Splitting a list into N parts of approximately equal length

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What is the best way to divide a list into roughly equal parts? For example, if the list has 7 elements and is split it into 2 parts, we want to get 3 elements in one part, and the other should have 4 elements.

I"m looking for something like even_split(L, n) that breaks L into n parts.

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

The code above gives chunks of 3, rather than 3 chunks. I could simply transpose (iterate over this and take the first element of each column, call that part one, then take the second and put it in part two, etc), but that destroys the ordering of the items.

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Splitting a list into N parts of approximately equal length split: Questions

How do you split a list into evenly sized chunks?

5 answers

jespern By jespern

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?

2632

Answer #1

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)]

2632

Answer #2

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

2632

Answer #3

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

from itertools import izip, chain, repeat

def grouper(n, iterable, padvalue=None):
    "grouper(3, "abcdefg", "x") --> ("a","b","c"), ("d","e","f"), ("g","x","x")"
    return izip(*[chain(iterable, repeat(padvalue, n-1))]*n)

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)

def grouper(n, iterable, padvalue=None):
    "grouper(3, "abcdefg", "x") --> ("a","b","c"), ("d","e","f"), ("g","x","x")"
    return zip_longest(*[iter(iterable)]*n, fillvalue=padvalue)

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.

We hope this article has helped you to resolve the problem. Apart from Splitting a list into N parts of approximately equal length, check other split-related topics.

Want to excel in Python? See our review of the best Python online courses 2022. If you are interested in Data Science, check also how to learn programming in R.

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Oliver Ungerschaft

Moscow | 2022-12-03

I was preparing for my coding interview, thanks for clarifying this - Splitting a list into N parts of approximately equal length in Python is not the simplest one. I just hope that will not emerge anymore

Dmitry Chamberlet

Prague | 2022-12-03

Thanks for explaining! I was stuck with Splitting a list into N parts of approximately equal length for some hours, finally got it done 🤗. I just hope that will not emerge anymore

Marie Chamberlet

Shanghai | 2022-12-03

I was preparing for my coding interview, thanks for clarifying this - Splitting a list into N parts of approximately equal length in Python is not the simplest one. Will use it in my bachelor thesis

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