## How do you split a list into evenly sized chunks?

### Question

I have a list of arbitrary length, and I need to split it up into equal size chunks and operate on it. There are some obvious ways to do this, like keeping a counter and two lists, and when the second list fills up, add it to the first list and empty the second list for the next round of data, but this is potentially extremely expensive.

I was wondering if anyone had a good solution to this for lists of any length, e.g. using generators.

I was looking for something useful in `itertools` but I couldn't find anything obviously useful. Might've missed it, though.

Related question: What is the most “pythonic” way to iterate over a list in chunks?

2017/05/23
1
2343
5/23/2017 11:55:11 AM

Here's a generator that yields the chunks you want:

``````def chunks(lst, n):
"""Yield successive n-sized chunks from lst."""
for i in range(0, len(lst), n):
yield lst[i:i + n]
``````

``````import pprint
pprint.pprint(list(chunks(range(10, 75), 10)))
[[10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
[20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
[30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
[40, 41, 42, 43, 44, 45, 46, 47, 48, 49],
[50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
[60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
[70, 71, 72, 73, 74]]
``````

If you're using Python 2, you should use `xrange()` instead of `range()`:

``````def chunks(lst, n):
"""Yield successive n-sized chunks from lst."""
for i in xrange(0, len(lst), n):
yield lst[i:i + n]
``````

Also you can simply use list comprehension instead of writing a function, though it's a good idea to encapsulate operations like this in named functions so that your code is easier to understand. Python 3:

``````[lst[i:i + n] for i in range(0, len(lst), n)]
``````

Python 2 version:

``````[lst[i:i + n] for i in xrange(0, len(lst), n)]
``````
2019/11/28
3281
11/28/2019 1:43:27 AM

If you want something super simple:

``````def chunks(l, n):
n = max(1, n)
return (l[i:i+n] for i in range(0, len(l), n))
``````

Use `xrange()` instead of `range()` in the case of Python 2.x

2020/01/06

Directly from the (old) Python documentation (recipes for itertools):

``````from itertools import izip, chain, repeat

"grouper(3, 'abcdefg', 'x') --> ('a','b','c'), ('d','e','f'), ('g','x','x')"
``````

The current version, as suggested by J.F.Sebastian:

``````#from itertools import izip_longest as zip_longest # for Python 2.x
from itertools import zip_longest # for Python 3.x
#from six.moves import zip_longest # for both (uses the six compat library)

"grouper(3, 'abcdefg', 'x') --> ('a','b','c'), ('d','e','f'), ('g','x','x')"
``````

I guess Guido's time machine works—worked—will work—will have worked—was working again.

These solutions work because `[iter(iterable)]*n` (or the equivalent in the earlier version) creates one iterator, repeated `n` times in the list. `izip_longest` then effectively performs a round-robin of "each" iterator; because this is the same iterator, it is advanced by each such call, resulting in each such zip-roundrobin generating one tuple of `n` items.

2017/09/21

I know this is kind of old but nobody yet mentioned `numpy.array_split`:

``````import numpy as np

lst = range(50)
np.array_split(lst, 5)
# [array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]),
#  array([10, 11, 12, 13, 14, 15, 16, 17, 18, 19]),
#  array([20, 21, 22, 23, 24, 25, 26, 27, 28, 29]),
#  array([30, 31, 32, 33, 34, 35, 36, 37, 38, 39]),
#  array([40, 41, 42, 43, 44, 45, 46, 47, 48, 49])]
``````
2019/10/14

I'm surprised nobody has thought of using `iter`'s two-argument form:

``````from itertools import islice

def chunk(it, size):
it = iter(it)
return iter(lambda: tuple(islice(it, size)), ())
``````

Demo:

``````>>> list(chunk(range(14), 3))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13)]
``````

This works with any iterable and produces output lazily. It returns tuples rather than iterators, but I think it has a certain elegance nonetheless. It also doesn't pad; if you want padding, a simple variation on the above will suffice:

``````from itertools import islice, chain, repeat

return iter(lambda: tuple(islice(it, size)), (padval,) * size)
``````

Demo:

``````>>> list(chunk_pad(range(14), 3))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13, None)]
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13, 'a')]
``````

Like the `izip_longest`-based solutions, the above always pads. As far as I know, there's no one- or two-line itertools recipe for a function that optionally pads. By combining the above two approaches, this one comes pretty close:

``````_no_padding = object()

it = iter(it)
sentinel = ()
else:
return iter(lambda: tuple(islice(it, size)), sentinel)
``````

Demo:

``````>>> list(chunk(range(14), 3))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13)]
>>> list(chunk(range(14), 3, None))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13, None)]
>>> list(chunk(range(14), 3, 'a'))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13, 'a')]
``````

I believe this is the shortest chunker proposed that offers optional padding.

As Tomasz Gandor observed, the two padding chunkers will stop unexpectedly if they encounter a long sequence of pad values. Here's a final variation that works around that problem in a reasonable way:

``````_no_padding = object()
it = iter(it)
chunker = iter(lambda: tuple(islice(it, size)), ())
yield from chunker
else:
for ch in chunker:
yield ch if len(ch) == size else ch + (padval,) * (size - len(ch))
``````

Demo:

``````>>> list(chunk([1, 2, (), (), 5], 2))
[(1, 2), ((), ()), (5,)]
>>> list(chunk([1, 2, None, None, 5], 2, None))
[(1, 2), (None, None), (5, None)]
``````
2018/11/17

Here is a generator that work on arbitrary iterables:

``````def split_seq(iterable, size):
it = iter(iterable)
item = list(itertools.islice(it, size))
while item:
yield item
item = list(itertools.islice(it, size))
``````

Example:

``````>>> import pprint
>>> pprint.pprint(list(split_seq(xrange(75), 10)))
[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
[20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
[30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
[40, 41, 42, 43, 44, 45, 46, 47, 48, 49],
[50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
[60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
[70, 71, 72, 73, 74]]
``````
2012/09/17