Python List Division/Splitting

Question

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

Hello,

I'm trying to find a simpler way to do the following:

def list_split(list, size):
  result = [[]]
  while len(list) > 0:
    if len(result[-1]) >= size: result.append([])
    result[-1].append(list.pop(0))
  return result

Example usage:

>>> list_split([0, 1, 2, 3, 4, 5, 6], 2)
[[0, 1], [2, 3], [4, 5], [6]]
>>> list_split([0, 1, 2, 3, 4, 5, 6], 3)
[[0, 1, 2], [3, 4, 5], [6]]

I can't tell if there's a built-in way to do this, possibly with slicing or something.

This is similar but not the same to the post at http://stackoverflow.com/questions/2235526/how-to-split-a-list-into-a-given-number-of-sub-lists-in-python-closed

Thanks

EDIT: As is commented on by Anurag Uniyal, this is a duplicate of http://stackoverflow.com/questions/312443/how-do-you-split-a-list-into-evenly-sized-chunks-in-python/312467#312467, and should be closed, which I cannot do.

Answer 1

def list_split(L, size):
    return [L[i*size:(i+1)*size] for i in range(1+((len(L)-1)//size))]

If you prefer a generator instead of a list, you can replace the brackets with parens, like so:

def list_split(L, size):
    return (L[i*size:(i+1)*size] for i in range(1+((len(L)-1)//size)))

Answer 2

from itertools import izip_longest

def list_split(L, size):
    return [[j for j in i if j is not None] for i in izip_longest(*[iter(L)]*size)]

---

>>> list_split([0, 1, 2, 3, 4, 5, 6], 2)
[[0, 1], [2, 3], [4, 5], [6]]
>>> list_split([0, 1, 2, 3, 4, 5, 6], 3)
[[0, 1, 2], [3, 4, 5], [6]]
>>> list_split([0, 1, 2, 3, 4, 5, 6], 4)
[[0, 1, 2, 3], [4, 5, 6]]
>>> list_split([0, 1, 2, 3, 4, 5, 6], 5)
[[0, 1, 2, 3, 4], [5, 6]]

Answer 3

You could use slices to get subsets of a list.

Example:

>>> L = [0, 1, 2, 3, 4, 5, 6]
>>> n = 3
>>> [L[i:i+n] for i in range(0, len(L), n)]
[[0, 1, 2], [3, 4, 5], [6]]
>>>

Answer 4

You have a simple functional solution in the itertools recipes (grouper):

http://docs.python.org/library/itertools.html#recipes

Whereas this function adds padding, you can easily write a non-padded implementation taking advantage of the (usually overlooked) iter built-in used this way: iter(callable, sentinel) -> iterator

def grouper(n, it):
  "grouper(3, 'ABCDEFG') --> ABC DEF G"
  return iter(lambda: list(itertools.islice(it, n)), [])

list(grouper(3, iter(mylist)))

These solutions are more generic because they both work with sequences and iterables (even if they are infinite).