Question

我问，给定一个装饰器的函数，是否可以在不调用装饰器调用的情况下运行该函数？

给定一个函数foo，是否可以选择打开或关闭装饰器？

给出

@decorator
def foo():
    //do_somthing

是否可以在foo关闭时运行decorator？

可能存在某些功能，您可能希望在有或没有装饰器的情况下运行它。例如（并且不是一个好的，因为它涉及有效的缓存）在factorial(n)函数中关闭基于装饰器的缓存。

我的问题与此问题类似Optionally use decorators on class methods。它讨论了装饰器切换ON / OFF的良好应用（作为api暴露）;

如果我必须使用一个函数，说goo并给出选项以运行goo有或没有装饰器，我尝试了一种原始的，hackish方式来实现这个可选的装饰器打开/关闭切换功能如下：

# this is the decorator class that executes the function `goo`
class deco(object):
    def __init__(self, attr):
        print "I am initialized"
        self.fn = None
        # some args you may wana pass
        self.attr = attr
        # lets say you want these attributes to persist
        self.cid = self.attr['cid']
        self.vid = 0

    def __call__(self, f):
        # the call executes and returns another inner wrapper
        def wrap(*args):
            # this executes main function - see closure 
            self.fn = f
            self.vid = args[0]
            self.closure(*args)
        return wrap

    def closure(self, *args):
        n = args[0]
        self.cid[n] = self.vid
        #goo = deco(fn, attr)
        print 'n',n
        # executes function `goo`
        self.fn(*args)


class gooClass(object):
    class that instantias and wraps `goo`around
    def __init__(self, attr, deco):
        '''
        @param:
              - attr: some mutable data structure
              - deco: True or False. Whether to run decorator or not
        '''
        self.attr = attr
        self.deco = deco

    def __call__(self, *args):
        if self.deco:
            # initiate deco class with passed args
            foo = deco(self.attr)
            # now pass the `goo` function to the wrapper inside foo.__class__.__call__
            foo = foo(self.goo)
            return foo(*args)
        else:
            # execute w/o decorator
            return self.goo(*args)                        

    def goo(self, n):
        # recursive goo
        if n>0:
            print 'in goo',n
            #print n
            # to recurse, I recreate the whole scene starting with the class 
            # because of this the args in `deco` Class init never persist
            too = gooClass(self.attr, self.deco)
            return too(n-1)
        else: return n


def Fn(n, decoBool):
    # this function is where to start running from
    attr = {}
    cid = [0]*(n+1)
    attr['cid'] = cid

    #following wud work too but defeat the purpose - have to define again! foo is goo actually
    #@deco(attr)
    #def foo(n):
    #    if n>0:
    #        print 'in foo',n
    #        #print n
    #        return foo(n-1)
    #    else: return n
    #return foo(n), attr
    # create the gooClass and execute `goo` method instance
    foo = gooClass(attr, decoBool)
    print foo(n)
    return foo



res = Fn(5, True)
print res.attr
print "="*10
res = Fn(5, False)
print res.attr

输出：

I am initialized
n 5
in goo 5
I am initialized
n 4
in goo 4
I am initialized
n 3
in goo 3
I am initialized
n 2
in goo 2
I am initialized
n 1
in goo 1
I am initialized
n 0
None
{'cid': [0, 1, 2, 3, 4, 5]}
==========
in goo 5
in goo 4
in goo 3
in goo 2
in goo 1
0
{'cid': [0, 0, 0, 0, 0, 0]}

技术上有效，但我认为这是一个自助式黑客攻击。不是pythonic。每次以递归方式创建新类时。

问题代表，我在这里找不到一个相关的答案，所以我创建了这个，有没有办法可以选择打开/关闭装饰器？

Answer 1

将未修饰的函数附加到装饰的函数，如unwrapped所说，然后从装饰器返回。

例如

def add42(fn):
    def wrap(i):
        return fn(i) + 42
    wrap.unwrapped = fn
    return wrap

@add42
def mult3(i):
    return i * 3

mult3(1) # 45
mult3.unwrapped(1) # 3

Answer 2

只需将foo装饰为另一个名称。然后，您可以选择使用任何一个，干净的foo或装饰的foo。

decorated_foo = decorator(foo)

如果您同时具有foo和decorator的参数：

decorated_foo = decorator(decorator_args)(foo)
#when using:
decorated_foo(foo_args)

如何使装饰器可选择打开或关闭

2 个答案: