jespern - 2 months ago 7
Python Question

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

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.

This should work:

``````l = range(1, 1000)
print chunks(l, 10) -> [ [ 1..10 ], [ 11..20 ], .., [ 991..999 ] ]
``````

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?

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

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

Also you can simply use list comprehension instead of write a function. Python 3:

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

Python 2 version:

``````[l[i:i + n] for i in xrange(0, len(l), n)]
``````