是否可以有条件地装饰函数.例如,我想用定时器函数(timeit
)装饰函数foo()
只有doing_performance_analysis是True
(见伪代码下面).
Is it possible to decorator a function conditionally. For example, I want to decorate the function foo()
with a timer function (timeit
) only doing_performance_analysis is True
(see the psuedo-code below).
if doing_performance_analysis:
@timeit
def foo():
"""
do something, timeit function will return the time it takes
"""
time.sleep(2)
else:
def foo():
time.sleep(2)
装饰器是简单的可调用函数,它返回一个替换,可选相同的函数、包装器或完全不同的东西.因此,您可以创建一个条件装饰器:
Decorators are simply callables that return a replacement, optionally the same function, a wrapper, or something completely different. As such, you could create a conditional decorator:
def conditional_decorator(dec, condition):
def decorator(func):
if not condition:
# Return the function unchanged, not decorated.
return func
return dec(func)
return decorator
现在你可以像这样使用它了:
Now you can use it like this:
@conditional_decorator(timeit, doing_performance_analysis)
def foo():
time.sleep(2)
装饰器也可以是一个类:
The decorator could also be a class:
class conditional_decorator(object):
def __init__(self, dec, condition):
self.decorator = dec
self.condition = condition
def __call__(self, func):
if not self.condition:
# Return the function unchanged, not decorated.
return func
return self.decorator(func)
这里的__call__
方法和第一个例子中返回的decorator()
嵌套函数的作用一样,封闭的dec
和 condition
参数在这里作为参数存储在实例上,直到应用装饰器.
Here the __call__
method plays the same role as the returned decorator()
nested function in the first example, and the closed-over dec
and condition
parameters here are stored as arguments on the instance until the decorator is applied.
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