Python generator that groups another iterable into groups of N

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When you want to group an iterator in chunks of n without padding the final group with a fill value, use iter(lambda: list(IT.islice(iterable, n)), []):

import itertools as IT

def grouper(n, iterable):
    """
    >>> list(grouper(3, 'ABCDEFG'))
    [['A', 'B', 'C'], ['D', 'E', 'F'], ['G']]
    """
    iterable = iter(iterable)
    return iter(lambda: list(IT.islice(iterable, n)), [])

seq = [1,2,3,4,5,6,7]
print(list(grouper(3, seq)))

yields

[[1, 2, 3], [4, 5, 6], [7]]

There is an explanation of how it works in the second half of this answer.

When you want to group an iterator in chunks of n and pad the final group with a fill value, use the grouper recipe zip_longest(*[iterator]*n):

For example, in Python2:

>>> list(IT.izip_longest(*[iter(seq)]*3, fillvalue="x"))
[(1, 2, 3), (4, 5, 6), (7, 'x', 'x')]

In Python3, what was izip_longest is now renamed zip_longest:

>>> list(IT.zip_longest(*[iter(seq)]*3, fillvalue="x"))
[(1, 2, 3), (4, 5, 6), (7, 'x', 'x')]

When you want to group a sequence in chunks of n you can use the chunks recipe:

def chunks(seq, n):
    # https://stackoverflow.com/a/312464/190597 (Ned Batchelder)
    """ Yield successive n-sized chunks from seq."""
    for i in xrange(0, len(seq), n):
        yield seq[i:i + n]

Note that, unlike iterators in general, sequences by definition have a length (i.e. __len__ is defined).

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