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Optional key to itertools.groupby #6377
Description
itertools.groupby has an @overload type signature but this seems makes Optional[Callable[[T], S]] not match, e.g.:
def fn(iterable: Iterable[T], key: Optional[Callable[[T], S]] = None):
iterator: Iterator[Tuple[S, Iterator[T]]] = itertools.groupby(iterable, key)
Fails with:
error: Argument 2 to "groupby" has incompatible type "Optional[Callable[[T], S]]"; expected "Callable[[T], S]"
The revealed type of itertools.groupby:
Overload(def [_T] (iterable: typing.Iterable[_T`-1]) -> typing.Iterator[Tuple[_T`-1, typing.Iterator[_T`-1]]], def [_T, _S] (iterable: typing.Iterable[_T`-1], key: def (_T`-1) -> _S`-2) -> typing.Iterator[Tuple[_S`-2, typing.Iterator[_T`-1]]])
It is legitimate to explicitly pass None as key, so is this expected behaviour?
You can do a similar @overload for myfunc to overcome the error.
Versions:
Python 3.7.2
MyPy 0.670
Activity
ilevkivskyi commentedon Feb 11, 2019
I just tried and
groupby([1, 2, 3, 3, 3], None)indeed doesn't crash, but according to typeshed this call is illegal. You can open an issue there, if you are sure this is actually intended/documented behavior, or you can also try something like this:achampion commentedon Feb 11, 2019
Thanks will do. Wasn't sure if this was a
mypyissue or atypeshedissue.With the
@overloadsforitertools.groupby(), shouldOptional[Callable[[T], S]]match?While you can do
if key is not None:the problem with that is you can't effectively describe the type ofiterator, e.g.:You get the error:
And you can't declare the types in each branch because you will get a redeclaration error
Name 'iterator' already defined on line ...ilevkivskyi commentedon Feb 11, 2019
No, with current definition in typeshed it is illegal,
Optionalmeans it can beNone(not that it is an optional argument), while with current definition neither of two overloads acceptsNone.