Breaking Python list into set-length list of lists

[Gerard Flanagan]

Jason wrote:
> Hey everyone--
> 
> I'm pretty new to Python, & I need to do something that's incredibly
> simple, but combing my Python Cookbook & googling hasn't helped me out
> too much yet, and my brain is very, very tired & flaccid @ the
> moment....
> 
> I have a list of objects, simply called "list".  I need to break it
> into an array (list of lists) wherein each sublist is the length of
> the variable "items_per_page".  So array[0] would go from array[0][0]
> to array[0][items_per_page], then bump up to array[1][0] - array[1]
> [items_per_page], until all the items in the original list were
> accounted for.
> 

If you need to pad the last item:

def chunk( seq, size, pad=None ):
     '''
     Slice a list into consecutive disjoint 'chunks' of
     length equal to size. The last chunk is padded if necessary.

     >>> list(chunk(range(1,10),3))
     [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
     >>> list(chunk(range(1,9),3))
     [[1, 2, 3], [4, 5, 6], [7, 8, None]]
     >>> list(chunk(range(1,8),3))
     [[1, 2, 3], [4, 5, 6], [7, None, None]]
     >>> list(chunk(range(1,10),1))
     [[1], [2], [3], [4], [5], [6], [7], [8], [9]]
     >>> list(chunk(range(1,10),9))
     [[1, 2, 3, 4, 5, 6, 7, 8, 9]]
     >>> for X in chunk([],3): print X
     >>>
     '''
     n = len(seq)
     mod = n % size
     for i in xrange(0, n-mod, size):
         yield seq[i:i+size]
     if mod:
         padding = [pad] * (size-mod)
         yield seq[-mod:] + padding

If you search the list archive, there is surely an example which will do 
the same for an arbitrary iterable (in general, you can't assume that 
`len(seq)` will work for an iterable that is not a list, set or tuple). 
But if you are just starting with Python, don't worry about this.