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Python3有条件地装饰?

沃梦达教程 Python问题

2022-01-01

Python3 decorating conditionally?(Python3有条件地装饰?)

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问题描述

是否可以根据条件装饰函数?

Is it possible to decorate a function based on a condition?

阿拉:

if she.weight() == duck.weight(): 
    @burn
def witch():
    pass

我只是想知道是否可以使用逻辑(当调用 witch 时?)来确定是否用 @burn<装饰 witch/代码>?


I'm just wondering if logic could be used (when witch is called?) to figure out whether or not to decorate witch with @burn?
如果不是，是否可以在装饰器中创建一个条件以达到相同的效果?(witch 被称为未装饰.)
If not, is it possible to create a condition within the decorator to the same effect? (witch being called undecorated.)
推荐答案
你可以创建一个'有条件'的装饰器:
You can create a 'conditionally' decorator: 
>>> def conditionally(dec, cond):
    def resdec(f):
        if not cond:
            return f
        return dec(f)
    return resdec

用法示例如下:
>>> def burn(f):
    def blah(*args, **kwargs):
        print 'hah'
        return f(*args, **kwargs)
    return blah

>>> @conditionally(burn, True)
def witch(): pass
>>> witch()
hah

>>> @conditionally(burn, False)
def witch(): pass
>>> witch()


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