Refactor any(item for item in items if item in container) into any(item in container for item in items)

[brettcannon]

Came up in a code review for any(p for p in parts if p in s). With no assignment expression or any need to know which or how many items match since the loop variant doesn't leak out of the generator expression, there's no need to condition the any() call on succeeding via which items matched.

[JonathanPlasse]

Hi, this is not equivalent if some items are falsy.

>>> items = [True, False]
>>> container = [False]
>>> any(item for item in items if item in container)
False
>>> any(item in container for item in items)
True

[brettcannon]

True, although I think you're asking for trouble in that situation. But I see why Ruff wouldn't want to risk the pathological case of being wrong here.