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.)
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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