how to split an iterable in constant-size chunks

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.

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 zip_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)]

As others have noted, the code you have given does exactly what you want. For another approach using itertools.islice you could see an example of following recipe:

from itertools import islice, chain

def batch(iterable, size):
    sourceiter = iter(iterable)
    while True:
        batchiter = islice(sourceiter, size)
        yield chain([batchiter.next()], batchiter)

More-itertools includes two functions that do what you need:

• chunked(iterable, n) returns an iterable of lists, each of length n (except the last one, which may be shorter);
• ichunked(iterable, n) is similar, but returns an iterable of iterables instead.

Solution for Python 3.8 if you are working with iterables that don't define a len function, and get exhausted:

from itertools import islice

def batcher(iterable, batch_size):
    iterator = iter(iterable)
    while batch := list(islice(iterator, batch_size)):
        yield batch

Example usage:

def my_gen():
    yield from range(10)
 
for batch in batcher(my_gen(), 3):
    print(batch)

>>> [0, 1, 2]
>>> [3, 4, 5]
>>> [6, 7, 8]
>>> [9]

Could of course be implemented without the walrus operator as well.