解压 Python 的类型注解

Question

I'm trying to generate some JavaScript based on the type annotations I have provided in on some Python functions by using the signature() function in the inspect module.

This part works as I expect when the type is a simple builtin class:

import inspect

def my_function() -> dict:
    pass

signature = inspect.signature(my_function)
signature.return_annotation is dict  # True

Though I'm not sure how to unwrap and inspect more complex annotations eg:

from typing import List
import inspect

def my_function() -> List[int]:
    pass

signature = inspect.signature(my_function)
signature.return_annotation is List[int]  # False

Again similar problem with forward referencing a custom class:

def my_function() -> List['User']:
    pass
...
signature.return_annotation  # typing.List[_ForwardRef('User')]

What I'm looking to get out is something like this - so I can branch appropriately while generating the JavaScript:

type = signature.return_annotation... # list
member_type = signature.return_annotation... # int / 'User'

Thanks.

Answer 1

List is not a map of types to GenericMeta , despite the syntax. Each access to it generates a new instance:

>>> [ id(List[str]) for i in range(3) ]
[33105112, 33106872, 33046936]

This means that even List[int] is not List[int] . To compare two instances, you have multiple options:

• Use == , ie, signature.return_annotation == List[int] .

• Store an instance of your type in a global variable and check against that, ie,

   a = List[int] def foo() -> a: pass inspect.signature(foo).return_annotation is a

• Use issubclass . The typing module defines that. Note that this might do more than you'd like, make sure to read the _TypeAlias documentation if you use this.

• Check against List only and read the contents yourself. Though the property is internal, it is unlikely that the implementation will change soon: List[int].__args__[0] contains the type argument starting from Python 3.5.2, and in earlier versions, its List[int].__parameters__[0] .

If you'd like to write generic code for your exporter, then the last option is probably best. If you only need to cover a specific use case, I'd personally go with using == .

Answer 2

Python 3.8 provides typing.get_origin() and typing.get_args() for this!

assert get_origin(Dict[str, int]) is dict
assert get_args(Dict[int, str]) == (int, str)

assert get_origin(Union[int, str]) is Union
assert get_args(Union[int, str]) == (int, str)

Seehttps://docs.python.org/3/library/typing.html#typing.get_origin

Answer 3

Take note, this applies to Python 3.5.1

For Python 3.5.2 take a look at phillip's answer.

You shouldn't be checking with the identity operator as Phillip stated, use equality to get this right.

To check if a hint is a subclass of a list you could use issubclass checks (even though you should take note that this can be quirky in certain cases and is currently worked on):

issubclass(List[int], list)  # True

To get the members of a type hint you generally have two watch out for the cases involved.

If it has a simple type, as in List[int] the value of the argument is located in the __parameters__ value:

signature.return_annotation.__parameters__[0] # int

Now, in more complex scenarios ie a class supplied as an argument with List[User] you must again extract the __parameter__[0] and then get the __forward_arg__ . This is because Python wraps the argument in a special ForwardRef class:

d = signature.return_annotation.__parameter__[0]
d.__forward_arg__ # 'User'

Take note , you don't need to actually use inspect here, typing has a helper function named get_type_hints that returns the type hints as a dictionary (it uses the function objects __annotations__ attribute).