Python 3.10 Generator Expression Exception Handling?

Question

The code below is from the book "Functional Python Programming, Steven F. Lott" however I can't handle the exception and run the code as expected.

How can I handle the "RuntimeError: generator raised StopIteration"?

Note: I tried googling and read all the results I found from stackoverflow.

Thanks.

from typing import Callable, Iterable, Tuple, Iterator, Any

def group_by_iter(n: int, items: Iterator) -> Iterator[Tuple]:
    """
    .>>> list( group_by_iter( 7, filter( lambda x: x%3==0 or x%5==0, range(1,50) ) ) )
    [(3, 5, 6, 9, 10, 12, 15), (18, 20, 21, 24, 25, 27, 30), (33, 35, 36, 39, 40, 42, 45), (48,)]
    """
    row = tuple(next(items) for i in range(n))
    while row:
        yield row
        row = tuple( next(items) for i in range(n) )

print( list( group_by_iter( 7, filter( lambda x: x%3==0 or x%5==0, range(1,50) ) ) ) )

Exception is "RuntimeError: generator raised StopIteration". — muc, Dec 06 '22 at 08:57
This code relies on old, pre-[PEP 479](https://peps.python.org/pep-0479/) behavior of `StopIteration` propagating out of generators. It won't work on Python 3.7 or up. — user2357112, Dec 06 '22 at 08:59
Thanks, you are right. This didn't work but is there any simple solution to work as expected. I tried the row = tuple ... inside try, exception but this didn't catch the last value. — muc, Dec 06 '22 at 09:00
@user2357112 Which is funny, considering that typing, used here, was introduced in Python 3.5. — 9769953, Dec 06 '22 at 09:00
@9769953: Sorry, got the Python version mixed up - the behavior change is behind a `__future__` import on Python 3.5. It only became the default in 3.7. — user2357112, Dec 06 '22 at 09:02
Does this answer your question? [How to get the n next values of a generator in a list (python)](https://stackoverflow.com/questions/4152376/how-to-get-the-n-next-values-of-a-generator-in-a-list-python) — gre_gor, Dec 06 '22 at 09:13
@gre_gor: the answer I am looking is the answer below, thanks. — muc, Dec 06 '22 at 11:01

score 1 · Accepted Answer · answered Dec 06 '22 at 10:06

The built-in next() function can take a second argument which is the default value returned when the iterator is exhausted. You could exploit this functionality to avoid the StopIteration error by returning None by default and filtering it from the result.

Change tuple(next(items) for _ in range(n)) to tuple(e for e in [next(items, None) for _ in range(n)] if e is not None) and the final tuple will be generated cleanly.

from typing import Callable, Iterable, Tuple, Iterator, Any

def group_by_iter(n: int, items: Iterator) -> Iterator[Tuple]:
    """
    .>>> list( group_by_iter( 7, filter( lambda x: x%3==0 or x%5==0, range(1,50) ) ) )
    [(3, 5, 6, 9, 10, 12, 15), (18, 20, 21, 24, 25, 27, 30), (33, 35, 36, 39, 40, 42, 45), (48,)]
    """
    row = tuple(next(items) for _ in range(n))
    while row:
        yield row
        row = tuple(e for e in [next(items, None) for _ in range(n)] if e is not None)

print( list( group_by_iter( 7, filter( lambda x: x%3==0 or x%5==0, range(1,50) ) ) ) )

Output:

[(3, 5, 6, 9, 10, 12, 15), (18, 20, 21, 24, 25, 27, 30), (33, 35, 36, 39, 40, 42, 45), (48,)]

Python 3.10 Generator Expression Exception Handling?

1 Answers1