We Keep Coding

In Python can isinstance() be used to detect a class method?

How to determine if an object is a class method? Isn't it best practice to use isinstance(), and how does one make that work?
class Foo:
class_var = 0
#classmethod
def bar(cls):
cls.class_var += 1
print("class variable value:", cls.class_var)
def wrapper(wrapped: classmethod):
"""
Call the wrapped method.
:param wrapped (classmethod, required)
"""
wrapped()
Foo.bar()
wrapper(Foo.bar)
print("the type is:", type(Foo.bar))
print("instance check success:", isinstance(Foo.bar, classmethod))
Output:
class variable value: 1
class variable value: 2
the type is: <class 'method'>
instance check success: False
Process finished with exit code 0

If you just want to tell class methods apart from regular methods and static methods, then you can check this with inspect.ismethod(f).
class A:
def method(self): pass
#classmethod
def class_method(cls): pass
#staticmethod
def static_method(): pass
In the REPL:
>>> from inspect import ismethod
>>> ismethod(A.method)
False
>>> ismethod(A.class_method)
True
>>> ismethod(A.static_method)
False
If you prefer to do this with isinstance, then that's possible using typing.types.MethodType:
>>> from typing import types
>>> isinstance(A.method, types.MethodType)
False
>>> isinstance(A.class_method, types.MethodType)
True
>>> isinstance(A.static_method, types.MethodType)
False
Note that these tests will incorrectly identify e.g. A().method because really we're just testing for a bound method as opposed to an unbound function. So the above solutions only work assuming that you are checking A.something where A is a class and something is either a regular method, a class method or a static method.

As you know Python fills the first parameter of the classmethods with a reference to the class itself and it doesn't matter if you call that method from the class or the instance of the class. A method object is a function which has an object bound to it.
That object can be retrieved by .__self__ attribute. So you can simply check that if the .__self__ attribute is a class or not. If it is a class , it's class is type.
One way of doing it:
class Foo:
#classmethod
def fn1(cls):
pass
def fn2(self):
pass
def is_classmethod(m):
first_parameter = getattr(m, '__self__', None)
if not first_parameter:
return False
type_ = type(first_parameter)
return type_ is type
print(is_classmethod(Foo.fn1))
print(is_classmethod(Foo().fn1))
print("-----------------------------------")
print(is_classmethod(Foo.fn2))
print(is_classmethod(Foo().fn2))
output:
True
True
-----------------------------------
False
False
There is a ismethod function in inspect module that specifically checks that if the object is a bound method. You can use this as well before checking for the type of the first parameter.
NOTE: There is a caveat with the above solution, I'll mention it at the end.
Solution number 2:
Your isinstance solution didn't work because classmethod is a descriptor. If you want to get the actual classmethod instance, you should check the Foo's namespace and get the methods from there.
class Foo:
#classmethod
def fn1(cls):
pass
def fn2(self):
pass
def is_classmethod(cls, m):
return isinstance(cls.__dict__[m.__name__], classmethod)
print(is_classmethod(Foo, Foo.fn1))
print(is_classmethod(Foo, Foo().fn1))
print("-----------------------------------")
print(is_classmethod(Foo, Foo.fn2))
print(is_classmethod(Foo, Foo().fn2))
Solution number 1 caveat: For example if you have a simple MethodType object whose bound object is a different class like int here, this solution isn't going to work. Because remember we just checked that if the first parameter is of type type:
from types import MethodType
class Foo:
def fn2(self):
pass
fn2 = MethodType(fn2, int)
#classmethod
def fn1(cls):
pass
Now only solution number 2 works.

Python: Check if a method uses #staticmethod [duplicate]

assume following class definition:
class A:
def f(self):
return 'this is f'
#staticmethod
def g():
return 'this is g'
a = A()
So f is a normal method and g is a static method.
Now, how can I check if the funcion objects a.f and a.g are static or not? Is there a "isstatic" funcion in Python?
I have to know this because I have lists containing many different function (method) objects, and to call them I have to know if they are expecting "self" as a parameter or not.

Lets experiment a bit:
>>> import types
>>> class A:
... def f(self):
... return 'this is f'
... #staticmethod
... def g():
... return 'this is g'
...
>>> a = A()
>>> a.f
<bound method A.f of <__main__.A instance at 0x800f21320>>
>>> a.g
<function g at 0x800eb28c0>
>>> isinstance(a.g, types.FunctionType)
True
>>> isinstance(a.f, types.FunctionType)
False
So it looks like you can use types.FunctionType to distinguish static methods.

Your approach seems a bit flawed to me, but you can check class attributes:
(in Python 2.7):
>>> type(A.f)
<type 'instancemethod'>
>>> type(A.g)
<type 'function'>
or instance attributes in Python 3.x
>>> a = A()
>>> type(a.f)
<type 'method'>
>>> type(a.g)
<type 'function'>

To supplement the answers here, in Python 3 the best way is like so:
import inspect
class Test:
#staticmethod
def test(): pass
isstatic = isinstance(inspect.getattr_static(Test, "test"), staticmethod)
We use getattr_static rather than getattr, since getattr will retrieve the bound method or function, not the staticmethod class object. You can do a similar check for classmethod types and property's (e.g. attributes defined using the #property decorator)
Note that even though it is a staticmethod, don't assume it was defined inside the class. The method source may have originated from another class. To get the true source, you can look at the underlying function's qualified name and module. For example:
class A:
#staticmethod:
def test(): pass
class B: pass
B.test = inspect.getattr_static(A, "test")
print("true source: ", B.test.__qualname__)
Technically, any method can be used as "static" methods, so long as they are called on the class itself, so just keep that in mind. For example, this will work perfectly fine:
class Test:
def test():
print("works!")
Test.test()
That example will not work with instances of Test, since the method will be bound to the instance and called as Test.test(self) instead.
Instance and class methods can be used as static methods as well in some cases, so long as the first arg is handled properly.
class Test:
def test(self):
print("works!")
Test.test(None)
Perhaps another rare case is a staticmethod that is also bound to a class or instance. For example:
class Test:
#classmethod
def test(cls): pass
Test.static_test = staticmethod(Test.test)
Though technically it is a staticmethod, it is really behaving like a classmethod. So in your introspection, you may consider checking the __self__ (recursively on __func__) to see if the method is bound to a class or instance.

I happens to have a module to solve this. And it's Python2/3 compatible solution. And it allows to test with method inherit from parent class.
Plus, this module can also test:
regular attribute
property style method
regular method
staticmethod
classmethod
For example:
class Base(object):
attribute = "attribute"
#property
def property_method(self):
return "property_method"
def regular_method(self):
return "regular_method"
#staticmethod
def static_method():
return "static_method"
#classmethod
def class_method(cls):
return "class_method"
class MyClass(Base):
pass
Here's the solution for staticmethod only. But I recommend to use the module posted here.
import inspect
def is_static_method(klass, attr, value=None):
"""Test if a value of a class is static method.
example::
class MyClass(object):
#staticmethod
def method():
...
:param klass: the class
:param attr: attribute name
:param value: attribute value
"""
if value is None:
value = getattr(klass, attr)
assert getattr(klass, attr) == value
for cls in inspect.getmro(klass):
if inspect.isroutine(value):
if attr in cls.__dict__:
bound_value = cls.__dict__[attr]
if isinstance(bound_value, staticmethod):
return True
return False

Why bother? You can just call g like you call f:
a = A()
a.f()
a.g()

Possible to hijack class definition with decorators?

Say I have a
class A:
def __init__(self, *args):
pass
and I want an decorator that copies A's definition and extend it with the new class.
def decorator(cls): # some decorator here
# make a new class which inherits from A
# return it while preserving the original A
Is that possible? (PS: This is to avoid maintainence problems.)

When you invoke a function using decorator syntax:
#my_decorator_function
class A:
pass
The decorator function's return value will replace the existing definition of A. So if you want it to create a new class and "return it while preserving the original A", you've got a tricky challenge. What you return will replace A, so you need to decide if that should be the original A or the new class. You can put the other one somewhere else.
For instance, this decorator would replace A with a subclass, and the subclass will make the original A class available as a class attribute named _orig:
def class_decorator(cls):
class SubClass(cls):
_orig = cls
# add other stuff here?
return SubClass
You can add extra logic to copy the original class's __name__ and __doc__ into the new class if you want to. You could also turn the logic around, and add SubClass as an attribute of cls before returning the otherwise unmodified cls.

Using #decorator is not the only possible syntax. You can put B = decorator(A) after the class definition.
class A:
...
B = decorator(A)
Now you still have a reference on the undecorated A, and you have a decorated version B.

The other answers have done a good job, but to make it crystal clear why you don't want to do this:
def dec(cls):
new_cls = type(cls.__name__, (cls,), {})
return new_cls
#dec
class A():
pass
Now inspect the method resolution order class A:
>>> A.__mro__
(<class '__main__.A'>, <class '__main__.A'>, <class 'object'>)
>>> classes = A.__mro__
>>> classes[0].__name__
'A'
>>> classes[1].__name__
'A'
TWO class As! Are they the same?
>>> classes[0] is classes[1]
False
Nope; different. The current variable A is pointing to the lowest one of course:
>>> A is classes[0]
True
But now you've lost name-access to the parent. That's usually not optimal.
In short: you are creating a metric ton of confusion and ambiguity for yourself a few months from now when you have forgotten all about what you did. Do something else.
If you really want to, here is an idea for spinning out new subclasses:
def add_babymaker(cls):
'''Adds a method for making new child classes.'''
def babymaker(name=None):
'''Creates a new child class based on the parent class.'''
name = name if name is not None else cls.__name__
new_cls = type(name, (cls,), {})
return new_cls
cls.babymaker = babymaker
return cls
#add_babymaker
class A():
pass
B = A.babymaker('B')
C = A.babymaker('C')
ANew = A.babymaker()

I think I have worked it out. That's not really a good idea.
def make_variant(cls):
suffix='VARIANT'
new = type(cls.__name__+suffix, (cls, ), {})
# new.__repr__ = lambda self: 'HELLO' # Just do whatever needed here
assert cls.__name__ + suffix not in globals()
globals()[cls.__name__+suffix] = new # Think twice about this line
return cls
#make_variant
class A:
def __init__(self):
pass
print(AVARIANT(), A())

Ways to set per instance __call__?

What I want to do is something like:
class Foo(object):
def __init__(self):
pass
def f(self):
print "f"
def g(self):
print "g"
# programatically set the "default" operation
fer=Foo()
fer.__call__=fer.f
# a different instance does something else as its
# default operation
ger=Foo()
ger.__call__=ger.g
fer() # invoke different functions on different
ger() # objects depending on how they were set up.
But as of 2.7 (which I'm currently using) I can't do this, the attempts at fer()
raise an exception.
Is there a way to, in effect, set a per instance __call__ method?

The normal stuff with types.MethodType unfortunately doesn't work here since __call__ is a special method.
From the data model:
Class instances are callable only when the class has a __call__() method; x(arguments) is a shorthand for x.__call__(arguments).
This is slightly ambiguous as to what is actually called, but it's clear that your class needs to have a __call__ method.
You'll need to create some sort of hack:
class Foo(object):
def __init__(self):
pass
def f(self):
print "f"
def g(self):
print "g"
def __call__(self):
return self.__call__()
f = Foo()
f.__call__ = f.f
f()
g = Foo()
g.__call__ = g.g
g()
Careful with this though, it'll result in an infinite recursion if you don't set a __call__ on an instance before you try to call it.
Note that I don't actually recommend calling the magic attribute that you rebind __call__. The point here is to demonstrate that python translates: f() into f.__class__.__call__(f) and so there's nothing you can do to change it on a per-instance basis. the class's __call__ will be called no matter what you do -- You just need to do something to change the behavior of the class's __call__ per-instance which is easily achieved.
You could use a setter type thing to actually create methods on your class (rather than simple functions) -- and of course that could be turned into a property:
import types
class Foo(object):
def __init__(self):
pass
def f(self):
print "f"
def g(self):
print "g"
def set_func(self,f):
self.func = types.MethodType(f,self)
def __call__(self,*args,**kwargs):
self.func(*args,**kwargs)
f = Foo()
f.set_func(Foo.f)
f()
def another_func(self,*args):
print args
f.set_func(another_func)
f(1,2,3,"bar")

You might be trying to solve the wrong problem.
Since python allows procedural creation of classes you could write code like that:
>>> def create_class(cb):
... class Foo(object):
... __call__ = cb
... return Foo
...
>>> Foo1 = create_class(lambda self: 42)
>>> foo1 = Foo1()
>>> foo1()
>>> Foo2 = create_class(lambda self: self.__class__.__name__)
>>> foo2 = Foo2()
>>> foo2()
Please note thought that Foo1 and Foo2 do not have a common base class in this case. So isinstance and issubclass will not work. If you need them to have a common base class I would go for the following code:
>>> class Foo(object):
... #classmethod
... def create_subclass(cls, cb):
... class SubFoo(cls):
... __call__ = cb
... return SubFoo
...
>>> Foo1 = Foo.create_subclass(lambda self: 42)
>>> foo1 = Foo1()
>>> foo1()
>>> Foo2 = Foo.create_subclass(lambda self: self.__class__.__name__)
>>> foo1 = Foo2()
>>> foo2()
'Foo'
>>> issubclass(Foo1, Foo)
True
>>> issubclass(Foo2, Foo)
True
I really like the second way as it provides a clean class hierarchy and looks quite clean to me.

Possible solution:
class Foo(object):
def __init__(self):
self._callable = lambda s: None
def f(self):
print "f"
def set_callable(self, func):
self._callable = func
def g(self):
print "g"
def __call__(self):
return self._callable()
d = Foo()
d.set_callable(d.g)

Python __str__: Magic Console

Suppose one decided (yes, this is horrible) to create handle input in the following manner: A user types in a command on the python console after importing your class, the command is actually a class name, the class name's __str__ function is actually a function with side effects (e.g. the command is "north" and the function changes some global variables and then returns text describing your current location). Obviously this is a stupid thing to do, but how would you do it (if possible)?
Note that the basic question is how to define the __str__ method for a class without creating an instance of the class, otherwise it would be simple (but still just as crazy:
class ff:
def __str__(self):
#do fun side effects
return "fun text string"
ginst = ff()
>>ginst

What you are looking for is the metaclass
class Magic(type):
def __str__(self):
return 'Something crazy'
def __repr__(self):
return 'Another craziness'
class Foo(object):
__metaclass__ = Magic
>>> print Foo
Something crazy
>>> Foo
Another craziness

in console you're getting representation of your object, which __repr__ is responsible for. __str__ used for printing:
>>> class A:
def __str__(self):
return 'spam'
>>> A()
<__main__.A object at 0x0107E3D0>
>>> print(A())
spam
>>> class B:
def __repr__(self):
return 'ham'
>>> B()
ham
>>> print(B())
ham
>>> class C:
def __str__(self):
return 'spam'
def __repr__(self):
return 'ham'
>>> C()
ham
>>> print(C())
spam

You could use instances of a class rather than classes themselves. Something like
class MagicConsole(object):
def __init__(self, f):
self.__f = f
def __repr__(self):
return self.__f()
north = MagicConsole(some_function_for_north)
south = MagicConsole(some_function_for_south)
# etc