如何注释返回类型取决于其参数的函数?
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【中文标题】如何注释返回类型取决于其参数的函数?【英文标题】:How do I annotate a function whose return type depends on its argument? 【发布时间】:2021-10-15 01:38:49 【问题描述】:在 Python 中,我经常编写过滤集合以查找特定子类型实例的函数。例如,我可能会在 DOM 中查找特定类型的节点或在日志中查找特定类型的事件:
def find_pre(soup: TagSoup) -> List[tags.pre]:
"""Find all <pre> nodes in `tag_soup`."""
…
def filter_errors(log: List[LogEvent]) -> List[LogError]:
"""Keep only errors from `log`."""
…
为这些函数编写类型很容易。但是这些函数的泛型版本需要一个参数来指定要返回的类型呢?
def find_tags(tag_soup: TagSoup, T: type) -> List[T]:
"""Find all nodes of type `T` in `tag_soup`."""
…
def filter_errors(log: List[LogEvent], T: type) -> List[T]:
"""Keep only events of type `T` from `log`."""
…
(上面的签名是错误的:我不能在返回类型中引用T
。)
这是一个相当常见的设计:docutils
有node.traverse(T: type)
,BeautifulSoup
有soup.find_all()
等。当然它可以变得任意复杂,但是Python 类型注释可以处理像上面这样的简单情况吗?
这里有一个 MWE 让它变得非常具体:
from dataclasses import dataclass
from typing import *
@dataclass
class Packet: pass
@dataclass
class Done(Packet): pass
@dataclass
class Exn(Packet):
exn: str
loc: Tuple[int, int]
@dataclass
class Message(Packet):
ref: int
msg: str
Stream = Callable[[], Union[Packet, None]]
def stream_response(stream: Stream, types) -> Iterator[??]:
while response := stream():
if isinstance(response, Done): return
if isinstance(response, types): yield response
def print_messages(stream: Stream):
for m in stream_response(stream, Message):
print(m.msg) # Error: Cannot access member "msg" for "Packet"
msgs = iter((Message(0, "hello"), Exn("Oops", (1, 42)), Done()))
print_messages(lambda: next(msgs))
Pyright 说:
29:17 - error: Cannot access member "msg" for type "Packet"
Member "msg" is unknown (reportGeneralTypeIssues)
在上面的例子中,有没有办法注解stream_response
,以便Python类型检查器接受print_messages
的定义?
【问题讨论】:
打字文档的this section 有帮助吗?TypeVar
似乎正是您所需要的。
@Kemp:它没有:在def f(x: T) -> List[T]
中,返回类型取决于x
的type。在def f(x: type) -> List[x]
(我想要/需要的)中,返回类型取决于x
的值。
@AlexWaygood:不,返回类型更精确:首先,它永远不是None
;其次,它保证是特定类型的数据包(以types
中传递的为准。
@AlexWaygood 后者最好,但作为最后的手段,前者也可以。
实际上,在我的情况下不会超过 10 个;我很想知道它会扮演什么角色!
【参考方案1】:
好的,我们开始吧。它通过了 MyPy --strict,但它并不漂亮。
这里发生了什么
对于给定的类A
,我们知道A
的实例类型将是A
(显然)。但是A
本身的类型是什么?从技术上讲,A
的类型是type
,因为所有不使用元类的python 类都是type
的实例。然而,用type
注释一个参数并不能告诉类型检查器太多。相反,用于在类型层次结构中“上一层”的 Python 类型检查语法是 Type[A]
。因此,如果我们有一个函数myfunc
,它返回一个作为参数输入的类的实例,我们可以相当简单地注释如下:
from typing import TypeVar, Type
T = TypeVar('T')
def myfunc(some_class: Type[T]) -> T:
# do some stuff
return some_class()
但是,您的情况要复杂得多。您可以输入一个类作为参数,或者您可以输入两个类或三个类......等等。我们可以使用typing.overload
解决这个问题,它允许我们为给定函数注册多个签名。这些签名在运行时被完全忽略;它们纯粹用于类型检查器;因此,这些函数的主体可以留空。通常,您只在用@overload
装饰的函数体中放置一个文档字符串或文字省略号...
。
我认为没有办法概括这些重载函数,这就是为什么可以传递给types
参数的最大元素数很重要。您必须繁琐地枚举函数的每个可能签名。如果您沿着这条路线走,您可能需要考虑将 @overload
签名移动到单独的 .pyi 存根文件中。
from dataclasses import dataclass
from typing import (
Callable,
Tuple,
Union,
Iterator,
overload,
TypeVar,
Type,
Sequence
)
@dataclass
class Packet: pass
P1 = TypeVar('P1', bound=Packet)
P2 = TypeVar('P2', bound=Packet)
P3 = TypeVar('P3', bound=Packet)
P4 = TypeVar('P4', bound=Packet)
P5 = TypeVar('P5', bound=Packet)
P6 = TypeVar('P6', bound=Packet)
P7 = TypeVar('P7', bound=Packet)
P8 = TypeVar('P8', bound=Packet)
P9 = TypeVar('P9', bound=Packet)
P10 = TypeVar('P10', bound=Packet)
@dataclass
class Done(Packet): pass
@dataclass
class Exn(Packet):
exn: str
loc: Tuple[int, int]
@dataclass
class Message(Packet):
ref: int
msg: str
Stream = Callable[[], Union[Packet, None]]
@overload
def stream_response(stream: Stream, types: Type[P1]) -> Iterator[P1]:
"""Signature if exactly one type is passed in for the `types` parameter"""
@overload
def stream_response(
stream: Stream,
types: Tuple[Type[P1], Type[P2]]
) -> Iterator[Union[P1, P2]]:
"""Signature if exactly two types are passed in for the `types` parameter"""
@overload
def stream_response(
stream: Stream,
types: Tuple[Type[P1], Type[P2], Type[P3]]
) -> Iterator[Union[P1, P2, P3]]:
"""Signature if exactly three types are passed in for the `types` parameter"""
@overload
def stream_response(
stream: Stream,
types: Tuple[Type[P1], Type[P2], Type[P3], Type[P4]]
) -> Iterator[Union[P1, P2, P3, P4]]:
"""Signature if exactly four types are passed in for the `types` parameter"""
@overload
def stream_response(
stream: Stream,
types: Tuple[Type[P1], Type[P2], Type[P3], Type[P4], Type[P5]]
) -> Iterator[Union[P1, P2, P3, P4, P5]]:
"""Signature if exactly five types are passed in for the `types` parameter"""
@overload
def stream_response(
stream: Stream,
types: Tuple[Type[P1], Type[P2], Type[P3], Type[P4], Type[P5], Type[P6]]
) -> Iterator[Union[P1, P2, P3, P4, P5, P6]]:
"""Signature if exactly six types are passed in for the `types` parameter"""
@overload
def stream_response(
stream: Stream,
types: Tuple[
Type[P1],
Type[P2],
Type[P3],
Type[P4],
Type[P5],
Type[P6],
Type[P7]
]
) -> Iterator[Union[P1, P2, P3, P4, P5, P6, P7]]:
"""Signature if exactly seven types are passed in for the `types` parameter"""
@overload
def stream_response(
stream: Stream,
types: Tuple[
Type[P1],
Type[P2],
Type[P3],
Type[P4],
Type[P5],
Type[P6],
Type[P7],
Type[P8]
]
) -> Iterator[Union[P1, P2, P3, P4, P5, P6, P7, P8]]:
"""Signature if exactly eight types are passed in for the `types` parameter"""
@overload
def stream_response(
stream: Stream,
types: Tuple[
Type[P1],
Type[P2],
Type[P3],
Type[P4],
Type[P5],
Type[P6],
Type[P7],
Type[P8],
Type[P9]
]
) -> Iterator[Union[P1, P2, P3, P4, P5, P6, P7, P8, P9]]:
"""Signature if exactly nine types are passed in for the `types` parameter"""
@overload
def stream_response(
stream: Stream,
types: Tuple[
Type[P1],
Type[P2],
Type[P3],
Type[P4],
Type[P5],
Type[P6],
Type[P7],
Type[P8],
Type[P9],
Type[P10]
]
) -> Iterator[Union[P1, P2, P3, P4, P5, P6, P7, P8, P9, P10]]:
"""Signature if exactly ten types are passed in for the `types` parameter"""
# We have to be more generic in our type-hinting for the concrete implementation
# Otherwise, MyPy struggles to figure out that it's a valid argument to `isinstance`
def stream_response(
stream: Stream,
types: Union[type, Tuple[type, ...]]
) -> Iterator[Packet]:
while response := stream():
if isinstance(response, Done): return
if isinstance(response, types): yield response
def print_messages(stream: Stream) -> None:
for m in stream_response(stream, Message):
print(m.msg)
msgs = iter((Message(0, "hello"), Exn("Oops", (1, 42)), Done()))
print_messages(lambda: next(msgs))
减少冗长的策略
如果您想让它更简洁,实现此目的的一种方法是为某些类型结构引入别名。这里的危险是类型提示的意图和含义变得很难阅读,但它确实使重载 7-10 看起来不那么可怕:
from dataclasses import dataclass
from typing import (
Callable,
Tuple,
Union,
Iterator,
overload,
TypeVar,
Type,
Sequence
)
@dataclass
class Packet: pass
P1 = TypeVar('P1', bound=Packet)
P2 = TypeVar('P2', bound=Packet)
P3 = TypeVar('P3', bound=Packet)
P4 = TypeVar('P4', bound=Packet)
P5 = TypeVar('P5', bound=Packet)
P6 = TypeVar('P6', bound=Packet)
P7 = TypeVar('P7', bound=Packet)
P8 = TypeVar('P8', bound=Packet)
P9 = TypeVar('P9', bound=Packet)
P10 = TypeVar('P10', bound=Packet)
_P = TypeVar('_P', bound=Packet)
S = Type[_P]
T7 = Tuple[S[P1], S[P2], S[P3], S[P4], S[P5], S[P6], S[P7]]
T8 = Tuple[S[P1], S[P2], S[P3], S[P4], S[P5], S[P6], S[P7], S[P8]]
T9 = Tuple[S[P1], S[P2], S[P3], S[P4], S[P5], S[P6], S[P7], S[P8], S[P9]]
T10 = Tuple[S[P1], S[P2], S[P3], S[P4], S[P5], S[P6], S[P7], S[P8], S[P9], S[P10]]
@dataclass
class Done(Packet): pass
@dataclass
class Exn(Packet):
exn: str
loc: Tuple[int, int]
@dataclass
class Message(Packet):
ref: int
msg: str
Stream = Callable[[], Union[Packet, None]]
@overload
def stream_response(stream: Stream, types: Type[P1]) -> Iterator[P1]:
"""Signature if exactly one type is passed in for the `types` parameter"""
@overload
def stream_response(
stream: Stream,
types: Tuple[Type[P1], Type[P2]]
) -> Iterator[Union[P1, P2]]:
"""Signature if exactly two types are passed in for the `types` parameter"""
@overload
def stream_response(
stream: Stream,
types: Tuple[Type[P1], Type[P2], Type[P3]]
) -> Iterator[Union[P1, P2, P3]]:
"""Signature if exactly three types are passed in for the `types` parameter"""
@overload
def stream_response(
stream: Stream,
types: Tuple[Type[P1], Type[P2], Type[P3], Type[P4]]
) -> Iterator[Union[P1, P2, P3, P4]]:
"""Signature if exactly four types are passed in for the `types` parameter"""
@overload
def stream_response(
stream: Stream,
types: Tuple[Type[P1], Type[P2], Type[P3], Type[P4], Type[P5]]
) -> Iterator[Union[P1, P2, P3, P4, P5]]:
"""Signature if exactly five types are passed in for the `types` parameter"""
@overload
def stream_response(
stream: Stream,
types: Tuple[Type[P1], Type[P2], Type[P3], Type[P4], Type[P5], Type[P6]]
) -> Iterator[Union[P1, P2, P3, P4, P5, P6]]:
"""Signature if exactly six types are passed in for the `types` parameter"""
@overload
def stream_response(
stream: Stream,
types: T7[P1, P2, P3, P4, P5, P6, P7]
) -> Iterator[Union[P1, P2, P3, P4, P5, P6, P7]]:
"""Signature if exactly seven types are passed in for the `types` parameter"""
@overload
def stream_response(
stream: Stream,
types: T8[P1, P2, P3, P4, P5, P6, P7, P8]
) -> Iterator[Union[P1, P2, P3, P4, P5, P6, P7, P8]]:
"""Signature if exactly eight types are passed in for the `types` parameter"""
@overload
def stream_response(
stream: Stream,
types: T9[P1, P2, P3, P4, P5, P6, P7, P8, P9]
) -> Iterator[Union[P1, P2, P3, P4, P5, P6, P7, P8, P9]]:
"""Signature if exactly nine types are passed in for the `types` parameter"""
@overload
def stream_response(
stream: Stream,
types: T10[P1, P2, P3, P4, P5, P6, P7, P8, P9, P10]
) -> Iterator[Union[P1, P2, P3, P4, P5, P6, P7, P8, P9, P10]]:
"""Signature if exactly ten types are passed in for the `types` parameter"""
# We have to be more generic in our type-hinting for the concrete implementation
# Otherwise, MyPy struggles to figure out that it's a valid argument to `isinstance`
def stream_response(
stream: Stream,
types: Union[type, Tuple[type, ...]]
) -> Iterator[Packet]:
while response := stream():
if isinstance(response, Done): return
if isinstance(response, types): yield response
def print_messages(stream: Stream) -> None:
for m in stream_response(stream, Message):
print(m.msg)
msgs = iter((Message(0, "hello"), Exn("Oops", (1, 42)), Done()))
print_messages(lambda: next(msgs))
【讨论】:
啊,在类型参数上使用边界的技巧非常聪明。元组重载令人不快,但至少一种情况还不错! @Clément,同意——当注释classmethod
s 返回类 ***.com/a/68283181/13990016 的实例时,单一类型的情况是一种常见模式
很好的捕捉,也有很好的答案! +1
非常感谢!
@Clément -- 我想到了一种方法,可以让这一点不那么冗长;已将其编辑到我的答案中。但是,它确实使注释的可读性更差!以上是关于如何注释返回类型取决于其参数的函数?的主要内容,如果未能解决你的问题,请参考以下文章
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