argument that is the object through which the method was invoked. You can call it anything you want (self is not a keyword like this in C++). The name self is convention rather than a syntactic rule.
Your method myFunction defines the variable self as a parameter so it works. There's no such variable at the class level so it's erroring out.
So much for the explanation. I'm not aware of a straightforward way for you to do what you want and I've never seen such requirement in Python. Can you detail why you want to do such a thing? Perhaps there's an assumption that you're making which can be handled in another way using Python.
self is just a name, your self in this case is a class variable and not this for the object using which it is called,
self is treated as a normal variable and it is not defined, where as the self in the function comes from the object used for calling.
you want to treat the object reference in self as a class variable which is not possible.
self isn't a keyword, it's just a convention. The methods are attributes of the class object (not the instance), but they receive the instance as their first argument. You could rename the argument to xyzzy if you wanted and it would still work the same way.
But (as should be obvious) you can't refer to a method argument outside the body of the method. Inside a class block but outside of any method, self is undefined. And the concept wouldn't even make sense -- at the time the class block is being evaluated, no instance of the class can possibly exist yet.
Because the name self is explicitly defined as part of the arguments to myFunction. The first argument to a method is the instance that the method was called on; in the class body, there isn't an "instance we're dealing with", because the class body deals with every possible instance of the class (including ones that don't necessarily exist yet) - so, there isn't a particular object that could be called self.
If you want to refer to the class itself, rather than some instance of it, this is spelled self.__class__ (or, for new-style classes in Py2 and all classes in Py3, type(self)) anywhere self exists. If you want to be able to deal with this in situations where self doesn't exist, then you may want to look at class methods which aren't associated with any particular instance, and so take the class itself in place of self. If you really need to do this in the class body (and, you probably don't), you'll just have to call it by name.
You can't refer to the class itself within the class body because the class doesn't exist at the time that the class body is executed. (If the previous sentence is confusing, reading up about metaclasses will either clear this up or make you more confused.)
Within an instance method, you can refer to the class of the instance with self.__class__, but be careful here. This will be the instance's actual class, which through the power of inheritance might not be the class in which the method was defined.
Within a class method, the class is passed in as the first argument, much like instances are the first argument to instance methods:
class MyClass(object):
#classmethod
def foo(cls):
print cls.__name__
MyClass.foo() # Should print "MyClass"
As with instance methods, the actual class might differ due to inheritance.
class OtherClass(MyClass):
pass
OtherClass.foo() # Should print "OtherClass"
If you really need to refer to MyClass within a method of MyClass, you're pretty much going to have to refer to it as MyClass unless you use magic. This sort of magic is more trouble than it is worth.