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Mon Dec 1 18:57:28 EST 2003 · https://mail.python.org/pipermail/python-dev/2003-December/040575.html

> I think this would be helpful for lazy coders if some function of
> itertools cover the use case: (`LIMIT' keyword of SQL)
> 
> >>> from groupby import groupby
> >>> alwaystrue = lambda n: True
> >>> for k, g in groupby(alwaystrue, range(20), 5):
> ...     print list(g)
> ...
> [0, 1, 2, 3, 4]
> [5, 6, 7, 8, 9]
> [10, 11, 12, 13, 14]
> [15, 16, 17, 18, 19]
> 
> prototype of this case is groupby(keyfunc, iterable, limit=None).
> Either, groupby(5, range(20)) is okay but 5 is not a sort of `key'. :)

I'd rather not weigh down groupby() with more options.  If you really
want only the first 5 of each group, you can use itertools.islice():

  for k, g in groupby(alwaystrue, range(20)):
     print list(itertools.islice(g, 0, 5))

--Guido van Rossum (home page: http://www.python.org/~guido/)

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[Python-Dev] "groupby" iterator