Does the #staticmethod decorator do anything? - python

I made these two classes:
class A:
#staticmethod
def f(x):
print("x is", x)
class B:
def f(x):
print("x is", x)
And used them like this:
>>> A.f(1)
x is 1
>>> B.f(1)
x is 1
It looks like f became a static method on B even without the decorator. Why would I need the decorator?

It used to matter more back in Python 2, where the instance-ness of instance methods was enforced more strongly:
>>> class B:
... def f(x):
... print("x is", x)
...
>>> B.f(1)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: unbound method f() must be called with B instance as first argument (
got int instance instead)
You had to mark static methods with #staticmethod back then.
These days, #staticmethod still makes it clearer that the method is static, which helps with code readability and documentation generation, and it lets you call the method on instances without the system trying to bind self.

Try these two classes, both having a cry method, one as a classmethod and another as a staticmethod with self passed on
class Cat:
def __init__(self):
self.sound = "meow"
def cry(self):
print(self.sound)
x = Cat()
x.cry()
meow
and with another class
class Dog:
def __init__(self):
self.sound = "ruff-ruff"
#staticmethod
def cry(self):
print(self.sound)
x = Dog()
x.cry()
TypeError: cry() missing 1 required positional argument: 'self'
and we can see the #staticmethod decorator basically removed the passed in self

Related

How to use python's getattr feature on a method of a class? [duplicate]

If I have a class ...
class MyClass:
def method(arg):
print(arg)
... which I use to create an object ...
my_object = MyClass()
... on which I call method("foo") like so ...
>>> my_object.method("foo")
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: method() takes exactly 1 positional argument (2 given)
... why does Python tell me I gave it two arguments, when I only gave one?
In Python, this:
my_object.method("foo")
... is syntactic sugar, which the interpreter translates behind the scenes into:
MyClass.method(my_object, "foo")
... which, as you can see, does indeed have two arguments - it's just that the first one is implicit, from the point of view of the caller.
This is because most methods do some work with the object they're called on, so there needs to be some way for that object to be referred to inside the method. By convention, this first argument is called self inside the method definition:
class MyNewClass:
def method(self, arg):
print(self)
print(arg)
If you call method("foo") on an instance of MyNewClass, it works as expected:
>>> my_new_object = MyNewClass()
>>> my_new_object.method("foo")
<__main__.MyNewClass object at 0x29045d0>
foo
Occasionally (but not often), you really don't care about the object that your method is bound to, and in that circumstance, you can decorate the method with the builtin staticmethod() function to say so:
class MyOtherClass:
#staticmethod
def method(arg):
print(arg)
... in which case you don't need to add a self argument to the method definition, and it still works:
>>> my_other_object = MyOtherClass()
>>> my_other_object.method("foo")
foo
In simple words
In Python you should add self as the first parameter to all defined methods in classes:
class MyClass:
def method(self, arg):
print(arg)
Then you can use your method according to your intuition:
>>> my_object = MyClass()
>>> my_object.method("foo")
foo
For a better understanding, you can also read the answers to this question: What is the purpose of self?
Something else to consider when this type of error is encountered:
I was running into this error message and found this post helpful. Turns out in my case I had overridden an __init__() where there was object inheritance.
The inherited example is rather long, so I'll skip to a more simple example that doesn't use inheritance:
class MyBadInitClass:
def ___init__(self, name):
self.name = name
def name_foo(self, arg):
print(self)
print(arg)
print("My name is", self.name)
class MyNewClass:
def new_foo(self, arg):
print(self)
print(arg)
my_new_object = MyNewClass()
my_new_object.new_foo("NewFoo")
my_bad_init_object = MyBadInitClass(name="Test Name")
my_bad_init_object.name_foo("name foo")
Result is:
<__main__.MyNewClass object at 0x033C48D0>
NewFoo
Traceback (most recent call last):
File "C:/Users/Orange/PycharmProjects/Chapter9/bad_init_example.py", line 41, in <module>
my_bad_init_object = MyBadInitClass(name="Test Name")
TypeError: object() takes no parameters
PyCharm didn't catch this typo. Nor did Notepad++ (other editors/IDE's might).
Granted, this is a "takes no parameters" TypeError, it isn't much different than "got two" when expecting one, in terms of object initialization in Python.
Addressing the topic: An overloading initializer will be used if syntactically correct, but if not it will be ignored and the built-in used instead. The object won't expect/handle this and the error is thrown.
In the case of the sytax error: The fix is simple, just edit the custom init statement:
def __init__(self, name):
self.name = name
Newcomer to Python, I had this issue when I was using the Python's ** feature in a wrong way. Trying to call this definition from somewhere:
def create_properties_frame(self, parent, **kwargs):
using a call without a double star was causing the problem:
self.create_properties_frame(frame, kw_gsp)
TypeError: create_properties_frame() takes 2 positional arguments but 3 were given
The solution is to add ** to the argument:
self.create_properties_frame(frame, **kw_gsp)
As mentioned in other answers - when you use an instance method you need to pass self as the first argument - this is the source of the error.
With addition to that,it is important to understand that only instance methods take self as the first argument in order to refer to the instance.
In case the method is Static you don't pass self, but a cls argument instead (or class_).
Please see an example below.
class City:
country = "USA" # This is a class level attribute which will be shared across all instances (and not created PER instance)
def __init__(self, name, location, population):
self.name = name
self.location = location
self.population = population
# This is an instance method which takes self as the first argument to refer to the instance
def print_population(self, some_nice_sentence_prefix):
print(some_nice_sentence_prefix +" In " +self.name + " lives " +self.population + " people!")
# This is a static (class) method which is marked with the #classmethod attribute
# All class methods must take a class argument as first param. The convention is to name is "cls" but class_ is also ok
#classmethod
def change_country(cls, new_country):
cls.country = new_country
Some tests just to make things more clear:
# Populate objects
city1 = City("New York", "East", "18,804,000")
city2 = City("Los Angeles", "West", "10,118,800")
#1) Use the instance method: No need to pass "self" - it is passed as the city1 instance
city1.print_population("Did You Know?") # Prints: Did You Know? In New York lives 18,804,000 people!
#2.A) Use the static method in the object
city2.change_country("Canada")
#2.B) Will be reflected in all objects
print("city1.country=",city1.country) # Prints Canada
print("city2.country=",city2.country) # Prints Canada
It occurs when you don't specify the no of parameters the __init__() or any other method looking for.
For example:
class Dog:
def __init__(self):
print("IN INIT METHOD")
def __unicode__(self,):
print("IN UNICODE METHOD")
def __str__(self):
print("IN STR METHOD")
obj = Dog("JIMMY", 1, 2, 3, "WOOF")
When you run the above programme, it gives you an error like that:
TypeError: __init__() takes 1 positional argument but 6 were given
How we can get rid of this thing?
Just pass the parameters, what __init__() method looking for
class Dog:
def __init__(self, dogname, dob_d, dob_m, dob_y, dogSpeakText):
self.name_of_dog = dogname
self.date_of_birth = dob_d
self.month_of_birth = dob_m
self.year_of_birth = dob_y
self.sound_it_make = dogSpeakText
def __unicode__(self, ):
print("IN UNICODE METHOD")
def __str__(self):
print("IN STR METHOD")
obj = Dog("JIMMY", 1, 2, 3, "WOOF")
print(id(obj))
If you want to call method without creating object, you can change method to static method.
class MyClass:
#staticmethod
def method(arg):
print(arg)
MyClass.method("i am a static method")
I get this error when I'm sleep-deprived, and create a class using def instead of class:
def MyClass():
def __init__(self, x):
self.x = x
a = MyClass(3)
-> TypeError: MyClass() takes 0 positional arguments but 1 was given
You should actually create a class:
class accum:
def __init__(self):
self.acc = 0
def accumulator(self, var2add, end):
if not end:
self.acc+=var2add
return self.acc
In my case, I forgot to add the ()
I was calling the method like this
obj = className.myMethod
But it should be is like this
obj = className.myMethod()

How to use a function outside a class as a property inside a class?

I'm having some problems. How we can define a function outside of a function that can be used in a class property? Also, how we can insert the self parameter into the function signature? I would like to visualize it like this:
>>> def a(self, x): #I thought maybe class will give "self" to this property function
... print(self)
...
>>> class aa:
... def __init__(self):
... pass
... #a
... def p():
... print('in it')
...
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 4, in aa
TypeError: a() missing 1 required positional argument: 'x'
I want to define a function outside but to use inside of a class. Like a class's method as a property. How can I do this?
It's not really clear what you want your out-of-class function to do. There are a bunch of possibilities, but you may not know the terminology yet to describe it to us.
Here's the three I think are most likely:
You may want your function to be a decorator. That means you can apply it to a method with #decorator syntax to other functions, including methods in a class.
For this to work, your function needs to be written to accept a function object as its only argument. Whatever it returns is what will replace the function or method it was being called on, so usually you want to return a callable, but you could instead return a descriptor like property does. Try something like this:
def decorator(func):
def wrapper(self, *args, **kwargs):
print("in the wrapper")
result = func(self, *args, **kwargs)
print("wrapper is done")
return result
return wrapper
class Foo:
#decorator
def foo(self, x):
print("in foo(), x is", x)
f = Foo()
f.foo(1) # prints three messages
When you call the foo method, you're actually going to be calling the wrapper method that the decorator returned after it was applied to the original method (func). Because of how we wrote the wrapper, it will call func so the original method prints out its message too.
You may want to use property (a descriptor type) to call your out-of-class function. This is a less common way of using property than applying it as a decorator on a method, but it's not impossible. You could even have two different functions, one to be called when requesting the attribute, the other than will be called when setting it (but I'll demonstrate with just the getter):
def getter(obj):
print("in the getter")
return 1
class Foo2:
foo = property(getter)
f2 = Foo2()
print(f2.foo) # prints a message from the getter function first, then prints 1
Note that you can't use #decorator syntax when building a property this way. That is only legal syntax immediately before a function definition, and we're not defining any functions that way inside our class.
You may just want to copy a function defined outside of the class into it, without any decorator or property nonsense. This is the easiest one to do, it's just a simple assignment:
def func(self, x):
print("x is", x)
class Foo3:
method = func # just assign the global to a name in the class body
func = func # you can even use the same name if you don't mind confusing people
f3 = Foo3()
f3.method(1)
f3.func(2)
If you want to create a property that uses a function defined outside your class, it would be something like this:
def myfunc(self):
return self._p
class Foo:
def __init__(self, p):
self._p = p
p = property(myfunc)
f = Foo("Alpha")
f.p # gives "Alpha"
property accepts a function as its (first) argument. The function should have self as a parameter, and should return the value that you want the property to evaluate to.

when is super(Baseclass, self).__init__ used

When should the following code be used in Python
(Assume that Baseclass inherits from Parent class and Parent class has some variables initiated in __init__() method)
class Baseclass(Parent):
def __init__(self, some_arg):
self.some_arg = some_arg
super(Baseclass, self).__init__()
Does this code makes all the local variables defined in __init__ method of Parent class accessible in Baseclass? What significance does it make?
super keeps your code from being repetitive; a complex __init__ needn't be c/p'ed into your inheriting classes. It also makes MRO work as it should, such that if you use multiple inheritance it will work correctly.
One reason to do this would be to ensure that all of your inheriting objects have certain attributes which they don't have from the parent. If you simply write a new __init__, they won't have them unless you repeat your code. For example:
>>> class A(object):
... def __init__(self, x):
... self.x = x
...
>>> class B(A):
... def __init__(self, y):
... self.y = y
...
>>> Stick = B(15)
>>> Stick.x
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
AttributeError: 'B' object has no attribute 'x'
>>>
Without calling super during the __init__ the entire method is simply overridden. A call to super here ensures that both variables exist in the inherited class.
>>> class C(A):
... def __init__(self, x, y):
... super(C, self).__init__(x)
... self.y = y
...
>>> Dave = C(15, 22)
>>> Dave.x
15
>>> Dave.y
22
>>>
Note that in the super call, x is passed to the __init__() call, but self is taken care of in the super(C, self) part of the code.
EDIT: TyrantWave also rightly points out that super is also quite useful outside of __init__. Take an object with a simple foo method for example.
class Parent(object):
def foo(self):
return "I say foo!"
The inherited class may want to just alter the output of this function instead of totally rewriting it. So instead of repeating ourselves and writing the same code over again, we just call super to get the parent's return value, then work with the data and return the child class's modified results.
class Child(Parent):
def foo(self):
parent_result = super(Child, self).foo()
return "I'm a child!! %s" % parent_result
In the above, the call to super returns the Parents value for foo() and then the Child goes on to work with the data further before returning it themselves.
>>> Alan = Parent()
>>> Stan = Child()
>>> Alan.foo()
'I say foo!'
>>> Stan.foo()
"I'm a child!! I say foo!"
>>>

override recursive method in python

When I call the base class recursive method from the derived class, the recursive call is done against the derived method, instead of the base class method. How can I avoid that without modifying base class implementation (in example class A)?
Here is an example
class A(object):
# recursive method
def f(self, x):
print x,
if x < 0:
self.f(x+1)
if x > 0:
self.f(x-1)
if x == 0:
print ""
class B(A):
# Override method
def f(self):
# do some pretty cool stuff
super(B, self).f(25)
if __name__ == "__main__":
A().f(5)
B().f()
I've got this output:
5 4 3 2 1 0
25
Traceback (most recent call last):
File "./test.py", line 19, in <module>
B().f()
File "./test.py", line 15, in f
super(B, self).f(25)
File "./test.py", line 9, in f
self.f(x-1)
TypeError: f() takes exactly 1 argument (2 given)
Thanks in advance,
Name mangling is the tool for this job. This would look like this in your case:
class A(object):
# recursive method
def f(self, x):
print x,
if x < 0:
self.__f(x+1)
if x > 0:
self.__f(x-1)
if x == 0:
print ""
__f = f
class B(A):
# Override method
def f(self):
# do some pretty cool stuff
super(B, self).f(25)
Explanation from the linked documentation:
Any identifier of the form __spam (at least two leading underscores,
at most one trailing underscore) is textually replaced with
_classname__spam, where classname is the current class name with
leading underscore(s) stripped.
In your second example, your problem is that the self you're passing along is an instance of B, not an instance of A, so when you attempt to call self.f you're calling B.f.
Unfortunately, the behavior you're seeing is really sort of how OO programming should work. Anything you do to work around this is going to be a bit of a hack around the OO paradigm. Another option which might be more explicit than using mangling, but is not necessarily "real recursion", would be to pass along the function you want to recurse on:
class A(object):
# recursive method
def f(self, x, func=None):
if func is None:
func = A.f
print x,
if x < 0:
func(self,x+1,func)
if x > 0:
func(self,x-1,func)
if x == 0:
print ""
class B(A):
# Override method
def f(self):
# do some pretty cool stuff
super(B, self).f(25)
if __name__ == "__main__":
A().f(5)
B().f()
This probably isn't the best way this could be written, but I think it gets the idea across. You could alternately try passing A.f in from your call in B.f.
I would suggest renaming the base classes f method to a private method called _f and having that recurse. You can then introduce a new f method to the base class which just calls _f. Then your free to change f in the subclass.
However it may not be considered good practice to change the method signature in a subclass.
class A(object):
def f(self, x):
return self._f(x)
# recursive method
def _f(self, x):
print x,
if x < 0:
self._f(x+1)
if x > 0:
self._f(x-1)
if x == 0:
print ""
class B(A):
# Override method
def f(self):
# do some pretty cool stuff
super(B, self).f(25)
if __name__ == "__main__":
A().f(5)
B().f()
If you can't modify A's implementation, you can take advantage of the difference in function signatures.
class B(A):
def f(self, x=None):
if x is None:
# do some pretty cool stuff
self.f(25)
else:
super(B, self).f(x)

Python, a function in a method

I have a method (__init__) in a class, and I want to use a function from the class in this method.
But when I want to run my program. I get: NameError: global name 'myfunction' is not defined
Someone, who knows what I have to do? :)
Thank you. But I have still a problem, because def myFunc(self, a): is a method and I wanted a function.
class Myclass(object):
def __init__(self, a):
self.a = self.myFunc(a)
def myFunc(self, a):
return a+1
Then you don't have a function call in the method, but you have a method call in it.
When creating a class you must specify the object when calling its methods:
>>> class A(object):
... def __init__(self, val):
... self.val = self._process(val)
... def _process(self, val):
... return val % 7
... process = _process #if you are outside methods then you don't
... #have to add "self.".
...
>>> a = A(5)
>>> a.process(3)
3
>>> a._process(6) #"a" is passed as the "self" parameter
6
As you can see in a class definition, but outside the methods you must specify the method name only, and not the "self.". Also you can't refer to a method not already defined:
>>> class B(object):
... def __init__(self):pass
... def method1(self):pass
... __call__ = method2 #method2 not defined!
...
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 4, in B
NameError: name 'method2' is not defined

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