how to split an iterable in constant-size chunks

Question

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

I am surprised I could not find a "batch" function that would take as input an iterable and return an iterable of iterables.

For example:

for i in batch(range(0,10), 1): print i [0] [1] ... [9] 

or:

for i in batch(range(0,10), 3): print i [0,1,2] [3,4,5] [6,7,8] [9] 

Now, I wrote what I thought was a pretty simple generator:

def batch(iterable, n = 1):    current_batch = []    for item in iterable:        current_batch.append(item)        if len(current_batch) == n:            yield current_batch            current_batch = []    if current_batch:        yield current_batch 

But the above does not give me what I would have expected:

for x in   batch(range(0,10),3): print x [0] [0, 1] [0, 1, 2] [3] [3, 4] [3, 4, 5] [6] [6, 7] [6, 7, 8] [9] 

So, I have missed something and this probably shows my complete lack of understanding of python generators. Anyone would care to point me in the right direction ?

[Edit: I eventually realized that the above behavior happens only when I run this within ipython rather than python itself]

Answer 1

This is probably more efficient (faster)

def batch(iterable, n=1):     l = len(iterable)     for ndx in range(0, l, n):         yield iterable[ndx:min(ndx + n, l)]  for x in batch(range(0, 10), 3):     print x 

---

Example using list

data = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] # list of data   for x in batch(data, 3):     print(x)  # Output  [0, 1, 2] [3, 4, 5] [6, 7, 8] [9, 10] 

It avoids building new lists.