Decorate methods per instance in Python

问题

Assume I have some simple class

class TestClass:
    def doSomething(self):
        print 'Did something'

I would like to decorate the doSomething method, for example to count the number of calls

class SimpleDecorator(object):
    def __init__(self,func):
        self.func=func
        self.count=0
    def __get__(self,obj,objtype=None):
        return MethodType(self,obj,objtype)
    def __call__(self,*args,**kwargs):
        self.count+=1
        return self.func(*args,**kwargs)

Now this counts the number of calls to the decorated method, however I would like to have per-instance counter, such that after

foo1=TestClass()
foo1.doSomething()
foo2=TestClass()

foo1.doSomething.count is 1 and foo2.doSomething.count is 0. From what I understand, this is not possible using decorators. Is there some way to achieve such behaviour?

回答1:

Utilize the fact that self (i.e. the object which the method is invoked on) is passed as a parameter to the method:

import functools

def counted(method):
    @functools.wraps(method)
    def wrapped(obj, *args, **kwargs):
        if hasattr(obj, 'count'): 
            obj.count += 1
        else:
            obj.count = 1
        return method(obj, *args, **kwargs)
    return wrapped

In above code, we intercept the object as obj parameter of the decorated version of method. Usage of the decorator is pretty straightforward:

class Foo(object):
    @counted
    def do_something(self): pass

回答2:

Wouldn't the first element of *args be the object the method is being invoked on? Can't you just store the count there?

来源：https://stackoverflow.com/questions/6356858/decorate-methods-per-instance-in-python

标签

python

decorator