This is a dumb example based on a more complex thing that I want to do:
from typing import Generator
def f() -> Generator[list[int], None, None]:
result = list()
result.append(1)
if len(result) == 2:
yield result
result = list()
result.append(2)
if len(result) == 2:
yield result
result = list()
result.append(3)
if len(result) == 2:
yield result
result = list()
result.append(4)
if len(result) == 2:
yield result
result = list()
print(list(f()))
The point here is that this bit is copied multiple times:
if len(result) == 2:
yield result
result = list()
Normally, I'd change it into something like this:
from typing import Generator
def f() -> Generator[list[int], None, None]:
def add_one(value: int) -> None:
nonlocal result
result.append(value)
if len(result) == 2:
nonlocal_yield result
result = list()
result = list()
add_one(1)
add_one(2)
add_one(3)
add_one(4)
print(list(f()))
Obviously, nonlocal_yield is not a thing. Is there an elegant way to achieve this?
I know that I can just create the full list of results, i.e., [[1, 2], [3, 4]], and then either return it or yield individual 2-element sublists. Something like this:
from typing import Generator
def f() -> list[list[int]]:
def add_one(value: int) -> None:
nonlocal current
current.append(value)
if len(current) == 2:
result.append(current)
current = list()
result = list()
current = list()
add_one(1)
add_one(2)
add_one(3)
add_one(4)
return result
print(list(f()))
However, this beats the purpose of a generator. I'll go for it in absence of a better solution, but I'm curious if there is a "pure" generator way to do it.
One possibility:
def f() -> Generator[list[int], None, None]:
def add_one(value: int) -> Generator[list[int], None, None]:
nonlocal result
result.append(value)
if len(result) == 2:
yield result
result = list()
result = list()
yield from add_one(1)
yield from add_one(2)
yield from add_one(3)
yield from add_one(4)