How to group elements in python by n elements

Question

Given a list [1,2,3,4,5,6,7,8,9,10,11,12] and a specified chunk size (say 3), how can I get a list of chunks [[1,2,3],[4,5,6],[7,8,9],[10,11,12]]?

Answer 1

Well, the brute force answer is:

subList = [theList[n:n+N] for n in range(0, len(theList), N)]

where N is the group size (3 in your case):

>>> theList = list(range(10))
>>> N = 3
>>> subList = [theList[n:n+N] for n in range(0, len(theList), N)]
>>> subList
[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]]

If you want a fill value, you can do this right before the list comprehension:

tempList = theList + [fill] * N
subList = [tempList[n:n+N] for n in range(0, len(theList), N)]

Example:

>>> fill = 99
>>> tempList = theList + [fill] * N
>>> subList = [tempList[n:n+N] for n in range(0, len(theList), N)]
>>> subList
[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 99, 99]]

Answer 2

You can use the grouper function from the recipes in the itertools documentation:

from itertools import zip_longest

def grouper(iterable, n, fillvalue=None):
    """Collect data into fixed-length chunks or blocks.

    >>> grouper('ABCDEFG', 3, 'x')
    ['ABC', 'DEF', 'Gxx']
    """
    args = [iter(iterable)] * n
    return zip_longest(*args, fillvalue=fillvalue)

Answer 3

How about

a = range(1,10)
n = 3
out = [a[k:k+n] for k in range(0, len(a), n)]

Answer 4

See examples at the bottom of the itertools docs: http://docs.python.org/library/itertools.html?highlight=itertools#module-itertools

You want the "grouper" method, or something like it.

Answer 5

answer = [L[3*i:(3*i)+3] for i in range((len(L)/3) +1)]
if not answer[-1]:
    answer = answer[:-1]