问题描述

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)  

推荐答案

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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