How to Split a List into Equally-Sized Chunks

How do I split a list into equally-sized chunks?

Here's a generator that yields evenly-sized chunks:

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]]

For Python 2, using 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]

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Below is a list comprehension one-liner. The method above is preferable, though, since using named functions makes code easier to understand. For Python 3:

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

For Python 2:

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

Splitting a list into N parts of approximately equal length

This code is broken due to rounding errors. Do not use it!!!

assert len(chunkIt([1,2,3], 10)) == 10  # fails

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Here's one that could work:

def chunkIt(seq, num):
    avg = len(seq) / float(num)
    out = []
    last = 0.0

    while last < len(seq):
        out.append(seq[int(last):int(last + avg)])
        last += avg

    return out

Testing:

>>> chunkIt(range(10), 3)
[[0, 1, 2], [3, 4, 5], [6, 7, 8, 9]]
>>> chunkIt(range(11), 3)
[[0, 1, 2], [3, 4, 5, 6], [7, 8, 9, 10]]
>>> chunkIt(range(12), 3)
[[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11]]

How do I split a list into equally-sized chunks?

Here's a generator that yields evenly-sized chunks:

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]]

For Python 2, using 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]

---

Below is a list comprehension one-liner. The method above is preferable, though, since using named functions makes code easier to understand. For Python 3:

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

For Python 2:

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

How to split an array into chunks of a given length in python?

Why don't you try out a list comprehension?

Example:

[ids[i:i+2] for i in range(0,len(ids),2)]

Output:

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

How to split python list into chunks of equal size?

>>> x = [1,2,3,4,5,6,7,8,9]
>>> zip(*[iter(x)]*3)
[(1, 2, 3), (4, 5, 6), (7, 8, 9)]

How does zip(*[iter(s)]*n) work in Python?

Split a python list into other sublists i.e smaller lists

I'd say

chunks = [data[x:x+100] for x in range(0, len(data), 100)]

If you are using python 2.x instead of 3.x, you can be more memory-efficient by using xrange(), changing the above code to:

chunks = [data[x:x+100] for x in xrange(0, len(data), 100)]

Splitting a list into n-sized chunks

I'm not sure if this is exactly what you want or not, but I would recommend:

def split_list(list, n):
    toReturn = []
    for i in range(len(list)//n):
        toReturn.append(list[i*n:i*n+n])
    toReturn.append(list[-n:])
    return toReturn

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