Lambda as RHS for operator always gets inferred as Any

When a lambda is given as the RHS in an operator (example uses >>, but also occurs with others), the lambda’s type is always inferred as Any. Whereas calling the operator function directly, or passing an explicit function, does not.

from typing import Callable, Generic, TypeVar

_T = TypeVar('_T')
_U = TypeVar('_U')

class A(Generic[_T]):
    _value: _T
    def __init__(self, val: _T) -> None: ...
    def __rshift__(self, f: Callable[[_T], _U]) -> _U: ...

def inc(x: int) -> int: ...

A(1) >> inc  # No error
A(1).__rshift__(lambda x: x + 1)  # No error
A(1) >> (lambda x: x + 1)
#        │         └ Expression has type "Any"
#        └ Expression type contains "Any" (has type Callable[[Any], Any])

• Python 3.7.0
• mypy 0.641
• mypy config:

  [mypy]
  check_untyped_defs = True
  disallow_any_decorated = True
  disallow_any_expr = True
  disallow_any_unimported = True
  disallow_incomplete_defs = True
  disallow_subclassing_any = True
  disallow_untyped_calls = True
  disallow_untyped_decorators = True
  disallow_untyped_defs = True
  warn_return_any = True
  warn_unused_ignores = True
  

提供された Python の再現コードを、記載された mypy 設定で実行し、演算子形式を直接のメソッド呼び出しおよび名前付き関数と比較します。そのエントリポイントから演算子式の型推論を追跡します。完了条件は、括弧で囲まれた lambda が Any を生成するのではなく、オペランドの型から推論されることです。