如何在python中做一个条件装饰器

Question

是否有可能有条件地装饰一个功能.例如,我想foo()用timer函数装饰函数(timeit)只有doing_performance_analysis True(参见下面的伪代码).

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)  

Answer 1

装饰器只是返回替换,可选择相同功能,包装器或完全不同的东西的callables.因此,您可以创建条件装饰器:

def conditional_decorator(dec, condition):
    def decorator(func):
        if not condition:
            # Return the function unchanged, not decorated.
            return func
        return dec(func)
    return decorator

现在您可以像这样使用它:

@conditional_decorator(timeit, doing_performance_analysis)
def foo():
    time.sleep(2)  

装饰者也可以是一个类:

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decorator()与第一个示例中返回的嵌套函数扮演相同的角色,此处的closed-over dec和conditionparameters作为参数存储在实例中,直到应用了decorator.

Answer 2

装饰器只是应用于另一个功能的函数.您可以手动应用它:

def foo():
   # whatever
   time.sleep(2)

if doing_performance_analysis:
    foo = timeit(foo)

Answer 3

怎么样：

def foo():
   ...

if doing_performance_analysis:
   foo = timeit(foo)

我想您甚至可以将其包装到一个带有布尔值标志的装饰器和另一个装饰器中，并且仅当将标志设置为时才应用后者True。

def cond_decorator(flag, dec):
   def decorate(fn):
      return dec(fn) if flag else fn
   return decorate

@cond_decorator(doing_performance_analysis, timeit)
def foo():
   ...