Python split list into n chunks

Question

I know this question has been covered many times but my requirement is different.

I have a list like: range(1, 26). I want to divide this list into a fi

Answer 1

If you want to have the chunks as evenly sized as possible:

def chunk_ranges(items: int, chunks: int) -> List[Tuple[int, int]]:
    """
    Split the items by best effort into equally-sized chunks.
    
    If there are fewer items than chunks, each chunk contains an item and 
    there are fewer returned chunk indices than the argument `chunks`.

    :param items: number of items in the batch.
    :param chunks: number of chunks
    :return: list of (chunk begin inclusive, chunk end exclusive)
    """
    assert chunks > 0, \
        "Unexpected non-positive chunk count: {}".format(chunks)

    result = []  # type: List[Tuple[int, int]]
    if items <= chunks:
        for i in range(0, items):
            result.append((i, i + 1))
        return result

    chunk_size, extras = divmod(items, chunks)

    start = 0
    for i in range(0, chunks):
        if i < extras:
            end = start + chunk_size + 1
        else:
            end = start + chunk_size

        result.append((start, end))
        start = end

    return result

Test case:

def test_chunk_ranges(self):
    self.assertListEqual(chunk_ranges(items=8, chunks=1),
                         [(0, 8)])

    self.assertListEqual(chunk_ranges(items=8, chunks=2),
                         [(0, 4), (4, 8)])

    self.assertListEqual(chunk_ranges(items=8, chunks=3),
                         [(0, 3), (3, 6), (6, 8)])

    self.assertListEqual(chunk_ranges(items=8, chunks=5),
                         [(0, 2), (2, 4), (4, 6), (6, 7), (7, 8)])

    self.assertListEqual(chunk_ranges(items=8, chunks=6),
                         [(0, 2), (2, 4), (4, 5), (5, 6), (6, 7), (7, 8)])

    self.assertListEqual(
        chunk_ranges(items=8, chunks=7),
        [(0, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7), (7, 8)])

    self.assertListEqual(
        chunk_ranges(items=8, chunks=9),
        [(0, 1), (1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7), (7, 8)])

Answer 2

x=[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]
chunk = len(x)/6

l=[]
i=0
while i<len(x):
    if len(l)<=4:
        l.append(x [i:i + chunk])
    else:
        l.append(x [i:])
        break
    i+=chunk   

print l

#output=[[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]]

Answer 3

Assuming you want to divide into n chunks:

n = 6
num = float(len(x))/n
l = [ x [i:i + int(num)] for i in range(0, (n-1)*int(num), int(num))]
l.append(x[(n-1)*int(num):])

This method simply divides the length of the list by the number of chunks and, in case the length is not a multiple of the number, adds the extra elements in the last list.

Answer 4

One way would be to make the last list uneven and the rest even. This can be done as follows:

>>> x
[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]
>>> m = len(x) // 6
>>> test = [x[i:i+m] for i in range(0, len(x), m)]
>>> test[-2:] = [test[-2] + test[-1]]
>>> test
[[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]]

Answer 5

Use numpy

>>> import numpy
>>> x = range(25)
>>> l = numpy.array_split(numpy.array(x),6)

or

>>> import numpy
>>> x = numpy.arange(25)
>>> l = numpy.array_split(x,6);

You can also use numpy.split but that one throws in error if the length is not exactly divisible.

Answer 6

arr1=[-20, 20, -10, 0, 4, 8, 10, 6, 15, 9, 18, 35, 40, -30, -90, 99]
n=4
final = [arr1[i * n:(i + 1) * n] for i in range((len(arr1) + n - 1) // n )]
print(final)

Output:

[[-20, 20, -10, 0], [4, 8, 10, 6], [15, 9, 18, 35], [40, -30, -90, 99]]