因此，这是一个元类冲突，因为在python 3.6中键入NamedTuple和Generic使用不同的元类（typing.NamedTupleMeta和typing.GenericMeta），而这是python无法处理的。恐怕除了从tuple值中继承和手动初始化值外，没有其他解决方案：
T1 = TypeVar("T1")
T2 = TypeVar("T2")
class Group(tuple, Generic[T1, T2]):
key: T1
group: List[T2]
def __new__(cls, key: T1, group: List[T2]):
self = tuple.__new__(cls, (key, group))
self.key = key
self.group = group
return self
def __repr__(self) -> str:
return f'Group(key={self.key}, group={self.group})'
Group(1, [""]) # --> Group(key=1, group=[""])
由于PEP560和563，此问题已在python 3.7中修复：
Python 3.7.0b2 (v3.7.0b2:b0ef5c979b, Feb 28 2018, 02:24:20) [MSC v.1912 64 bit (AMD64)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> from __future__ import annotations
>>> from typing import *
>>> T1 = TypeVar("T1")
>>> T2 = TypeVar("T2")
>>> class Group(NamedTuple, Generic[T1, T2]):
... key: T1
... group: List[T2]
...
>>> g: Group[int, str] = Group(1, [""])
>>> g
Group(key=1, group=[''])
当然，在python 3.7中，您只能使用轻量级（且易变）但用途相似的数据类。
from dataclasses import dataclass, astuple
from typing import Generic, TypeVar, List
T1 = TypeVar('T1')
T2 = TypeVar('T2')
@dataclass
class Group(Generic[T1, T2]):
key: T1
group: List[T2]
# if you want to be able to unpack like a tuple...
def __iter__(self):
yield from astuple(self)
g: Group[int, str] = Group(1, ['hello', 'world'])
k, v = g
print(g)
尽管我没有检查过，类型检查器在python 3.7中如何处理我的解决方案/您的解决方案。我怀疑这可能不是无缝的。
import typing
from typing import *
class NamedTupleGenericMeta(typing.NamedTupleMeta, typing.GenericMeta):
pass
class Group(NamedTuple, Generic[T1,T2], Metaclass=NamedTupleGenericMeta):
key: T1
group: List[T2]
Group(1, ['']) # --> Group(key=1, group=[''])
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