Question

Possible Duplicate:
How do you split a list into evenly sized chunks in Python?

Hi,

I would like to split a list in many list of a length of x elements, like:


a = (1, 2, 3, 4, 5)


and get :


b = (
(1,2),
(3,4),
(5,)
)


if the length is set to 2 or :


b = (
(1,2,3),
(4,5)
)


if the length is equal to 3 …

Is there a nice way to write this ? Otherwise I think the best way is to write it using an iterator …

Answer 1

The itertools module documentation. Read it, learn it, love it.

Specifically, from the recipes section:

import itertools

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

Which gives:

>>> tuple(grouper(3, (1,2,3,4,5)))
((1, 2, 3), (4, 5, None))

which isn’t quite what you want…you don’t want the None in there…so. a quick fix:

>>> tuple(tuple(n for n in t if n) for t in grouper(3, (1,2,3,4,5)))
((1, 2, 3), (4, 5))

If you don’t like typing the list comprehension every time, we can move its logic into the function:

def my_grouper(n, iterable):
  "my_grouper(3, 'ABCDEFG') --> ABC DEF G"
  args = [iter(iterable)] * n
  return tuple(tuple(n for n in t if n)
       for t in itertools.izip_longest(*args))

Which gives:

>>> tuple(my_grouper(3, (1,2,3,4,5)))
((1, 2, 3), (4, 5))

Done.