getting n items at a time from a generator

[Raymond Hettinger]

> > Also consider this solution from O'Reilly's Python Cookbook (2nd Ed.) p705
>
> >     def chop(iterable, length=2):
> >         return izip(*(iter(iterable),) * length)
>
> However, chop ignores the remainder of the data in the example.

There is a recipe in the itertools docs which handles the odd-length
data at the end:

def grouper(n, iterable, padvalue=None):
    "grouper(3, 'abcdefg', 'x') --> ('a','b','c'), ('d','e','f'),
('g','x','x')"
    return izip(*[chain(iterable, repeat(padvalue, n-1))]*n)


Raymond