How to iterate over a list in chunks

Question

I have a Python script which takes as input a list of integers, which I need to work with four integers at a time. Unfortunately, I don't have control of the input, or I'd have it passed in as a list of four-element tuples. Currently, I'm iterating over it this way:

for i in range(0, len(ints), 4):
    # dummy op for example code
    foo += ints[i] * ints[i + 1] + ints[i + 2] * ints[i + 3]

It looks a lot like "C-think", though, which makes me suspect there's a more pythonic way of dealing with this situation. The list is discarded after iterating, so it needn't be preserved. Perhaps something like this would be better?

while ints:
    foo += ints[0] * ints[1] + ints[2] * ints[3]
    ints[0:4] = []

Still doesn't quite "feel" right, though. :-/

Related question: How do you split a list into evenly sized chunks in Python?

Answer 1

def chunker(seq, size):
    return (seq[pos:pos + size] for pos in range(0, len(seq), size))

Works with any sequence:

text = "I am a very, very helpful text"

for group in chunker(text, 7):
   print(repr(group),)
# 'I am a ' 'very, v' 'ery hel' 'pful te' 'xt'

print('|'.join(chunker(text, 10)))
# I am a ver|y, very he|lpful text

animals = ['cat', 'dog', 'rabbit', 'duck', 'bird', 'cow', 'gnu', 'fish']

for group in chunker(animals, 3):
    print(group)
# ['cat', 'dog', 'rabbit']
# ['duck', 'bird', 'cow']
# ['gnu', 'fish']

Answer 2

Modified from the Recipes section of Python's itertools docs:

from itertools import zip_longest

def grouper(iterable, n, fillvalue=None):
    args = [iter(iterable)] * n
    return zip_longest(*args, fillvalue=fillvalue)

Example

grouper('ABCDEFGHIJ', 3, 'x')  # --> 'ABC' 'DEF' 'GHI' 'Jxx'

Note: on Python 2 use izip_longest instead of zip_longest.

Answer 3

chunk_size = 4
for i in range(0, len(ints), chunk_size):
    chunk = ints[i:i+chunk_size]
    # process chunk of size <= chunk_size

Answer 4

Since Python 3.8 you can use the walrus := operator and itertools.islice.

from itertools import islice

list_ = [i for i in range(10, 100)]

def chunker(it, size):
    iterator = iter(it)
    while chunk := list(islice(iterator, size)):
        print(chunk)

In [2]: chunker(list_, 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, 75, 76, 77, 78, 79]
[80, 81, 82, 83, 84, 85, 86, 87, 88, 89]
[90, 91, 92, 93, 94, 95, 96, 97, 98, 99]

Answer 5

import itertools
def chunks(iterable,size):
    it = iter(iterable)
    chunk = tuple(itertools.islice(it,size))
    while chunk:
        yield chunk
        chunk = tuple(itertools.islice(it,size))

# though this will throw ValueError if the length of ints
# isn't a multiple of four:
for x1,x2,x3,x4 in chunks(ints,4):
    foo += x1 + x2 + x3 + x4

for chunk in chunks(ints,4):
    foo += sum(chunk)

Another way:

import itertools
def chunks2(iterable,size,filler=None):
    it = itertools.chain(iterable,itertools.repeat(filler,size-1))
    chunk = tuple(itertools.islice(it,size))
    while len(chunk) == size:
        yield chunk
        chunk = tuple(itertools.islice(it,size))

# x2, x3 and x4 could get the value 0 if the length is not
# a multiple of 4.
for x1,x2,x3,x4 in chunks2(ints,4,0):
    foo += x1 + x2 + x3 + x4

Answer 6

If you don't mind using an external package you could use iteration_utilities.grouper from iteration_utilties ¹. It supports all iterables (not just sequences):

from iteration_utilities import grouper
seq = list(range(20))
for group in grouper(seq, 4):
    print(group)

which prints:

(0, 1, 2, 3)
(4, 5, 6, 7)
(8, 9, 10, 11)
(12, 13, 14, 15)
(16, 17, 18, 19)

In case the length isn't a multiple of the groupsize it also supports filling (the incomplete last group) or truncating (discarding the incomplete last group) the last one:

from iteration_utilities import grouper
seq = list(range(17))
for group in grouper(seq, 4):
    print(group)
# (0, 1, 2, 3)
# (4, 5, 6, 7)
# (8, 9, 10, 11)
# (12, 13, 14, 15)
# (16,)

for group in grouper(seq, 4, fillvalue=None):
    print(group)
# (0, 1, 2, 3)
# (4, 5, 6, 7)
# (8, 9, 10, 11)
# (12, 13, 14, 15)
# (16, None, None, None)

for group in grouper(seq, 4, truncate=True):
    print(group)
# (0, 1, 2, 3)
# (4, 5, 6, 7)
# (8, 9, 10, 11)
# (12, 13, 14, 15)

---

Benchmarks

I also decided to compare the run-time of a few of the mentioned approaches. It's a log-log plot grouping into groups of "10" elements based on a list of varying size. For qualitative results: Lower means faster:

At least in this benchmark the iteration_utilities.grouper performs best. Followed by the approach of Craz.

The benchmark was created with simple_benchmark¹. The code used to run this benchmark was:

import iteration_utilities
import itertools
from itertools import zip_longest

def consume_all(it):
    return iteration_utilities.consume(it, None)

import simple_benchmark
b = simple_benchmark.BenchmarkBuilder()

@b.add_function()
def grouper(l, n):
    return consume_all(iteration_utilities.grouper(l, n))

def Craz_inner(iterable, n, fillvalue=None):
    args = [iter(iterable)] * n
    return zip_longest(*args, fillvalue=fillvalue)

@b.add_function()
def Craz(iterable, n, fillvalue=None):
    return consume_all(Craz_inner(iterable, n, fillvalue))

def nosklo_inner(seq, size):
    return (seq[pos:pos + size] for pos in range(0, len(seq), size))

@b.add_function()
def nosklo(seq, size):
    return consume_all(nosklo_inner(seq, size))

def SLott_inner(ints, chunk_size):
    for i in range(0, len(ints), chunk_size):
        yield ints[i:i+chunk_size]

@b.add_function()
def SLott(ints, chunk_size):
    return consume_all(SLott_inner(ints, chunk_size))

def MarkusJarderot1_inner(iterable,size):
    it = iter(iterable)
    chunk = tuple(itertools.islice(it,size))
    while chunk:
        yield chunk
        chunk = tuple(itertools.islice(it,size))

@b.add_function()
def MarkusJarderot1(iterable,size):
    return consume_all(MarkusJarderot1_inner(iterable,size))

def MarkusJarderot2_inner(iterable,size,filler=None):
    it = itertools.chain(iterable,itertools.repeat(filler,size-1))
    chunk = tuple(itertools.islice(it,size))
    while len(chunk) == size:
        yield chunk
        chunk = tuple(itertools.islice(it,size))

@b.add_function()
def MarkusJarderot2(iterable,size):
    return consume_all(MarkusJarderot2_inner(iterable,size))

@b.add_arguments()
def argument_provider():
    for exp in range(2, 20):
        size = 2**exp
        yield size, simple_benchmark.MultiArgument([[0] * size, 10])

r = b.run()

---

¹ Disclaimer: I'm the author of the libraries iteration_utilities and simple_benchmark.

Answer 7

The more-itertools package has chunked method which does exactly that:

import more_itertools
for s in more_itertools.chunked(range(9), 4):
    print(s)

Prints

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

chunked returns the items in a list. If you'd prefer iterables, use ichunked.

Answer 8

The ideal solution for this problem works with iterators (not just sequences). It should also be fast.

This is the solution provided by the documentation for itertools:

def grouper(n, iterable, fillvalue=None):
    #"grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx"
    args = [iter(iterable)] * n
    return itertools.izip_longest(fillvalue=fillvalue, *args)

Using ipython's %timeit on my mac book air, I get 47.5 us per loop.

However, this really doesn't work for me since the results are padded to be even sized groups. A solution without the padding is slightly more complicated. The most naive solution might be:

def grouper(size, iterable):
    i = iter(iterable)
    while True:
        out = []
        try:
            for _ in range(size):
                out.append(i.next())
        except StopIteration:
            yield out
            break
        
        yield out

Simple, but pretty slow: 693 us per loop

The best solution I could come up with uses islice for the inner loop:

def grouper(size, iterable):
    it = iter(iterable)
    while True:
        group = tuple(itertools.islice(it, None, size))
        if not group:
            break
        yield group

With the same dataset, I get 305 us per loop.

Unable to get a pure solution any faster than that, I provide the following solution with an important caveat: If your input data has instances of filldata in it, you could get wrong answer.

def grouper(n, iterable, fillvalue=None):
    #"grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx"
    args = [iter(iterable)] * n
    # itertools.zip_longest on Python 3
    for x in itertools.izip_longest(*args, fillvalue=fillvalue):
        if x[-1] is fillvalue:
            yield tuple(v for v in x if v is not fillvalue)
        else:
            yield x

I really don't like this answer, but it is significantly faster. 124 us per loop

Answer 9

I needed a solution that would also work with sets and generators. I couldn't come up with anything very short and pretty, but it's quite readable at least.

def chunker(seq, size):
    res = []
    for el in seq:
        res.append(el)
        if len(res) == size:
            yield res
            res = []
    if res:
        yield res

List:

>>> list(chunker([i for i in range(10)], 3))
[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]]

Set:

>>> list(chunker(set([i for i in range(10)]), 3))
[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]]

Generator:

>>> list(chunker((i for i in range(10)), 3))
[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]]

Answer 10

from itertools import izip_longest

def chunker(iterable, chunksize, filler):
    return izip_longest(*[iter(iterable)]*chunksize, fillvalue=filler)

Answer 11

Similar to other proposals, but not exactly identical, I like doing it this way, because it's simple and easy to read:

it = iter([1, 2, 3, 4, 5, 6, 7, 8, 9])
for chunk in zip(it, it, it, it):
    print chunk

>>> (1, 2, 3, 4)
>>> (5, 6, 7, 8)

This way you won't get the last partial chunk. If you want to get (9, None, None, None) as last chunk, just use izip_longest from itertools.

Answer 12

Since nobody's mentioned it yet here's a zip() solution:

>>> def chunker(iterable, chunksize):
...     return zip(*[iter(iterable)]*chunksize)

It works only if your sequence's length is always divisible by the chunk size or you don't care about a trailing chunk if it isn't.

Example:

>>> s = '1234567890'
>>> chunker(s, 3)
[('1', '2', '3'), ('4', '5', '6'), ('7', '8', '9')]
>>> chunker(s, 4)
[('1', '2', '3', '4'), ('5', '6', '7', '8')]
>>> chunker(s, 5)
[('1', '2', '3', '4', '5'), ('6', '7', '8', '9', '0')]

Or using itertools.izip to return an iterator instead of a list:

>>> from itertools import izip
>>> def chunker(iterable, chunksize):
...     return izip(*[iter(iterable)]*chunksize)

Padding can be fixed using @ΤΖΩΤΖΙΟΥ's answer:

>>> from itertools import chain, izip, repeat
>>> def chunker(iterable, chunksize, fillvalue=None):
...     it   = chain(iterable, repeat(fillvalue, chunksize-1))
...     args = [it] * chunksize
...     return izip(*args)

Answer 13

As of Python 3.12, the itertools module gains a batched function that specifically covers iterating over batches of an input iterable, where the final batch may be incomplete (each batch is a tuple). Per the example code given in the docs:

>>> for batch in batched('ABCDEFG', 3):
...     print(batch)
...
('A', 'B', 'C')
('D', 'E', 'F')
('G',)

---

Performance notes:

The implementation of batched, like all itertools functions to date, is at the C layer, so it's capable of optimizations Python level code cannot match, e.g.

• On each pull of a new batch, it proactively allocates a tuple of precisely the correct size (for all but the last batch), instead of building the tuple up element by element with amortized growth causing multiple reallocations (the way a solution calling tuple on an islice does)
• It only needs to look up the .__next__ function of the underlying iterator once per batch, not n times per batch (the way a zip_longest((iter(iterable),) * n)-based approach does)
• The check for the end case is a simple C level NULL check (trivial, and required to handle possible exceptions anyway)
• Handling the end case is a C goto followed by a direct realloc (no making a copy into a smaller tuple) down to the already known final size, since it's tracking how many elements it has successfully pulled (no complex "create sentinel for use as fillvalue and do Python level if/else checks for each batch to see if it's empty, with the final batch requiring a search for where the fillvalue appeared last, to create the cut-down tuple" required by zip_longest-based solutions).

Between all these advantages, it should massively outperform any Python-level solution (even highly optimized ones that push most or all of the per-item work to the C layer), regardless of whether the input iterable is long or short, and regardless of whether the batch size and the size of the final (possibly incomplete) batch (zip_longest-based solutions using guaranteed unique fillvalues for safety are the best possible solution for almost all cases when itertools.batched is not available, but they can suffer in pathological cases of "few large batches, with final batch mostly, not completely, filled", especially pre-3.10 when bisect can't be used to optimize slicing off the fillvalues from O(n) linear search down to O(log n) binary search, but batched avoids that search entirely, so it won't experience pathological cases at all).

Answer 14

Another approach would be to use the two-argument form of iter:

from itertools import islice

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

This can be adapted easily to use padding (this is similar to Markus Jarderot’s answer):

from itertools import islice, chain, repeat

def group_pad(it, size, pad=None):
    it = chain(iter(it), repeat(pad))
    return iter(lambda: tuple(islice(it, size)), (pad,) * size)

These can even be combined for optional padding:

_no_pad = object()
def group(it, size, pad=_no_pad):
    if pad == _no_pad:
        it = iter(it)
        sentinel = ()
    else:
        it = chain(iter(it), repeat(pad))
        sentinel = (pad,) * size
    return iter(lambda: tuple(islice(it, size)), sentinel)

Answer 15

Using little functions and things really doesn't appeal to me; I prefer to just use slices:

data = [...]
chunk_size = 10000 # or whatever
chunks = [data[i:i+chunk_size] for i in xrange(0,len(data),chunk_size)]
for chunk in chunks:
    ...

Answer 16

If the list is large, the highest-performing way to do this will be to use a generator:

def get_chunk(iterable, chunk_size):
    result = []
    for item in iterable:
        result.append(item)
        if len(result) == chunk_size:
            yield tuple(result)
            result = []
    if len(result) > 0:
        yield tuple(result)

for x in get_chunk([1,2,3,4,5,6,7,8,9,10], 3):
    print x

(1, 2, 3)
(4, 5, 6)
(7, 8, 9)
(10,)

Answer 17

Using map() instead of zip() fixes the padding issue in J.F. Sebastian's answer:

>>> def chunker(iterable, chunksize):
...   return map(None,*[iter(iterable)]*chunksize)

Example:

>>> s = '1234567890'
>>> chunker(s, 3)
[('1', '2', '3'), ('4', '5', '6'), ('7', '8', '9'), ('0', None, None)]
>>> chunker(s, 4)
[('1', '2', '3', '4'), ('5', '6', '7', '8'), ('9', '0', None, None)]
>>> chunker(s, 5)
[('1', '2', '3', '4', '5'), ('6', '7', '8', '9', '0')]

Answer 18

One-liner, adhoc solution to iterate over a list x in chunks of size 4 -

for a, b, c, d in zip(x[0::4], x[1::4], x[2::4], x[3::4]):
    ... do something with a, b, c and d ...

Answer 19

To avoid all conversions to a list import itertools and:

>>> for k, g in itertools.groupby(xrange(35), lambda x: x/10):
...     list(g)

Produces:

... 
0 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
1 [10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
2 [20, 21, 22, 23, 24, 25, 26, 27, 28, 29]
3 [30, 31, 32, 33, 34]
>>> 

I checked groupby and it doesn't convert to list or use len so I (think) this will delay resolution of each value until it is actually used. Sadly none of the available answers (at this time) seemed to offer this variation.

Obviously if you need to handle each item in turn nest a for loop over g:

for k,g in itertools.groupby(xrange(35), lambda x: x/10):
    for i in g:
       # do what you need to do with individual items
    # now do what you need to do with the whole group

My specific interest in this was the need to consume a generator to submit changes in batches of up to 1000 to the gmail API:

    messages = a_generator_which_would_not_be_smart_as_a_list
    for idx, batch in groupby(messages, lambda x: x/1000):
        batch_request = BatchHttpRequest()
        for message in batch:
            batch_request.add(self.service.users().messages().modify(userId='me', id=message['id'], body=msg_labels))
        http = httplib2.Http()
        self.credentials.authorize(http)
        batch_request.execute(http=http)

Answer 20

Unless I misses something, the following simple solution with generator expressions has not been mentioned. It assumes that both the size and the number of chunks are known (which is often the case), and that no padding is required:

def chunks(it, n, m):
    """Make an iterator over m first chunks of size n.
    """
    it = iter(it)
    # Chunks are presented as tuples.
    return (tuple(next(it) for _ in range(n)) for _ in range(m))

Answer 21

With NumPy it's simple:

ints = array([1, 2, 3, 4, 5, 6, 7, 8])
for int1, int2 in ints.reshape(-1, 2):
    print(int1, int2)

output:

1 2
3 4
5 6
7 8

Answer 22

def chunker(iterable, n):
    """Yield iterable in chunk sizes.

    >>> chunks = chunker('ABCDEF', n=4)
    >>> chunks.next()
    ['A', 'B', 'C', 'D']
    >>> chunks.next()
    ['E', 'F']
    """
    it = iter(iterable)
    while True:
        chunk = []
        for i in range(n):
            try:
                chunk.append(next(it))
            except StopIteration:
                yield chunk
                raise StopIteration
        yield chunk

if __name__ == '__main__':
    import doctest

    doctest.testmod()

Answer 23

In your second method, I would advance to the next group of 4 by doing this:

ints = ints[4:]

However, I haven't done any performance measurement so I don't know which one might be more efficient.

Having said that, I would usually choose the first method. It's not pretty, but that's often a consequence of interfacing with the outside world.

Answer 24

I never want my chunks padded, so that requirement is essential. I find that the ability to work on any iterable is also requirement. Given that, I decided to extend on the accepted answer, https://stackoverflow.com/a/434411/1074659.

Performance takes a slight hit in this approach if padding is not wanted due to the need to compare and filter the padded values. However, for large chunk sizes, this utility is very performant.

#!/usr/bin/env python3
from itertools import zip_longest


_UNDEFINED = object()


def chunker(iterable, chunksize, fillvalue=_UNDEFINED):
    """
    Collect data into chunks and optionally pad it.

    Performance worsens as `chunksize` approaches 1.

    Inspired by:
        https://docs.python.org/3/library/itertools.html#itertools-recipes

    """
    args = [iter(iterable)] * chunksize
    chunks = zip_longest(*args, fillvalue=fillvalue)
    yield from (
        filter(lambda val: val is not _UNDEFINED, chunk)
        if chunk[-1] is _UNDEFINED
        else chunk
        for chunk in chunks
    ) if fillvalue is _UNDEFINED else chunks

Answer 25

Yet another answer, the advantages of which are:

1) Easily understandable
2) Works on any iterable, not just sequences (some of the above answers will choke on filehandles)
3) Does not load the chunk into memory all at once
4) Does not make a chunk-long list of references to the same iterator in memory
5) No padding of fill values at the end of the list

That being said, I haven't timed it so it might be slower than some of the more clever methods, and some of the advantages may be irrelevant given the use case.

def chunkiter(iterable, size):
  def inneriter(first, iterator, size):
    yield first
    for _ in xrange(size - 1): 
      yield iterator.next()
  it = iter(iterable)
  while True:
    yield inneriter(it.next(), it, size)

In [2]: i = chunkiter('abcdefgh', 3)
In [3]: for ii in i:                                                
          for c in ii:
            print c,
          print ''
        ...:     
        a b c 
        d e f 
        g h 

Update:
A couple of drawbacks due to the fact the inner and outer loops are pulling values from the same iterator:
1) continue doesn't work as expected in the outer loop - it just continues on to the next item rather than skipping a chunk. However, this doesn't seem like a problem as there's nothing to test in the outer loop.
2) break doesn't work as expected in the inner loop - control will wind up in the inner loop again with the next item in the iterator. To skip whole chunks, either wrap the inner iterator (ii above) in a tuple, e.g. for c in tuple(ii), or set a flag and exhaust the iterator.

Answer 26

def group_by(iterable, size):
    """Group an iterable into lists that don't exceed the size given.

    >>> group_by([1,2,3,4,5], 2)
    [[1, 2], [3, 4], [5]]

    """
    sublist = []

    for index, item in enumerate(iterable):
        if index > 0 and index % size == 0:
            yield sublist
            sublist = []

        sublist.append(item)

    if sublist:
        yield sublist

Answer 27

You can use partition or chunks function from funcy library:

from funcy import partition

for a, b, c, d in partition(4, ints):
    foo += a * b * c * d

These functions also has iterator versions ipartition and ichunks, which will be more efficient in this case.

You can also peek at their implementation.

Answer 28

About solution gave by J.F. Sebastian here:

def chunker(iterable, chunksize):
    return zip(*[iter(iterable)]*chunksize)

It's clever, but has one disadvantage - always return tuple. How to get string instead?
Of course you can write ''.join(chunker(...)), but the temporary tuple is constructed anyway.

You can get rid of the temporary tuple by writing own zip, like this:

class IteratorExhausted(Exception):
    pass

def translate_StopIteration(iterable, to=IteratorExhausted):
    for i in iterable:
        yield i
    raise to # StopIteration would get ignored because this is generator,
             # but custom exception can leave the generator.

def custom_zip(*iterables, reductor=tuple):
    iterators = tuple(map(translate_StopIteration, iterables))
    while True:
        try:
            yield reductor(next(i) for i in iterators)
        except IteratorExhausted: # when any of iterators get exhausted.
            break

Then

def chunker(data, size, reductor=tuple):
    return custom_zip(*[iter(data)]*size, reductor=reductor)

Example usage:

>>> for i in chunker('12345', 2):
...     print(repr(i))
...
('1', '2')
('3', '4')
>>> for i in chunker('12345', 2, ''.join):
...     print(repr(i))
...
'12'
'34'

Answer 29

Here is a chunker without imports that supports generators:

def chunks(seq, size):
    it = iter(seq)
    while True:
        ret = tuple(next(it) for _ in range(size))
        if len(ret) == size:
            yield ret
        else:
            raise StopIteration()

Example of use:

>>> def foo():
...     i = 0
...     while True:
...         i += 1
...         yield i
...
>>> c = chunks(foo(), 3)
>>> c.next()
(1, 2, 3)
>>> c.next()
(4, 5, 6)
>>> list(chunks('abcdefg', 2))
[('a', 'b'), ('c', 'd'), ('e', 'f')]

Answer 30

I like this approach. It feels simple and not magical and supports all iterable types and doesn't require imports.

def chunk_iter(iterable, chunk_size):
it = iter(iterable)
while True:
    chunk = tuple(next(it) for _ in range(chunk_size))
    if not chunk:
        break
    yield chunk

Answer 31

Quite pythonic here (you may also inline the body of the split_groups function)

import itertools
def split_groups(iter_in, group_size):
    return ((x for _, x in item) for _, item in itertools.groupby(enumerate(iter_in), key=lambda x: x[0] // group_size))

for x, y, z, w in split_groups(range(16), 4):
    foo += x * y + z * w

Answer 32

here is my go works on lists,iters and range ... lazily :

def chunker(it,size):
    rv = [] 
    for i,el in enumerate(it,1) :   
        rv.append(el)
        if i % size == 0 : 
            yield rv
            rv = []
    if rv : yield rv        

almost made it one-liner ;(

In [95]: list(chunker(range(9),2) )                                                                                                                                          
Out[95]: [[0, 1], [2, 3], [4, 5], [6, 7], [8]]

In [96]: list(chunker([1,2,3,4,5],2) )                                                                                                                                       
Out[96]: [[1, 2], [3, 4], [5]]

In [97]: list(chunker(iter(range(9)),2) )                                                                                                                                    
Out[97]: [[0, 1], [2, 3], [4, 5], [6, 7], [8]]

In [98]: list(chunker(range(9),25) )                                                                                                                                         
Out[98]: [[0, 1, 2, 3, 4, 5, 6, 7, 8]]

In [99]: list(chunker(range(9),1) )                                                                                                                                          
Out[99]: [[0], [1], [2], [3], [4], [5], [6], [7], [8]]

In [101]: %timeit list(chunker(range(101),2) )                                                                                                                               
11.3 µs ± 68.2 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

Answer 33

At first, I designed it to split strings into substrings to parse string containing hex.
Today I turned it into complex, but still simple generator.

def chunker(iterable, size, reductor, condition):
    it = iter(iterable)
    def chunk_generator():
        return (next(it) for _ in range(size))
    chunk = reductor(chunk_generator())
    while condition(chunk):
        yield chunk
        chunk = reductor(chunk_generator())

Arguments:

Obvious ones

• iterable is any iterable / iterator / generator containg / generating / iterating over input data,
• size is, of course, size of chunk you want get,

More interesting

• reductor is a callable, which receives generator iterating over content of chunk.
  I'd expect it to return sequence or string, but I don't demand that.

  You can pass as this argument for example list, tuple, set, frozenset,
  or anything fancier. I'd pass this function, returning string
  (provided that iterable contains / generates / iterates over strings):

  def concatenate(iterable):
      return ''.join(iterable)
  

  ^{Note that reductor can cause closing generator by raising exception.}

• condition is a callable which receives anything what reductor returned.
  It decides to approve & yield it (by returning anything evaluating to True),
  or to decline it & finish generator's work (by returning anything other or raising exception).

  When number of elements in iterable is not divisible by size, when it gets exhausted, reductor will receive generator generating less elements than size.
  Let's call these elements lasts elements.

  I invited two functions to pass as this argument:

  • lambda x:x - the lasts elements will be yielded.

  • lambda x: len(x)==<size> - the lasts elements will be rejected.
    ^{replace <size> using number equal to size}

Answer 34

It is easy to make itertools.groupby work for you to get an iterable of iterables, without creating any temporary lists:

groupby(iterable, (lambda x,y: (lambda z: x.next()/y))(count(),100))

Don't get put off by the nested lambdas, outer lambda runs just once to put count() generator and the constant 100 into the scope of the inner lambda.

I use this to send chunks of rows to mysql.

for k,v in groupby(bigdata, (lambda x,y: (lambda z: x.next()/y))(count(),100))):
    cursor.executemany(sql, v)

Answer 35

This answer splits a list of strings, f.ex. to achieve PEP8-line length compliance:

def split(what, target_length=79):
    '''splits list of strings into sublists, each 
    having string length at most 79'''
    out = [[]]
    while what:
        if len("', '".join(out[-1])) + len(what[0]) < target_length:
            out[-1].append(what.pop(0))
        else:
            if not out[-1]: # string longer than target_length
                out[-1] = [what.pop(0)]
            out.append([])
    return out

Use as

>>> split(['deferred_income', 'long_term_incentive', 'restricted_stock_deferred', 'shared_receipt_with_poi', 'loan_advances', 'from_messages', 'other', 'director_fees', 'bonus', 'total_stock_value', 'from_poi_to_this_person', 'from_this_person_to_poi', 'restricted_stock', 'salary', 'total_payments', 'exercised_stock_options'], 75)
[['deferred_income', 'long_term_incentive', 'restricted_stock_deferred'], ['shared_receipt_with_poi', 'loan_advances', 'from_messages', 'other'], ['director_fees', 'bonus', 'total_stock_value', 'from_poi_to_this_person'], ['from_this_person_to_poi', 'restricted_stock', 'salary', 'total_payments'], ['exercised_stock_options']]

Answer 36

I am hoping that by turning an iterator out of a list i am not simply copying a slice of the list. Generators can be sliced and they will automatically still be a generator, while lists will be sliced into huge chunks of 1000 entries, which is less efficient.

def iter_group(iterable, batch_size:int):
    length = len(iterable)
    start = batch_size*-1
    end = 0
    while(end < length):
        start += batch_size
        end += batch_size
        if type(iterable) == list:
            yield (iterable[i] for i in range(start,min(length-1,end)))
        else:
            yield iterable[start:end]

Usage:

items = list(range(1,1251))

for item_group in iter_group(items, 1000):
    for item in item_group:
        print(item)

Answer 37

In my special case, I needed to pad the items to repeat the last element until it reaches the size, so I changed this answer to suit my needs.

Example of needed input output with size 4:

Input = [1,2,3,4,5,6,7,8]
Output= [[1,2,3,4], [5,6,7,8]]

Input = [[1,2,3,4,5,6,7]]
Output= [[1,2,3,4], [5,6,7,7]]

Input = [1,2,3,4,5]
Output= [[1,2,3,4], [5,5,5,5]]

def chunker(seq, size):
    res = []
    for el in seq:
        res.append(el)
        if len(res) == size:
            yield res
            res = []
    if res:
        res = res + (size - len(res)) * [res[-1]] 
        yield res

Answer 38

If the lists are the same size, you can combine them into lists of 4-tuples with zip(). For example:

# Four lists of four elements each.

l1 = range(0, 4)
l2 = range(4, 8)
l3 = range(8, 12)
l4 = range(12, 16)

for i1, i2, i3, i4 in zip(l1, l2, l3, l4):
    ...

Here's what the zip() function produces:

>>> print l1
[0, 1, 2, 3]
>>> print l2
[4, 5, 6, 7]
>>> print l3
[8, 9, 10, 11]
>>> print l4
[12, 13, 14, 15]
>>> print zip(l1, l2, l3, l4)
[(0, 4, 8, 12), (1, 5, 9, 13), (2, 6, 10, 14), (3, 7, 11, 15)]

If the lists are large, and you don't want to combine them into a bigger list, use itertools.izip(), which produces an iterator, rather than a list.

from itertools import izip

for i1, i2, i3, i4 in izip(l1, l2, l3, l4):
    ...

Answer 39

There doesn't seem to be a pretty way to do this. Here is a page that has a number of methods, including:

def split_seq(seq, size):
    newseq = []
    splitsize = 1.0/size*len(seq)
    for i in range(size):
        newseq.append(seq[int(round(i*splitsize)):int(round((i+1)*splitsize))])
    return newseq

Answer 40

Why not use list comprehension

l = [1 , 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
n = 4
filler = 0
fills = len(l) % n
chunks = ((l + [filler] * fills)[x * n:x * n + n] for x in range(int((len(l) + n - 1)/n)))
print(chunks)

[[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 0]]