What are the differences between importing a class from a file and using super() in python?
Import: # I understand that this will import all the functions and init from a class to another class in a file when trigger the below code;
from something import Some
super(): # I understand that this also will inherit all the functions and init from a class to another class in a file. **Ain't this technically the same as importing?
super(xx, self).__init__() or super().__init__() or super()
Can any python experts shows some example? I have recently read some codes online and they used both of this, my question is since both do almost the same thing(to my understanding) then why dont just use either one for all files and why need to use both? If possible, can share some examples?
You'll need to understand a bit of Object-oriented programming principles to understand super.
In general, when you are importing a module, that module becomes available for your current function/class.
By using Super you satisfy inheritance; facilitating your current class's this to have references to all attributes of its parent(s). If you are just developing some functions, you will not need super. This is strictly an object-oriented paradigm.
from xyz import abc
def some_function():
# here you can use abc's attributes by explicit invocation.
x = abc.x
# Or call functions.
abc.some_function()
When you're dealing with classes, super does the magic of facilitating inheritance.
class Parent_Class(object):
def __init__(self):
super(Parent_Class, self).__init__()
def parent_method(self):
print('This is a parent method')
class Child_Class(Parent_Class):
def __init__(self):
super(Child_Class, self).__init__()
def child_method(self):
# Because of super, self of child class is aware of all its parent's methods and properties.
self.parent_method()
import is used for importing module
__init__() is a class method, which is invoked when a new instance of a class is instantiated
This is 2 different thing
EDIT: I've made this edit to clarify the answer for you, since I assume you are not familiar with OOP. TL;DR: These two terms are completely different, they relate to two separate things. super() call a class's parent method, while import import a module
we can still import the module form file B into file A and in file A
we can still use the methods from the file B in file A right?
Right.
what makes it different than super() which calls the methods from file B when used in file A
First, you only encounter super() in subclass (lets say class B) of a class(lets say class A). You access A's method through B's instance via super(). Let see this example: You have 2 file, A.py and B.py, each define A and B:
In A.py, you define class A
class A(object):
def __init__(self):
print("Instantiate class A")
def printA(self):
print("AAAAAAAAAAAAAA")
In B.py, you define class B:
from A import A # you need this import for `B.py` to know what is `A`
class B(A): # Mean B inherit A
def __init__(self):
super(B, self).__init__() # This line all its parent's method (`__init__()`)
def printB(self):
print("BBBBBBBBBBBB")
super(B, self).printA() # This line all its parent's method (`printA`)
# And you instantiate a `B` instance, then call a method of it, which call its super method `printA`
b = B()
b.printB()
Output is (note that `init():
Instantiate class A
BBBBBBBBBBBB
AAAAAAAAAAAAAA
To conclude, I suggest you to read some material about OOP. You will clearly see the differences btw these two things. Good luck. I hope this help!
The import statement is used to import modules into modules. The super callable is used to delegate from subclass implementation to superclass implementation.
Notably, import connects between fixed equals whereas super connects in a dynamic hierarchy.
The point of import is to make a specific module or its content available in another module.
# moda.py
def foo(who):
print(who, "called '%s.foo'" % __name__)
# modb.py
import moda
moda.foo(__name__)
Note that importing uses a qualified name that uniquely identifies the target. At any time, the name used in an import can refer to only one specific entity. That means that import couples strongly - the target is fully identified and cannot change after the import.
The point of super is to access the closest superclass implementation in a subclass implementation.
class A:
def call(self, who):
print(who, "called '%s.call'" % __class__, "on self of class", self.__class__)
class Alpha(A):
def call(self, who):
super(Alpha, self).call(who)
print("... via '%s.call'" % __class__)
Note that super only work with local information - super does only identify the source, not the target. As such, the target is dynamically bound and not uniquely identifiable. That means that super couples weakly - the target is relatively defined and can only be resolved at runtime.
class Alpher(A):
def call(self, who):
print("'%s.call' sneakily intercepts the call..." % __class__)
super(Alpher, self).call(who)
class Better(Alpha, Alpher):
pass
The above injects Alpher into the relation from Alpha to A without modifying either of the two.
>>> Better().call('Saul')
'<class '__main__.Alpher'>.call' sneakily intercepts the call...
Saul called <class '__main__.A'>.call on self of class <class '__main__.Better'>
...via '<class '__main__.Alpha'>.call'
Related
There are two main ways for a derived class to call a base class's methods.
Base.method(self):
class Derived(Base):
def method(self):
Base.method(self)
...
or super().method():
class Derived(Base):
def method(self):
super().method()
...
Suppose I now do this:
obj = Derived()
obj.method()
As far as I know, both Base.method(self) and super().method() do the same thing. Both will call Base.method with a reference to obj. In particular, super() doesn't do the legwork to instantiate an object of type Base. Instead, it creates a new object of type super and grafts the instance attributes from obj onto it, then it dynamically looks up the right attribute from Base when you try to get it from the super object.
The super() method has the advantage of minimizing the work you need to do when you change the base for a derived class. On the other hand, Base.method uses less magic and may be simpler and clearer when a class inherits from multiple base classes.
Most of the discussions I've seen recommend calling super(), but is this an established standard among Python coders? Or are both of these methods widely used in practice? For example, answers to this stackoverflow question go both ways, but generally use the super() method. On the other hand, the Python textbook I am teaching from this semester only shows the Base.method approach.
Using super() implies the idea that whatever follows should be delegated to the base class, no matter what it is. It's about the semantics of the statement. Referring explicitly to Base on the other hand conveys the idea that Base was chosen explicitly for some reason (perhaps unknown to the reader), which might have its applications too.
Apart from that however there is a very practical reason for using super(), namely cooperative multiple inheritance. Suppose you've designed the following class hierarchy:
class Base:
def test(self):
print('Base.test')
class Foo(Base):
def test(self):
print('Foo.test')
Base.test(self)
class Bar(Base):
def test(self):
print('Bar.test')
Base.test(self)
Now you can use both Foo and Bar and everything works as expected. However these two classes won't work together in a multiple inheritance schema:
class Test(Foo, Bar):
pass
Test().test()
# Output:
# Foo.test
# Base.test
That last call to test skips over Bar's implementation since Foo didn't specify that it wants to delegate to the next class in method resolution order but instead explicitly specified Base. Using super() resolves this issue:
class Base:
def test(self):
print('Base.test')
class Foo(Base):
def test(self):
print('Foo.test')
super().test()
class Bar(Base):
def test(self):
print('Bar.test')
super().test()
class Test(Foo, Bar):
pass
Test().test()
# Output:
# Foo.test
# Bar.test
# Base.test
In C++, given a class hierarchy, the most derived class's ctor calls its base class ctor which then initialized the base part of the object, before the derived part is instantiated. In Python I want to understand what's going on in a case where I have the requirement, that Derived subclasses a given class Base which takes a callable in its __init__ method which it then later invokes. The callable features some parameters which I pass in Derived class's __init__, which is where I also define the callable function. My idea then was to pass the Derived class itself to its Base class after having defined the __call__ operator
class Derived(Base):
def __init__(self, a, b):
def _process(c, d):
do_something with a and b
self.__class__.__call__ = _process
super(Derived, self).__init__(self)
Is this a pythonic way of dealing with this problem?
What is the exact order of initialization here? Does one needs to call super as a first instruction in the __init__ method or is it ok to do it the way I did?
I am confused whether it is considered good practice to use super with or without arguments in python > 3.6
What is the exact order of initialization here?
Well, very obviously the one you can see in your code - Base.__init__() is only called when you explicitely ask for it (with the super() call). If Base also has parents and everyone in the chain uses super() calls, the parents initializers will be invoked according to the mro.
Basically, Python is a "runtime language" - except for the bytecode compilation phase, everything happens at runtime - so there's very few "black magic" going on (and much of it is actually documented and fully exposed for those who want to look under the hood or do some metaprogramming).
Does one needs to call super as a first instruction in the init method or is it ok to do it the way I did?
You call the parent's method where you see fit for the concrete use case - you just have to beware of not using instance attributes (directly or - less obvious to spot - indirectly via a method call that depends on those attributes) before they are defined.
I am confused whether it is considered good practice to use super with or without arguments in python > 3.6
If you don't need backward compatibily, use super() without params - unless you want to explicitely skip some class in the MRO, but then chances are there's something debatable with your design (but well - sometimes we can't afford to rewrite a whole code base just to avoid one very special corner case, so that's ok too as long as you understand what you're doing and why).
Now with your core question:
class Derived(Base):
def __init__(self, a, b):
def _process(c, d):
do_something with a and b
self.__class__.__call__ = _process
super(Derived, self).__init__(self)
self.__class__.__call__ is a class attribute and is shared by all instances of the class. This means that you either have to make sure you are only ever using one single instance of the class (which doesn't seem to be the goal here) or are ready to have totally random results, since each new instance will overwrite self.__class__.__call__ with it's own version.
If what you want is to have each instance's __call__ method to call it's own version of process(), then there's a much simpler solution - just make _process an instance attribute and call it from __call__ :
class Derived(Base):
def __init__(self, a, b):
def _process(c, d):
do_something with a and b
self._process = _process
super(Derived, self).__init__(self)
def __call__(self, c, d):
return self._process(c, d)
Or even simpler:
class Derived(Base):
def __init__(self, a, b):
super(Derived, self).__init__(self)
self._a = a
self._b = b
def __call__(self, c, d):
do_something_with(self._a, self._b)
EDIT:
Base requires a callable in ins init method.
This would be better if your example snippet was closer to your real use case.
But when I call super().init() the call method of Derived should not have been instantiated yet or has it?
Now that's a good question... Actually, Python methods are not what you think they are. What you define in a class statement's body using the def statement are still plain functions, as you can see by yourself:
class Foo:
... def bar(self): pass
...
Foo.bar
"Methods" are only instanciated when an attribute lookup resolves to a class attribute that happens to be a function:
Foo().bar
main.Foo object at 0x7f3cef4de908>>
Foo().bar
main.Foo object at 0x7f3cef4de940>>
(if you wonder how this happens, it's documented here)
and they actually are just thin wrappers around a function, instance and class (or function and class for classmethods), which delegate the call to the underlying function, injecting the instance (or class) as first argument. In CS terms, a Python method is the partial application of a function to an instance (or class).
Now as I mentionned upper, Python is a runtime language, and both def and class are executable statements. So by the time you define your Derived class, the class statement creating the Base class object has already been executed (else Base wouldn't exist at all), with all the class statement block being executed first (to define the functions and other class attributes).
So "when you call super().__init()__", the __call__ function of Base HAS been instanciated (assuming it's defined in the class statement for Base of course, but that's by far the most common case).
I am trying to watch a video on python tricks shown here. He is trying to demonstrate when using a higher level class when calling a method provided by the developer extending the class, that he can assure the method is written. He then wrote two classes in two different files
-- library.py
class Base():
def foo(self):
return self.bar()
-- user.py
from library import Base
class Derived(Base):
def bar(self):
return 'bar'
Now, I am confused why this would even work since the class Base, cannot call methods of anyone that inherits from it. For example, the following code will produce an error.
class Higher():
def higher_method(self):
print("higher method")
class Lower(Higher):
def lower_method(self):
print("lower method")
higher = Higher()
higher.lower_method()
How is the method called when he executes the code?
When you call Derived().foo(), the object bound to self in Base.foo is not an instance of Base; it's an instance of Derived, and so its lookup will succeed for bar. Base().foo() would fail by the same logic: the instance of Base bound to self would not find a definition for bar.
In your Higher/Lower example, the same reasoning applies: an instance of Higher simply does not have access to a attribute named lower_method. The only difference is you are failing immediately, where your instance of Higher is bound to the global name higher rather than bound to the name self in a method that is defined by Higher.
Names are scoped statically, but lookups occur dynamically.
I am quite new to python, so pardon me for basic question. I tried google for past few days but could not make it in my program.
Can anyone show me a good example how can I use method from One class to another in python and what is significance of __init__ while defining class.
I am using python2.7
Thanks in anticipation.
To use a method defined in one class inside of another class, you have several options:
Create an instance of B from within one of A's methods, then call B's method:
class A:
def methodInA():
b = B()
b.methodInB()
If appropriate, use the concept of inheritance (one of the defining concepts of object-oriented design) to create a subclass of the original class whose method(s) you wish to use:
class B(A):
...
__init__() is a class initializer. Whenever you instantiate an object you are invoking __init__() whether or not it is explicitly defined. It's main purpose is to initialize class data members:
class C:
def __init__(self, name):
self.name = name
def printName(self):
print self.name
c = C("George")
c.printName() # outputs George
With __init__() defined, in particular with the additional argument name in this example, you are able to differentiate between would-be generically constructed instances by allowing for different initial states from instance to instance.
There are 2 issues here:
First: Using method of class A in class B, both classes in different files
class A:
def methodOfA(self):
print "method Of A"
let the above class be in file a.py Now the class B is supposed to be in b.py. Both a.py and b.py are assumed to be on the same level or in the same location. Then b.py would look like:
import a
class B:
def methodOfB(self):
print "Method of B"
a.A().methodOfA()
You can also do this by inherting A in B
import a
class B(a.A):
def methodOfB(self):
print "Method of B"
self.methodOfA()
there are several other ways to use A in B. I will leave it to you to explore.
Now to your second question. The use of __init__ in a class. __init__ is not a constructor, as popularly believed and explained above. It is, as the name suggests, an initialization function. It is called only after the object has already been constructed and it is implicitly passed the object instance as the first argument, as signified by self in its argument list.
The actual constructor in python is called __new__, which does not need a object to call it. This is actually a specialized Static method, which receives the class instance as the first argument. __new__ is exposed for overwriting only if the class inherits form the object base class of python
Whatever other arguments are passed while creating an object of a class, first go to __new__ and then are passed with the object instance to the __init__, if it accepts them.
The init function is what is called a constructor function. When you create an instance of a class object = myClass(), init is the function that is automatically called. i.e.
That being said, to call a function from one class to another, you need to call an instance of the second class inside the first one, or vice versa. for eg.
class One():
def func(self):
#does sometthing here
class Two():
def __init__(self):
self.anotherClass = One()
#Now you can access the functions of the first class by using anotherClass followed by dot operator
self.anotherClass.func()
#When you call the main class. This is the time the __init__ function is automatically called
mainClass = Two()
Another way to access from another class is the use of oop concept called Inheritance.
class One():
def __init__(self):
print('Class One Called')
def func(self):
print('func1 Called')
class Two(One):
def __init__(self):
One.__init__(self,) #This basically creates One's instance
print('Main Called')
c= Two()
c.func()
The output for this is:
Class One Called
Main Called
func1 Called
I was looking into Python's super method and multiple inheritance. I read along something like when we use super to call a base method which has implementation in all base classes, only one class' method will be called even with variety of arguments. For example,
class Base1(object):
def __init__(self, a):
print "In Base 1"
class Base2(object):
def __init__(self):
print "In Base 2"
class Child(Base1, Base2):
def __init__(self):
super(Child, self).__init__('Intended for base 1')
super(Child, self).__init__()# Intended for base 2
This produces TyepError for the first super method. super would call whichever method implementation it first recognizes and gives TypeError instead of checking for other classes down the road. However, this will be much more clear and work fine when we do the following:
class Child(Base1, Base2):
def __init__(self):
Base1.__init__(self, 'Intended for base 1')
Base2.__init__(self) # Intended for base 2
This leads to two questions:
Is __init__ method a static method or a class method?
Why use super, which implicitly choose the method on it's own rather than explicit call to the method like the latter example? It looks lot more cleaner than using super to me. So what is the advantage of using super over the second way(other than writing the base class name with the method call)
super() in the face of multiple inheritance, especially on methods that are present on object can get a bit tricky. The general rule is that if you use super, then every class in the hierarchy should use super. A good way to handle this for __init__ is to make every method take **kwargs, and always use keyword arguments everywhere. By the time the call to object.__init__ occurs, all arguments should have been popped out!
class Base1(object):
def __init__(self, a, **kwargs):
print "In Base 1", a
super(Base1, self).__init__()
class Base2(object):
def __init__(self, **kwargs):
print "In Base 2"
super(Base2, self).__init__()
class Child(Base1, Base2):
def __init__(self, **kwargs):
super(Child, self).__init__(a="Something for Base1")
See the linked article for way more explanation of how this works and how to make it work for you!
Edit: At the risk of answering two questions, "Why use super at all?"
We have super() for many of the same reasons we have classes and inheritance, as a tool for modularizing and abstracting our code. When operating on an instance of a class, you don't need to know all of the gritty details of how that class was implemented, you only need to know about its methods and attributes, and how you're meant to use that public interface for the class. In particular, you can be confident that changes in the implementation of a class can't cause you problems as a user of its instances.
The same argument holds when deriving new types from base classes. You don't want or need to worry about how those base classes were implemented. Here's a concrete example of how not using super might go wrong. suppose you've got:
class Foo(object):
def frob(self):
print "frobbign as a foo"
class Bar(object):
def frob(self):
print "frobbign as a bar"
and you make a subclass:
class FooBar(Foo, Bar):
def frob(self):
Foo.frob(self)
Bar.frob(self)
Everything's fine, but then you realize that when you get down to it,
Foo really is a kind of Bar, so you change it
class Foo(Bar):
def frob(self):
print "frobbign as a foo"
Bar.frob(self)
Which is all fine, except that in your derived class, FooBar.frob() calls Bar.frob() twice.
This is the exact problem super() solves, it protects you from calling superclass implementations more than once (when used as directed...)
As for your first question, __init__ is neither a staticmethod nor a classmethod; it is an ordinary instance method. (That is, it receives the instance as its first argument.)
As for your second question, if you want to explicitly call multiple base class implementations, then doing it explicitly as you did is indeed the only way. However, you seem to be misunderstanding how super works. When you call super, it does not "know" if you have already called it. Both of your calls to super(Child, self).__init__ call the Base1 implementation, because that is the "nearest parent" (the most immediate superclass of Child).
You would use super if you want to call just this immediate superclass implementation. You would do this if that superclass was also set up to call its superclass, and so on. The way to use super is to have each class call only the next implementation "up" in the class hierarchy, so that the sequence of super calls overall calls everything that needs to be called, in the right order. This type of setup is often called "cooperative inheritance", and you can find various articles about it online, including here and here.