I'd definitely recommend hiding this trick inside a function. Perhaps something like (using Michael's function name):
from itertools import izip, repeat, chain
def partition(seq, part_len): return izip(*((iter(seq),) * part_len))
def padded_partition(seq, part_len, pad_val=None): itr = iter(seq) if (len(seq) % part_len != 0): padding = repeat(pad_val, part_len) itr = chain(itr, padding) return izip(*((itr,) * part_len))
I think you can write that second one so that it works for iterables without a __len__:
py> def padded_partition(iterable, part_len, pad_val=None): ... itr = itertools.chain( ... iter(iterable), itertools.repeat(pad_val, part_len - 1)) ... return itertools.izip(*[itr]*part_len) ... py> list(padded_partition(itertools.islice(itertools.count(), 10), 2)) [(0, 1), (2, 3), (4, 5), (6, 7), (8, 9)] py> list(padded_partition(itertools.islice(itertools.count(), 10), 3)) [(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, None, None)]
I just unconditionally pad the iterable with 1 less than the partition size... I think that works right, but I haven't tested it any more than what's shown.
Steve -- http://mail.python.org/mailman/listinfo/python-list
