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

Here is my solution:

def chunks(l, amount):
    if amount < 1:
        raise ValueError('amount must be positive integer')
    chunk_len = len(l) // amount
    leap_parts = len(l) % amount
    remainder = amount // 2  # make it symmetrical
    i = 0
    while i < len(l):
        remainder += leap_parts
        end_index = i + chunk_len
        if remainder >= amount:
            remainder -= amount
            end_index += 1
        yield l[i:end_index]
        i = end_index

Produces

    >>> list(chunks([1, 2, 3, 4, 5, 6, 7], 3))
    [[1, 2], [3, 4, 5], [6, 7]]

Answer 2

Here's another variant that spreads the "remaining" elements evenly among all the chunks, one at a time until there are none left. In this implementation, the larger chunks occur at the beginning the process.

def chunks(l, k):
  """ Yield k successive chunks from l."""
  if k < 1:
    yield []
    raise StopIteration
  n = len(l)
  avg = n/k
  remainders = n % k
  start, end = 0, avg
  while start < n:
    if remainders > 0:
      end = end + 1
      remainders = remainders - 1
    yield l[start:end]
    start, end = end, end+avg

For example, generate 4 chunks from a list of 14 elements:

>>> list(chunks(range(14), 4))
[[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10], [11, 12, 13]]
>>> map(len, list(chunks(range(14), 4)))
[4, 4, 3, 3]

Answer 3

Have a look at numpy.split:

>>> a = numpy.array([1,2,3,4])
>>> numpy.split(a, 2)
[array([1, 2]), array([3, 4])]

Answer 4

Rounding the linspace and using it as an index is an easier solution than what amit12690 proposes.

function chunks=chunkit(array,num)

index = round(linspace(0,size(array,2),num+1));

chunks = cell(1,num);

for x = 1:num
chunks{x} = array(:,index(x)+1:index(x+1));
end
end

Answer 5

say you want to split into 5 parts:

p1, p2, p3, p4, p5 = np.split(df, 5)

Answer 6

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

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

---

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]]