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]

1
63
5/23/2017 12:03:02 PM

Accepted Answer

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

It avoids building new lists.

88
10/2/2015 9:52:23 AM

FWIW, the recipes in the itertools module provides this example:

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

It works like this:

>>> list(grouper(3, range(10)))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, None, None)]

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