implementing chop() using iterators

Question

Current version:

def chop(ar,size):
    p=len(ar)/size
    for i in xrange(p):
        yield ar[(i*size):((i+1)*size)]

ar is type of list().

What i want is that chop() takes iterator and return iterator.

for i in chop(xrange(9),3):
    for j in i:
       print j,
    print

prints

0 1 2
3 4 5
6 7 8

Answer 1

There's an implementation in the itertools documentation:

def grouper(n, iterable, fillvalue=None):
    "grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx"
    args = [iter(iterable)] * n
    return izip_longest(fillvalue=fillvalue, *args)