i don't understand this error in Python. I read about "self" and "__init__" in this previous question , where says that Python does not pass transparently the instance to the constructor. So I tried a simple class definition and then declare a new instance.
#Basic class
class Testing:
atr1 = 33
def __init__():
pass
def sayHi():
print("Hello world")
When I try to declare a new instance of this class, Jupyter throws this error:
t1 = Testing()
TypeError Traceback (most recent call last)
<ipython-input-6-0019e8f92b90> in <module>
----> 1 t1 = Testing()
TypeError: __init__() takes 0 positional arguments but 1 was given
So for me, this error doesn't make sense, otherwise, Python would be actually passing the instance itself as an argument when initializes the new instance and because i'm not giving an explicit argument.
The solution is quite simple: just write "self" as an argument of init method, but i'm still confused about the error.
Hope somebody can explain me this weird error message u.u
The first parameter to an object method is a reference to the object itself. Traditionally its called self but really you could name it anything you want. In the end, a method is really just a function assigned to a class. That's what happened when you did
class Testing:
def __init__():
pass
The def caused python to compile a function and assign it to __init__. Because __init__ is in the Testing class namespace, it assigned it to the class. You could just as easily have done
class Testing:
pass
def whatever():
pass
Testing.__init__ = whatever
So, the idea of python just magically creating the self parameter on methods doesn't work. It would be a crazy rule for regular functions.
__init__ is an initializer, not a constructor. The object has been constructed to the point that it has a functioning self by the time __init__ has been called. Classes also have a __new__ that can be used to construct the object.
I'm trying to call an external function via a class variable. The following is a simplification of my real code:
def func(arg):
print(arg)
class MyClass(object):
func_ref = None
#classmethod
def setUpClass(cls):
#MyClass.func_ref = func
cls.func_ref = func
#staticmethod
def func_override(arg):
print("override printing arg...")
MyClass.func_ref(arg)
if __name__ == "__main__":
print(type(func))
print(type(MyClass.func_ref))
MyClass.setUpClass()
print(type(MyClass.func_ref))
MyClass.func_override("hello!")
The above code produces the following output:
[~]$ python tmp.py
<type 'function'>
<type 'NoneType'>
<type 'instancemethod'>
override printing arg...
Traceback (most recent call last):
File "tmp.py", line 20, in <module>
MyClass.func_override("hello!")
TypeError: func_override() takes exactly 2 arguments (1 given)
The situation seems to be unchanged if I use MyClass in place of cls within the classmethod setUpClass().
I would expect the type of MyClass.func_ref to be function after the assignment in setUpClass() which explains the TypeError I get when I try to call it. Why is the type of func_ref being changed to instancemethod when the value I assigned to it is of type function?
This only seems to be an issue in Python 2. Python 3 behaves as I would expect.
How do I get calls to the static method MyClass.func_override() to call func()?
UPDATE
I was able to get the above to work by applying the following patch:
## -14,7 +14,7 ## class MyClass(object):
def func_override(arg):
print("override printing arg...")
func(arg)
- MyClass.func_ref.__func__(arg)
+ MyClass.func_ref(arg)
if __name__ == "__main__":
print(type(func))
While the above works, its not at all clear to me why I needed to do this. I still don't understand why the type of func_ref ends up an instancemethod when I assigned to it a value of type function.
Just put the function through a staticmethod as follows:
#classmethod
def setUpClass(cls):
#MyClass.func_ref = func
cls.func_ref = staticmethod(func)
There's no need to play with #-based decorators in this case as you want to modify how the method is bound to MyClass, not the general definition of func.
Why is this necessary? Because, when you assign a method to class, Python assumes you'll want to refer to an instance (via self) or the class (via cls). self, unlike this in JS, is only a naming convention, so when it sees arg it assumes it got an instance, but you passed a string in your call.
So, as as Python cares, you might have as well have written def func(self):. Which is why the message says unbound method func() must be called with MyClass 👉instance👈 as first argument.
staticmethod means, "please leave this alone and don't assume an instance or a class in the first variable".
You can even dispense with the setUpClass entirely:
class MyClass(object):
func_ref = staticmethod(func)
BTW: In 2021, 16 months past EOL, Python 2.7 has all the subtle fagrance of moldy gym socks. Except less safe, virologically-speaking.
When func_ref is called, it's expecting a self argument, just like any other normal (instance) class method (see this question and answers for discussions why). You can either add a self argument to func or make func a static method:
#staticmethod
def func(arg):
print(arg)
>>> MyClass.setUpClass()
>>> MyClass.func_override("hello!")
override printing arg...
hello!
Note that in either case func is now not normally callable as a regular function:
>>> func('what does this do?')
TypeError: 'staticmethod' object is not callable
If you need func to be usable as a regular function, you can wrap it with another, qualifying function and use the wrapper in MyClass:
def func(arg):
print(arg)
#staticmethod
def func_wrapper(arg):
func(arg)
class MyClass(object):
#classmethod
def setUpClass(cls):
cls.func_ref = func_wrapper # use wrapper function
>>> MyClass.setUpClass()
>>> MyClass.func_override("success!")
override printing arg...
success!
This is how we make static functions in Python:
class A:
#staticmethod
def fun():
print 'hello'
A.fun()
This works as expected and prints hello.
If it is a member function instead of a static one, we use self:
class A:
def fun(self):
print 'hello'
A().fun()
which also works as expected and prints hello.
My confusion is with the following case:
class A:
def fun():
print 'hello'
In the above case, there is no staticmethod, nor self. Python interpreter is okay with this definition. However, we cannot call it either of the above methods, namely:
A.fun()
A().fun()
both gives errors.
My question is: Is there any way that I can call this function? If not, why Python do not give me a syntax error in the first place?
Python doesn't give you a syntax error, because the binding of a method (which takes care of passing in self) is a runtime action.
Only when you look up a method on a class or instance, is a method being bound (because functions are descriptors they produce a method when looked up this way). This is done via the descriptor.__get__() method, which is called by the object.__getattribute__() method, which Python called when you tried to access the fun attribute on the A class or A() instance.
You can always 'unwrap' the bound method and reach for the un-wrapped function underneath to call it directly:
A.fun.__func__()
Incidentally, that's exactly what staticmethod does; it is there to 'intercept' the descriptor binding and return the raw function object instead of a bound method. In other words, staticmethod undoes the normal runtime method binding:
Demo:
>>> class A(object): pass
...
>>> def fun(): print 'hello!'
...
>>> fun.__get__(None, A) # binding to a class
<unbound method A.fun>
>>> fun.__get__(None, A)() # calling a bound function, fails as there is no first argument
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: unbound method fun() must be called with A instance as first argument (got nothing instead)
>>> fun.__get__(None, A).__func__ # access the wrapped function
<function fun at 0x100ba8378>
>>> staticmethod(fun).__get__(None, A) # staticmethod object just returns the function
<function fun at 0x100ba8378>
>>> staticmethod(fun).__get__(None, A)() # so calling it works
hello!
I am working in a dynamic programming environment where I might need to define (or redefine) a class function. So consider this for example:
def func(self):
print("hello2 \n")
class ManClass:
def __init__(self):
pass
def func1(self):
print("hello1\n")
a = ManClass()
a.func1()
hello1
a.func2 = func
>>> a.func2()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: func() takes exactly 1 argument (0 given)
If func2() had been defined inside the class - a.func2() would have been interpreted as ManClass.func2(a) - but now that I am assigning it outside, it seems to expect an argument. How do I fix this, but more importantly, why this difference in how the two definitions are interpereted ?
You didn't add func to the class, you added it to an instance. Try ManClass.func2 = func instead.
a.func2 = func adds func to the a instance of the class as an instance attribute named func2, not as an instance member method (which is really just special handling for callable members on the underlying class object).
Alternatively, you can also add a member method to a single instance using MethodType, as #jonrsharpe points out in his answer.
This is the difference between a function and a bound method, where "bound" refers to the instance self. To fix your problem, you need to make the standalone function MethodType:
from types import MethodType
a.func2 = MethodType(func, a)
This binds the func to the ManClass instance a, allowing it to access any instance attributes. Note that this only affects a, other ManClass instances will retain the original class definition unless similarly patched.
When you simply attach the function
a.func2 = func
you can still access it:
a.func2(None) # will print "hello2 \n"
But it doesn't get the implicit object instance self parameter and just treats it as a standard positional argument.
What is the difference between the following class methods?
Is it that one is static and the other is not?
class Test(object):
def method_one(self):
print "Called method_one"
def method_two():
print "Called method_two"
a_test = Test()
a_test.method_one()
a_test.method_two()
In Python, there is a distinction between bound and unbound methods.
Basically, a call to a member function (like method_one), a bound function
a_test.method_one()
is translated to
Test.method_one(a_test)
i.e. a call to an unbound method. Because of that, a call to your version of method_two will fail with a TypeError
>>> a_test = Test()
>>> a_test.method_two()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: method_two() takes no arguments (1 given)
You can change the behavior of a method using a decorator
class Test(object):
def method_one(self):
print "Called method_one"
#staticmethod
def method_two():
print "Called method two"
The decorator tells the built-in default metaclass type (the class of a class, cf. this question) to not create bound methods for method_two.
Now, you can invoke static method both on an instance or on the class directly:
>>> a_test = Test()
>>> a_test.method_one()
Called method_one
>>> a_test.method_two()
Called method_two
>>> Test.method_two()
Called method_two
Methods in Python are a very, very simple thing once you understood the basics of the descriptor system. Imagine the following class:
class C(object):
def foo(self):
pass
Now let's have a look at that class in the shell:
>>> C.foo
<unbound method C.foo>
>>> C.__dict__['foo']
<function foo at 0x17d05b0>
As you can see if you access the foo attribute on the class you get back an unbound method, however inside the class storage (the dict) there is a function. Why's that? The reason for this is that the class of your class implements a __getattribute__ that resolves descriptors. Sounds complex, but is not. C.foo is roughly equivalent to this code in that special case:
>>> C.__dict__['foo'].__get__(None, C)
<unbound method C.foo>
That's because functions have a __get__ method which makes them descriptors. If you have an instance of a class it's nearly the same, just that None is the class instance:
>>> c = C()
>>> C.__dict__['foo'].__get__(c, C)
<bound method C.foo of <__main__.C object at 0x17bd4d0>>
Now why does Python do that? Because the method object binds the first parameter of a function to the instance of the class. That's where self comes from. Now sometimes you don't want your class to make a function a method, that's where staticmethod comes into play:
class C(object):
#staticmethod
def foo():
pass
The staticmethod decorator wraps your class and implements a dummy __get__ that returns the wrapped function as function and not as a method:
>>> C.__dict__['foo'].__get__(None, C)
<function foo at 0x17d0c30>
Hope that explains it.
When you call a class member, Python automatically uses a reference to the object as the first parameter. The variable self actually means nothing, it's just a coding convention. You could call it gargaloo if you wanted. That said, the call to method_two would raise a TypeError, because Python is automatically trying to pass a parameter (the reference to its parent object) to a method that was defined as having no parameters.
To actually make it work, you could append this to your class definition:
method_two = staticmethod(method_two)
or you could use the #staticmethod function decorator.
>>> class Class(object):
... def __init__(self):
... self.i = 0
... def instance_method(self):
... self.i += 1
... print self.i
... c = 0
... #classmethod
... def class_method(cls):
... cls.c += 1
... print cls.c
... #staticmethod
... def static_method(s):
... s += 1
... print s
...
>>> a = Class()
>>> a.class_method()
1
>>> Class.class_method() # The class shares this value across instances
2
>>> a.instance_method()
1
>>> Class.instance_method() # The class cannot use an instance method
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: unbound method instance_method() must be called with Class instance as first argument (got nothing instead)
>>> Class.instance_method(a)
2
>>> b = 0
>>> a.static_method(b)
1
>>> a.static_method(a.c) # Static method does not have direct access to
>>> # class or instance properties.
3
>>> Class.c # a.c above was passed by value and not by reference.
2
>>> a.c
2
>>> a.c = 5 # The connection between the instance
>>> Class.c # and its class is weak as seen here.
2
>>> Class.class_method()
3
>>> a.c
5
method_two won't work because you're defining a member function but not telling it what the function is a member of. If you execute the last line you'll get:
>>> a_test.method_two()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: method_two() takes no arguments (1 given)
If you're defining member functions for a class the first argument must always be 'self'.
Accurate explanation from Armin Ronacher above, expanding on his answers so that beginners like me understand it well:
Difference in the methods defined in a class, whether static or instance method(there is yet another type - class method - not discussed here so skipping it), lay in the fact whether they are somehow bound to the class instance or not. For example, say whether the method receives a reference to the class instance during runtime
class C:
a = []
def foo(self):
pass
C # this is the class object
C.a # is a list object (class property object)
C.foo # is a function object (class property object)
c = C()
c # this is the class instance
The __dict__ dictionary property of the class object holds the reference to all the properties and methods of a class object and thus
>>> C.__dict__['foo']
<function foo at 0x17d05b0>
the method foo is accessible as above. An important point to note here is that everything in python is an object and so references in the dictionary above are themselves pointing to other objects. Let me call them Class Property Objects - or as CPO within the scope of my answer for brevity.
If a CPO is a descriptor, then python interpretor calls the __get__() method of the CPO to access the value it contains.
In order to determine if a CPO is a descriptor, python interpretor checks if it implements the descriptor protocol. To implement descriptor protocol is to implement 3 methods
def __get__(self, instance, owner)
def __set__(self, instance, value)
def __delete__(self, instance)
for e.g.
>>> C.__dict__['foo'].__get__(c, C)
where
self is the CPO (it could be an instance of list, str, function etc) and is supplied by the runtime
instance is the instance of the class where this CPO is defined (the object 'c' above) and needs to be explicity supplied by us
owner is the class where this CPO is defined(the class object 'C' above) and needs to be supplied by us. However this is because we are calling it on the CPO. when we call it on the instance, we dont need to supply this since the runtime can supply the instance or its class(polymorphism)
value is the intended value for the CPO and needs to be supplied by us
Not all CPO are descriptors. For example
>>> C.__dict__['foo'].__get__(None, C)
<function C.foo at 0x10a72f510>
>>> C.__dict__['a'].__get__(None, C)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
AttributeError: 'list' object has no attribute '__get__'
This is because the list class doesnt implement the descriptor protocol.
Thus the argument self in c.foo(self) is required because its method signature is actually this C.__dict__['foo'].__get__(c, C) (as explained above, C is not needed as it can be found out or polymorphed)
And this is also why you get a TypeError if you dont pass that required instance argument.
If you notice the method is still referenced via the class Object C and the binding with the class instance is achieved via passing a context in the form of the instance object into this function.
This is pretty awesome since if you chose to keep no context or no binding to the instance, all that was needed was to write a class to wrap the descriptor CPO and override its __get__() method to require no context.
This new class is what we call a decorator and is applied via the keyword #staticmethod
class C(object):
#staticmethod
def foo():
pass
The absence of context in the new wrapped CPO foo doesnt throw an error and can be verified as follows:
>>> C.__dict__['foo'].__get__(None, C)
<function foo at 0x17d0c30>
Use case of a static method is more of a namespacing and code maintainability one(taking it out of a class and making it available throughout the module etc).
It maybe better to write static methods rather than instance methods whenever possible, unless ofcourse you need to contexualise the methods(like access instance variables, class variables etc). One reason is to ease garbage collection by not keeping unwanted reference to objects.
that is an error.
first of all, first line should be like this (be careful of capitals)
class Test(object):
Whenever you call a method of a class, it gets itself as the first argument (hence the name self) and method_two gives this error
>>> a.method_two()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: method_two() takes no arguments (1 given)
The second one won't work because when you call it like that python internally tries to call it with the a_test instance as the first argument, but your method_two doesn't accept any arguments, so it wont work, you'll get a runtime error.
If you want the equivalent of a static method you can use a class method.
There's much less need for class methods in Python than static methods in languages like Java or C#. Most often the best solution is to use a method in the module, outside a class definition, those work more efficiently than class methods.
The call to method_two will throw an exception for not accepting the self parameter the Python runtime will automatically pass it.
If you want to create a static method in a Python class, decorate it with the staticmethod decorator.
Class Test(Object):
#staticmethod
def method_two():
print "Called method_two"
Test.method_two()
Please read this docs from the Guido First Class everything Clearly explained how Unbound, Bound methods are born.
Bound method = instance method
Unbound method = static method.
The definition of method_two is invalid. When you call method_two, you'll get TypeError: method_two() takes 0 positional arguments but 1 was given from the interpreter.
An instance method is a bounded function when you call it like a_test.method_two(). It automatically accepts self, which points to an instance of Test, as its first parameter. Through the self parameter, an instance method can freely access attributes and modify them on the same object.
Unbound Methods
Unbound methods are methods that are not bound to any particular class instance yet.
Bound Methods
Bound methods are the ones which are bound to a specific instance of a class.
As its documented here, self can refer to different things depending on the function is bound, unbound or static.
Take a look at the following example:
class MyClass:
def some_method(self):
return self # For the sake of the example
>>> MyClass().some_method()
<__main__.MyClass object at 0x10e8e43a0># This can also be written as:>>> obj = MyClass()
>>> obj.some_method()
<__main__.MyClass object at 0x10ea12bb0>
# Bound method call:
>>> obj.some_method(10)
TypeError: some_method() takes 1 positional argument but 2 were given
# WHY IT DIDN'T WORK?
# obj.some_method(10) bound call translated as
# MyClass.some_method(obj, 10) unbound method and it takes 2
# arguments now instead of 1
# ----- USING THE UNBOUND METHOD ------
>>> MyClass.some_method(10)
10
Since we did not use the class instance — obj — on the last call, we can kinda say it looks like a static method.
If so, what is the difference between MyClass.some_method(10) call and a call to a static function decorated with a #staticmethod decorator?
By using the decorator, we explicitly make it clear that the method will be used without creating an instance for it first. Normally one would not expect the class member methods to be used without the instance and accesing them can cause possible errors depending on the structure of the method.
Also, by adding the #staticmethod decorator, we are making it possible to be reached through an object as well.
class MyClass:
def some_method(self):
return self
#staticmethod
def some_static_method(number):
return number
>>> MyClass.some_static_method(10) # without an instance
10
>>> MyClass().some_static_method(10) # Calling through an instance
10
You can’t do the above example with the instance methods. You may survive the first one (as we did before) but the second one will be translated into an unbound call MyClass.some_method(obj, 10) which will raise a TypeError since the instance method takes one argument and you unintentionally tried to pass two.
Then, you might say, “if I can call static methods through both an instance and a class, MyClass.some_static_method and MyClass().some_static_method should be the same methods.” Yes!