20

如果我有一个功能:


@aDecorator
def myfunc1():
  # do something here

if __name__ = "__main__":
  # this will call the function and will use the decorator @aDecorator
  myfunc1() 
  # now I want the @aDecorator to be replaced with the decorator @otherDecorator
  # so that when this code executes, the function no longer goes through
  # @aDecorator, but instead through @otherDecorator. How can I do this?
  myfunc1()

是否可以在运行时替换装饰器?

4

7 回答 7

17

正如 Miya 所提到的,您可以在解释器到达该函数声明之前的任何时候用另一个函数替换装饰器。但是,一旦将装饰器应用于函数,我认为没有办法用不同的装饰器动态替换装饰器。例如:

@aDecorator
def myfunc1():
    pass

# Oops! I didn't want that decorator after all!

myfunc1 = bDecorator(myfunc1)

不起作用,因为 myfunc1 不再是您最初定义的函数;它已经被包裹了。这里最好的方法是手动应用装饰器,oldskool 风格,即:

def myfunc1():
    pass

myfunc2 = aDecorator(myfunc1)
myfunc3 = bDecorator(myfunc1)

编辑:或者,更清楚一点,

def _tempFunc():
    pass

myfunc1 = aDecorator(_tempFunc)
myfunc1()
myfunc1 = bDecorator(_tempFunc)
myfunc1()
于 2009-03-13T13:55:56.350 回答
5

我不知道是否有办法在应用装饰器后“替换”它,但我想可能没有,因为功能已经改变。

无论如何,您可能会根据某些条件在运行时应用装饰器:

#!/usr/bin/env python

class PrintCallInfo:
    def __init__(self,f):
        self.f = f
    def __call__(self,*args,**kwargs):
        print "-->",self.f.__name__,args,kwargs
        r = self.f(*args,**kwargs)
        print "<--",self.f.__name__,"returned: ",r
        return r

# the condition to modify the function...
some_condition=True

def my_decorator(f):
    if (some_condition): # modify the function
        return PrintCallInfo(f)
    else: # leave it as it is
        return f

@my_decorator
def foo():
    print "foo"

@my_decorator
def bar(s):
    print "hello",s
    return s

@my_decorator
def foobar(x=1,y=2):
    print x,y
    return x + y

foo()
bar("world")
foobar(y=5)
于 2009-03-13T14:55:06.770 回答
3

这是一个很棒的食谱,可以帮助您入门。基本上,这个想法是将一个类实例传递给装饰器。然后,您可以在类实例上设置属性(如果您愿意,可以将其设为 Borg)并使用它来控制装饰器本身的行为。

这是一个例子:

class Foo:
    def __init__(self, do_apply):
        self.do_apply = do_apply

def dec(foo):
    def wrap(f):
        def func(*args, **kwargs):
            if foo.do_apply:
                # Do something!
                pass 
            return f(*args, **kwargs)
        return func
    return wrap

foo = Foo(False)
@dec(foo)
def bar(x):
    return x

bar('bar') 
foo.do_apply = True 
# Decorator now active!
bar('baz')

当然,你也可以合并“装饰器装饰器”来保存签名等。

于 2009-03-13T13:59:40.710 回答
2

如果要显式更改装饰器,不妨选择更显式的方法,而不是创建装饰函数:

deco1(myfunc1, arg1, arg2)
deco2(myfunc1, arg2, arg3)

deco1() 和 deco2() 将应用装饰器提供的功能并使用参数调用 myfunc1()。

于 2009-03-13T13:53:41.440 回答
1

当然——你可以得到函数对象并用它做任何你想做的事情:

# Bypass a decorator

import types

class decorator_test(object):

    def __init__(self, f):
        self.f = f

    def __call__(self):
        print "In decorator ... entering: ", self.f.__name__
        self.f()
        print "In decorator ... exiting: ", self.f.__name__


@decorator_test
def func1():
    print "inside func1()"

print "\nCalling func1 with decorator..."
func1()

print "\nBypassing decorator..."
for value in func1.__dict__.values():
    if isinstance(value, types.FunctionType) and value.func_name == "func1":
        value.__call__()
于 2010-11-16T23:26:59.363 回答
-1

我知道这是一个旧线程,但我玩得很开心

def change_deco(name, deco, placeholder='    #'):
with open(name + '.py', 'r') as file:
    lines = file.readlines()
for idx, string in enumerate(lines):
    if placeholder in string and repr(placeholder) not in string:
        lines[idx] = f'    @{deco}\r\n'
exec(''.join(lines))
return locals()[name]
于 2017-11-18T04:42:22.673 回答
-3

如果装饰器是一个函数,只需替换它。

aDecorator = otherDecorator
于 2009-03-13T13:43:58.673 回答