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Inheritance method overwrite in some conditions [duplicate]

When creating a simple object hierarchy in Python, I'd like to be able to invoke methods of the parent class from a derived class. In Perl and Java, there is a keyword for this (super). In Perl, I might do this:
package Foo;
sub frotz {
return "Bamf";
}
package Bar;
#ISA = qw(Foo);
sub frotz {
my $str = SUPER::frotz();
return uc($str);
}
In Python, it appears that I have to name the parent class explicitly from the child.
In the example above, I'd have to do something like Foo::frotz().
This doesn't seem right since this behavior makes it hard to make deep hierarchies. If children need to know what class defined an inherited method, then all sorts of information pain is created.
Is this an actual limitation in python, a gap in my understanding or both?

Use the super() function:
class Foo(Bar):
def baz(self, **kwargs):
return super().baz(**kwargs)
For Python < 3, you must explicitly opt in to using new-style classes and use:
class Foo(Bar):
def baz(self, arg):
return super(Foo, self).baz(arg)

Python also has super as well:
super(type[, object-or-type])
Return a proxy object that delegates method calls to a parent or sibling class of type.
This is useful for accessing inherited methods that have been overridden in a class.
The search order is same as that used by getattr() except that the type itself is skipped.
Example:
class A(object): # deriving from 'object' declares A as a 'new-style-class'
def foo(self):
print "foo"
class B(A):
def foo(self):
super(B, self).foo() # calls 'A.foo()'
myB = B()
myB.foo()

ImmediateParentClass.frotz(self)
will be just fine, whether the immediate parent class defined frotz itself or inherited it. super is only needed for proper support of multiple inheritance (and then it only works if every class uses it properly). In general, AnyClass.whatever is going to look up whatever in AnyClass's ancestors if AnyClass doesn't define/override it, and this holds true for "child class calling parent's method" as for any other occurrence!

Python 3 has a different and simpler syntax for calling parent method.
If Foo class inherits from Bar, then from Bar.__init__ can be invoked from Foo via super().__init__():
class Foo(Bar):
def __init__(self, *args, **kwargs):
# invoke Bar.__init__
super().__init__(*args, **kwargs)

Many answers have explained how to call a method from the parent which has been overridden in the child.
However
"how do you call a parent class's method from child class?"
could also just mean:
"how do you call inherited methods?"
You can call methods inherited from a parent class just as if they were methods of the child class, as long as they haven't been overwritten.
e.g. in python 3:
class A():
def bar(self, string):
print("Hi, I'm bar, inherited from A"+string)
class B(A):
def baz(self):
self.bar(" - called by baz in B")
B().baz() # prints out "Hi, I'm bar, inherited from A - called by baz in B"
yes, this may be fairly obvious, but I feel that without pointing this out people may leave this thread with the impression you have to jump through ridiculous hoops just to access inherited methods in python. Especially as this question rates highly in searches for "how to access a parent class's method in Python", and the OP is written from the perspective of someone new to python.
I found:
https://docs.python.org/3/tutorial/classes.html#inheritance
to be useful in understanding how you access inherited methods.

Here is an example of using super():
#New-style classes inherit from object, or from another new-style class
class Dog(object):
name = ''
moves = []
def __init__(self, name):
self.name = name
def moves_setup(self):
self.moves.append('walk')
self.moves.append('run')
def get_moves(self):
return self.moves
class Superdog(Dog):
#Let's try to append new fly ability to our Superdog
def moves_setup(self):
#Set default moves by calling method of parent class
super(Superdog, self).moves_setup()
self.moves.append('fly')
dog = Superdog('Freddy')
print dog.name # Freddy
dog.moves_setup()
print dog.get_moves() # ['walk', 'run', 'fly'].
#As you can see our Superdog has all moves defined in the base Dog class

There's a super() in Python too. It's a bit wonky, because of Python's old- and new-style classes, but is quite commonly used e.g. in constructors:
class Foo(Bar):
def __init__(self):
super(Foo, self).__init__()
self.baz = 5

I would recommend using CLASS.__bases__
something like this
class A:
def __init__(self):
print "I am Class %s"%self.__class__.__name__
for parentClass in self.__class__.__bases__:
print " I am inherited from:",parentClass.__name__
#parentClass.foo(self) <- call parents function with self as first param
class B(A):pass
class C(B):pass
a,b,c = A(),B(),C()

If you don't know how many arguments you might get, and want to pass them all through to the child as well:
class Foo(bar)
def baz(self, arg, *args, **kwargs):
# ... Do your thing
return super(Foo, self).baz(arg, *args, **kwargs)
(From: Python - Cleanest way to override __init__ where an optional kwarg must be used after the super() call?)

There is a super() in python also.
Example for how a super class method is called from a sub class method
class Dog(object):
name = ''
moves = []
def __init__(self, name):
self.name = name
def moves_setup(self,x):
self.moves.append('walk')
self.moves.append('run')
self.moves.append(x)
def get_moves(self):
return self.moves
class Superdog(Dog):
#Let's try to append new fly ability to our Superdog
def moves_setup(self):
#Set default moves by calling method of parent class
super().moves_setup("hello world")
self.moves.append('fly')
dog = Superdog('Freddy')
print (dog.name)
dog.moves_setup()
print (dog.get_moves())
This example is similar to the one explained above.However there is one difference that super doesn't have any arguments passed to it.This above code is executable in python 3.4 version.

In this example cafec_param is a base class (parent class) and abc is a child class. abc calls the AWC method in the base class.
class cafec_param:
def __init__(self,precip,pe,awc,nmonths):
self.precip = precip
self.pe = pe
self.awc = awc
self.nmonths = nmonths
def AWC(self):
if self.awc<254:
Ss = self.awc
Su = 0
self.Ss=Ss
else:
Ss = 254; Su = self.awc-254
self.Ss=Ss + Su
AWC = Ss + Su
return self.Ss
def test(self):
return self.Ss
#return self.Ss*4
class abc(cafec_param):
def rr(self):
return self.AWC()
ee=cafec_param('re',34,56,2)
dd=abc('re',34,56,2)
print(dd.rr())
print(ee.AWC())
print(ee.test())
Output
56
56
56

In Python 2, I didn't have a lot luck with super(). I used the answer from
jimifiki on this SO thread how to refer to a parent method in python?.
Then, I added my own little twist to it, which I think is an improvement in usability (Especially if you have long class names).
Define the base class in one module:
# myA.py
class A():
def foo( self ):
print "foo"
Then import the class into another modules as parent:
# myB.py
from myA import A as parent
class B( parent ):
def foo( self ):
parent.foo( self ) # calls 'A.foo()'

class department:
campus_name="attock"
def printer(self):
print(self.campus_name)
class CS_dept(department):
def overr_CS(self):
department.printer(self)
print("i am child class1")
c=CS_dept()
c.overr_CS()

If you want to call the method of any class, you can simply call Class.method on any instance of the class. If your inheritance is relatively clean, this will work on instances of a child class too:
class Foo:
def __init__(self, var):
self.var = var
def baz(self):
return self.var
class Bar(Foo):
pass
bar = Bar(1)
assert Foo.baz(bar) == 1

class a(object):
def my_hello(self):
print "hello ravi"
class b(a):
def my_hello(self):
super(b,self).my_hello()
print "hi"
obj = b()
obj.my_hello()

This is a more abstract method:
super(self.__class__,self).baz(arg)

Naming conventions for class method vs instance method

I have two methods, one for the individual Instance, and one for every Instance in that class:
class MasterMatches(models.Model):
#classmethod
def update_url_if_any_matches_has_one(cls):
# apply to all instances, call instance method.
def update_url_if_any_matches_has_one(self):
# do something
Should I name these the same? Or, what is a good naming convention here?

The question of using the same names can be clarified by understanding how decorators work.
#dec
def foo(x):
print(x)
translates to
def foo(x):
print(x)
foo = dec(foo)
In your example the decorator syntax can be expanded to
class MasterMatches(models.Model):
def update_url_if_any_matches_has_one(cls):
# apply to all instances, call instance method.
update_url_if_any_matches_has_one = classmethod(update_url_if_any_matches_has_one)
def update_url_if_any_matches_has_one(self):
# do something
The former implementation of update_url_if_any_matches_has_one will be overwritten by the latter.

Usually use self declaration style. #classmethod use only if method not works with class instance fields.
Function decorated as #classmethod takes the first argument is the class type, while normal method takes instance of object.
class A:
#classmethod
def a(cls):
print(cls)
def b(self):
print(self)
a = A()
a.a()
a.b()
# Output:
# <class '__main__.A'>
# <__main__.A object at 0x03FC5DF0>
It can be useful if you have a static class fields. The to access therm you don't need explicitly specify the class name. But you don't get access to instance fields. Example:
class A:
field = 1
#classmethod
def a(cls):
print(cls.field)
def b(self):
self.field = 2
print(self.field, A.field)
a = A()
a.a()
a.b()
# Outputs:
# 1
# 2 1

How can I access a classmethod from inside a class in Python

I would like to create a class in Python that manages above all static members. These members should be initiliazed during definition of the class already. Due to the fact that there will be the requirement to reinitialize the static members later on I would put this code into a classmethod.
My question: How can I call this classmethod from inside the class?
class Test():
# static member
x = None
# HERE I WOULD LOVE TO CALL SOMEHOW static_init!
# initialize static member in classmethod, so that it can be
#reinitialized later on again
#classmethod
def static_init(cls):
cls.x = 10
Any help is appreciated!
Thanks in advance,
Volker

At the time that x=10 is executed in your example, not only does the class not exist, but the classmethod doesn't exist either.
Execution in Python goes top to bottom. If x=10 is above the classmethod, there is no way you can access the classmethod at that point, because it hasn't been defined yet.
Even if you could run the classmethod, it wouldn't matter, because the class doesn't exist yet, so the classmethod couldn't refer to it. The class is not created until after the entire class block runs, so while you're inside the class block, there's no class.
If you want to factor out some class initialization so you can re-run it later in the way you describe, use a class decorator. The class decorator runs after the class is created, so it can call the classmethod just fine.
>>> def deco(cls):
... cls.initStuff()
... return cls
>>> #deco
... class Foo(object):
... x = 10
...
... #classmethod
... def initStuff(cls):
... cls.x = 88
>>> Foo.x
88
>>> Foo.x = 10
>>> Foo.x
10
>>> Foo.initStuff() # reinitialize
>>> Foo.x
88

You call a class method by appending the class name likewise:
class.method
In your code something like this should suffice:
Test.static_init()
You could also do this:
static_init(Test)
To call it inside your class, have your code do this:
Test.static_init()
My working code:
class Test(object):
#classmethod
def static_method(cls):
print("Hello")
def another_method(self):
Test.static_method()
and Test().another_method() returns Hello

You can't call a classmethod in the class definition because the class hasn't been fully defined yet, so there's nothing to pass the method as its first cls argument...a classic chicken-and-egg problem. However you can work around this limitation by overloading the __new__() method in a metaclass, and calling the classmethod from there after the class has been created as illustrated below:
class Test(object):
# nested metaclass definition
class __metaclass__(type):
def __new__(mcl, classname, bases, classdict):
cls = type.__new__(mcl, classname, bases, classdict) # creates class
cls.static_init() # call the classmethod
return cls
x = None
#classmethod
def static_init(cls): # called by metaclass when class is defined
print("Hello")
cls.x = 10
print Test.x
Output:
Hello
10

After re-reading your question carefully this time I can think of two solutions. The first one is to apply the Borg design pattern. The second one is to discard the class method and use a module level function instead. This appears to solve your problem:
def _test_static_init(value):
return value, value * 2
class Test:
x, y = _test_static_init(20)
if __name__ == "__main__":
print Test.x, Test.y
Old, incorrect answer:
Here's an example, I hope it helps:
class Test:
x = None
#classmethod
def set_x_class(cls, value):
Test.x = value
def set_x_self(self):
self.__class__.set_x_class(10)
if __name__ == "__main__":
obj = Test()
print Test.x
obj.set_x_self()
print Test.x
obj.__class__.set_x_class(15)
print Test.x
Anyway, NlightNFotis's answer is a better one: use the class name when accessing the class methods. It makes your code less obscure.

This seems like a reasonable solution:
from __future__ import annotations
from typing import ClassVar, Dict
import abc
import string
class Cipher(abc.ABC):
#abc.abstractmethod
def encrypt(self, plaintext: str) -> str:
pass
#abc.abstractmethod
def decrypt(self, ciphertext: str) -> str:
pass
class RotateCipher(Cipher, abc.ABC):
#staticmethod
def rotate(n: int) -> str:
return string.ascii_uppercase[n:] + string.ascii_uppercase[:n]
class VigenereCipher(RotateCipher):
_TABLE: ClassVar[Dict[str, str]] = dict({(chr(i + ord("A")), RotateCipher.rotate(i)) for i in range(26)})
def encrypt(self, plaintext: str) -> str:
pass
def decrypt(self, plaintext: str) -> str:
pass
vc = VigenereCipher()
The method is now a static method of the cipher, nothing outside the classes is referenced. You could opt to name RotateCipher _RotateCipher instead, if you don't want people using it by itself.
Note: I removed the Final, as I ran this on 3.7, but after reading the documentation on Final, I don't think it would affect the solution? Also added an import for string which the question was missing. And finally added an implementation for the abstract methods, alternatively, could have let VigenereCipher inherit from abc.ABC as well.

If your classmethod is not used very often do a lazy evaluation
class A() {
# this does not work: x=A.initMe()
#classmethod
def initMe(cls) {
if not hasattr(cls,"x"):
# your code her
cls.x=# your result
pass
#classmethod
def f1(cls) {
# needs initMe
cls.initMe()
# more code using cls.x
}
}

"Programmatically" add stuff to a class?

I'm writing a class that has a dict containing int to method mappings. However setting the values in this dict results in the dict being populated with unbound functions.
class A:
def meth_a: ...
def meth_b: ...
...
map = {1: meth_a, 2: meth_b, ...}
for int in ...:
map[int] = meth_x
This doesn't work for a few reasons:
The methods aren't bound when the class is initialized because they're not in the class dict?
I can't bind the methods manually using __get__ because the class name isn't bound to any namespace yet.
So:
How can I do this?
Do I have to drop out of the class and define the dict after the class has been initialized?
Is it really necessary to call __get__ on the methods to bind them?
Update0
The methods will be called like this:
def func(self, int):
return self.map[int]()
Also regarding the numeric indices/list: Not all indices will be present. I'm not aware that one can do list([1]=a, [2]=b, [1337]=leet) in Python, is there an equivalent? Should I just allocate a arbitrary length list and set specific values? The only interest I have here is in minimizing the lookup time, would it really be that different to the O(1) hash that is {}? I've ignored this for now as premature optimization.

I'm not sure exactly why you're doing what you're doing, but you certainly can do it right in the class definition; you don't need __init__.
class A:
def meth_a(self): pass
m = {1: meth_a}
def foo(self, number):
self.m[number](self)
a = A()
a.foo(1)
An "unbound" instance method simply needs you to pass it an instance of the class manually, and it works fine.
Also, please don't use int as the name of a variable, either, it's a builtin too.
A dictionary is absolutely the right type for this kind of thing.
Edit: This will also work for staticmethods and classmethods if you use new-style classes.

First of all Don't use variable "map" since build in python function map will be fetched.
You need to have init method and initialize your dictionary in the init method using self. The dictionary right now is only part of the class, and not part of instances of the class. If you want instances of the class to have the dictionary as well you need to make an init method and initialize your dictionary there. So you need to do this:
def __init__(self):
self.mymap[int] = self.meth_x
or if you want the dictionary to be a class variable, then this:
def __init__(self):
A.mymap[int] = self.meth_x

It's not totally clear just what you're trying to do. I suspect you want to write code something like
class Foo(object):
def __init__(self, name):
self.name = name
def method_1(self, bar):
print self.name, bar
# ... something here
my_foo = Foo('baz')
my_foo.methods[1]('quux')
# prints "baz quux"
so, that methods attribute needs to return a bound instance method somehow, but without being called directly. This is a good opportunity to use a descriptor. We need to do something that will return a special object when accessed through an instance, and we need that special object to return a bound method when indexed. Let's start from the inside and work our way out.
>>> import types
>>> class BindMapping(object):
... def __init__(self, instance, mapping):
... self.instance, self.mapping = instance, mapping
...
... def __getitem__(self, key):
... func = self.mapping[key]
... if isinstance(func, types.MethodType):
... return types.MethodType(func.im_func, self.instance, func.im_class)
... else:
... return types.MethodType(func, self.instance, type(self))
...
We're just implementing the barest minimum of the mapping protocol, and deferring completely to an underlying collection. here we make use of types.MethodType to get a real instance method when needed, including binding something that's already an instance method. We'll see how that's useful in a minute.
We could implement a descriptor directly, but for the purposes here, property already does everything we need out of a descriptor, so we'll just define one that returns a properly constructed BindMapping instance.
>>> class Foo(object):
... def method_1(self):
... print "1"
... def method_2(self):
... print "2"
...
... _mapping = [method_1, method_2]
...
... #property
... def mapping(self):
... return BindMapping(self, self._mapping)
...
Just for kicks, we also throw in some extra methods outside the class body. Notice how the the methods added inside the class body are functions, just like functions added outside; methods added outside the class body are actual instance methods (although unbound).
>>> def method_3(self):
... print "3"
...
>>> Foo._mapping.append(method_3)
>>> Foo._mapping.append(Foo.method_1)
>>> map(type, Foo._mapping)
[<type 'function'>, <type 'function'>, <type 'function'>, <type 'instancemethod'>]
And it works as advertised:
>>> f = Foo()
>>> for i in range(len(f._mapping)):
... f.mapping[i]()
...
1
2
3
1
>>>

This seems kind of convoluted to me. What is the ultimate goal?
If you really want do to this, you can take advantage of the fact that the methods are alreaday contained in a mapping (__dict__).
class A(object):
def meth_1(self):
print("method 1")
def meth_2(self):
print("method 2")
def func(self, i):
return getattr(self, "meth_{}".format(i))()
a = A()
a.func(2)
This pattern is found in some existing library modules.

Get name of current class?

How do I get the name of the class I am currently in?
Example:
def get_input(class_name):
[do things]
return class_name_result
class foo():
input = get_input([class name goes here])
Due to the nature of the program I am interfacing with (vistrails), I cannot use __init__() to initialize input.

obj.__class__.__name__ will get you any objects name, so you can do this:
class Clazz():
def getName(self):
return self.__class__.__name__
Usage:
>>> c = Clazz()
>>> c.getName()
'Clazz'

Within the body of a class, the class name isn't defined yet, so it is not available. Can you not simply type the name of the class? Maybe you need to say more about the problem so we can find a solution for you.
I would create a metaclass to do this work for you. It's invoked at class creation time (conceptually at the very end of the class: block), and can manipulate the class being created. I haven't tested this:
class InputAssigningMetaclass(type):
def __new__(cls, name, bases, attrs):
cls.input = get_input(name)
return super(MyType, cls).__new__(cls, name, bases, newattrs)
class MyBaseFoo(object):
__metaclass__ = InputAssigningMetaclass
class foo(MyBaseFoo):
# etc, no need to create 'input'
class foo2(MyBaseFoo):
# etc, no need to create 'input'

PEP 3155 introduced __qualname__, which was implemented in Python 3.3.
For top-level functions and classes, the __qualname__ attribute is equal to the __name__ attribute. For nested classes, methods, and nested functions, the __qualname__ attribute contains a dotted path leading to the object from the module top-level.
It is accessible from within the very definition of a class or a function, so for instance:
class Foo:
print(__qualname__)
will effectively print Foo.
You'll get the fully qualified name (excluding the module's name), so you might want to split it on the . character.
However, there is no way to get an actual handle on the class being defined.
>>> class Foo:
... print('Foo' in globals())
...
False

You can access it by the class' private attributes:
cls_name = self.__class__.__name__
EDIT:
As said by Ned Batchelder, this wouldn't work in the class body, but it would in a method.

EDIT: Yes, you can; but you have to cheat: The currently running class name is present on the call stack, and the traceback module allows you to access the stack.
>>> import traceback
>>> def get_input(class_name):
... return class_name.encode('rot13')
...
>>> class foo(object):
... _name = traceback.extract_stack()[-1][2]
... input = get_input(_name)
...
>>>
>>> foo.input
'sbb'
However, I wouldn't do this; My original answer is still my own preference as a solution. Original answer:
probably the very simplest solution is to use a decorator, which is similar to Ned's answer involving metaclasses, but less powerful (decorators are capable of black magic, but metaclasses are capable of ancient, occult black magic)
>>> def get_input(class_name):
... return class_name.encode('rot13')
...
>>> def inputize(cls):
... cls.input = get_input(cls.__name__)
... return cls
...
>>> #inputize
... class foo(object):
... pass
...
>>> foo.input
'sbb'
>>>

#Yuval Adam answer using #property
class Foo():
#property
def name(self):
return self.__class__.__name__
f = Foo()
f.name # will give 'Foo'

I think, it should be like this:
class foo():
input = get_input(__qualname__)

import sys
def class_meta(frame):
class_context = '__module__' in frame.f_locals
assert class_context, 'Frame is not a class context'
module_name = frame.f_locals['__module__']
class_name = frame.f_code.co_name
return module_name, class_name
def print_class_path():
print('%s.%s' % class_meta(sys._getframe(1)))
class MyClass(object):
print_class_path()

I'm using python3.8 and below is example to get your current class name.
class MyObject():
#classmethod
def print_class_name(self):
print(self.__name__)
MyObject.print_class_name()
Or without #classmethod you can use
class ClassA():
def sayhello(self):
print(self.getName())
def getName(self):
return self.__class__.__name__
ClassA().sayhello()
Hope that helps others !!!