Splitting a list into N parts of approximately equal length

Question

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 o

Answer 1

If you divide n elements into roughly k chunks you can make n % k chunks 1 element bigger than the other chunks to distribute the extra elements.

The following code will give you the length for the chunks:

[(n // k) + (1 if i < (n % k) else 0) for i in range(k)]

Example: n=11, k=3 results in [4, 4, 3]

You can then easily calculate the start indizes for the chunks:

[i * (n // k) + min(i, n % k) for i in range(k)]

Example: n=11, k=3 results in [0, 4, 8]

Using the i+1th chunk as the boundary we get that the ith chunk of list l with len n is

l[i * (n // k) + min(i, n % k):(i+1) * (n // k) + min(i+1, n % k)]

As a final step create a list from all the chunks using list comprehension:

[l[i * (n // k) + min(i, n % k):(i+1) * (n // k) + min(i+1, n % k)] for i in range(k)]

Example: n=11, k=3, l=range(n) results in [range(0, 4), range(4, 8), range(8, 11)]

Answer 2

I tried most part of solutions, but they didn't work for my case, so I make a new function that work for most of cases and for any type of array:

import math

def chunkIt(seq, num):
    seqLen = len(seq)
    total_chunks = math.ceil(seqLen / num)
    items_per_chunk = num
    out = []
    last = 0

    while last < seqLen:
        out.append(seq[last:(last + items_per_chunk)])
        last += items_per_chunk

    return out

Answer 3

#!/usr/bin/python


first_names = ['Steve', 'Jane', 'Sara', 'Mary','Jack','Bob', 'Bily', 'Boni', 'Chris','Sori', 'Will', 'Won','Li']

def chunks(l, n):
for i in range(0, len(l), n):
    # Create an index range for l of n items:
    yield l[i:i+n]

result = list(chunks(first_names, 5))
print result

Picked from this link, and this was what helped me. I had a pre-defined list.

Answer 4

You can write it fairly simply as a list generator:

def split(a, n):
    k, m = divmod(len(a), n)
    return (a[i * k + min(i, m):(i + 1) * k + min(i + 1, m)] for i in range(n))

Example:

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

Answer 5

Using list comprehension:

def divide_list_to_chunks(list_, n):
    return [list_[start::n] for start in range(n)]

Answer 6

def chunk_array(array : List, n: int) -> List[List]:
    chunk_size = len(array) // n 
    chunks = []
    i = 0
    while i < len(array):
        # if less than chunk_size left add the remainder to last element
        if len(array) - (i + chunk_size + 1) < 0:
            chunks[-1].append(*array[i:i + chunk_size])
            break
        else:
            chunks.append(array[i:i + chunk_size])
            i += chunk_size
    return chunks

here's my version (inspired from Max's)