Python : Assert that variable is instance method?

问题:

How can one check if a variable is an instance method or not? I'm using python 2.5.

Something like this:

class Test:
    def method(self):
        pass

assert is_instance_method(Test().method)

回答1:

inspect.ismethod is what you want to find out if you definitely have a method, rather than just something you can call.

import inspect

def foo(): pass

class Test(object):
    def method(self): pass

print inspect.ismethod(foo) # False
print inspect.ismethod(Test) # False
print inspect.ismethod(Test.method) # True
print inspect.ismethod(Test().method) # True

print callable(foo) # True
print callable(Test) # True
print callable(Test.method) # True
print callable(Test().method) # True

callable is true if the argument if the argument is a method, a function (including lambdas), an instance with __call__ or a class.

Methods have different properties than functions (like im_class and im_self). So you want

assert inspect.ismethod(Test().method)  

回答2:

If you want to know if it is precisely an instance method use the following function. (It considers methods that are defined on a metaclass and accessed on a class class methods, although they could also be considered instance methods)

import types
def is_instance_method(obj):
    """Checks if an object is a bound method on an instance."""
    if not isinstance(obj, types.MethodType):
        return False # Not a method
    if obj.im_self is None:
        return False # Method is not bound
    if issubclass(obj.im_class, type) or obj.im_class is types.ClassType:
        return False # Method is a classmethod
    return True

Usually checking for that is a bad idea. It is more flexible to be able to use any callable() interchangeably with methods.