Re: [Python-Dev] "groupby" iterator

Dec. 1, 2003


      ...
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/)