Ana*_*and 11 python type-hinting python-typing
我有一个装饰器,它接受一个函数并返回具有一些附加属性的相同函数:
import functools
from typing import *
def decorator(func: Callable) -> Callable:
func.attr1 = "spam"
func.attr2 = "eggs"
return func
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如何键入提示的返回值decorator?我希望类型提示传达两条信息:
Callableattr1和attr2如果我写一个协议,
class CallableWithAttrs(Protocol):
attr1: str
attr2: str
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那我输了Callable。显然我不能让协议继承自Callable;
class CallableWithAttrs(Callable, Protocol):
attr1: str
attr2: str
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mypy 说:
error: Invalid base class "Callable"
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另一方面,如果我只使用Callable,我会丢失有关添加属性的信息。
这在引入类型变量时可能更加复杂,即当装饰器必须返回与给定函数相同类型的可调用对象时func,正如 MisterMiyagi 在评论中指出的那样。
import functools
from typing import *
C = TypeVar('C', bound=Callable)
def decorator(func: C) -> C:
func.attr1 = "spam"
func.attr2 = "eggs"
return func
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现在我该怎么办?我不能从类型变量继承:
class CallableWithAttrs(C, Protocol):
attr1: str
attr2: str
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error: Invalid base class "C"
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由于typing.Callable对应于collections.abc.Callable,您可以定义一个Protocol实现__call__:
class CallableWithAttrs(Protocol):
attr1: str
attr2: str
def __call__(self, *args, **kwargs): pass
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人们可以参数化一个Protocol由Callable:
from typing import Callable, TypeVar, Protocol
C = TypeVar('C', bound=Callable) # placeholder for any Callable
class CallableObj(Protocol[C]): # Protocol is parameterised by Callable C ...
attr1: str
attr2: str
__call__: C # ... which defines the signature of the protocol
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这会创建Protocol自身与任意的交集Callable。
C因此,接受任何可调用对象的函数可以返回CallableObj[C]具有所需属性的相同签名的可调用对象:
def decorator(func: C) -> CallableObj[C]: ...
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MyPy 正确识别签名和属性:
def dummy(arg: str) -> int: ...
reveal_type(decorator(dummy)) # CallableObj[def (arg: builtins.str) -> builtins.int]'
reveal_type(decorator(dummy)('Hello')) # int
reveal_type(decorator(dummy).attr1) # str
decorator(dummy)(b'Fail') # error: Argument 1 to "dummy" has incompatible type "bytes"; expected "str"
decorator(dummy).attr3 # error: "CallableObj[Callable[[str], int]]" has no attribute "attr3"; maybe "attr2"?
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