我如何成功地使用装饰器函数来解决我的问题？

Question

我尝试对装饰器使用派"@“语法，但我似乎无法使其适用于我的特定示例。

这样我就能达到我想要的效果。

import pandas as pd

def prepare_todays_campaigns(decorated):
  
  campaigns_outgoing = pd.DataFrame({"test":[1,2,3]})

  n_campaigns = len(campaigns_outgoing)

  if n_campaigns > 0:

      print(
          f"{n_campaigns} to process."
      )

      decorated(campaigns_outgoing)
      
def dec_test(campaigns_outgoing):
  for _, row in campaigns_outgoing.iterrows():
    print(row)
    

prepare_todays_campaigns(dec_test)

即

​

​

但是当我尝试使用@语法糖时，我似乎得不到预期的结果……

def prepare_todays_campaigns(decorated):
  
  def wrapper(*args, **kwargs):
    # creatives x campaign
    campaigns_outgoing = pd.DataFrame({"test":[1,2,3]})

    n_campaigns = len(campaigns_outgoing)

    if n_campaigns > 0:

        print(
            f"{n_campaigns} to process."
        )

        decorated(*args, **kwargs)

    
  return wrapper

@prepare_todays_campaigns
def dec_test(campaigns_outgoing):
  for _, row in campaigns_outgoing.iterrows():
    print(row)
    
dec_test()

结果：

​

​

有没有人对如何解决这个问题有什么建议？

Answer 1

看一下方法签名def dec_test(campaigns_outgoing):

当你调用这个方法时：

dec_test()

您不需要传入campaigns_outgoing参数。