如何正确编写返回类型被装饰器修改的函数的类型?
简单的例子:
def example_decorator(fn):
def wrapper(data):
res = fn(data)
return ', '.join(res)
return wrapper
@example_decorator
def func(data: list): # -> ???? str ? list ?
new_data = data
new_data.append('XYZ')
return new_data
# When we type func -> list
def test() -> str:
result = func(['ABC', 'EFG'])
print(type(result)) # <class 'str'>
return result # Incompatible return type [7]: Expected str but got list.
test()
可能是类型检查器问题。解决方案如下。 有了mypy,它还可以,但pycharm和柴堆检查正在抱怨。
from typing import *
def example_decorator(
fn: Callable[[List[str]], List[str]]
) -> Callable[[List[str]], str]:
def wrapper(data: List[str]) -> str:
res = fn(data)
return ', '.join(res)
return wrapper
@example_decorator
def func(data: List[str]) -> List[str]:
data.append('XYZ')
return data
def test() -> str:
result = func(['ABC', 'EFG'])
print(type(result)) # <class 'str'>
return result
test()
该函数返回一个列表,如下所示:
@example_decorator
def func(data: list) -> list:
new_data = data
new_data.append('XYZ')
return new_data