How to split python list into chunks of equal size? [duplicate]

Answer 1

>>> 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?

Answer 2

If you really want the sub elements to be lists vs tuples:

In [9]: [list(t) for t in zip(*[iter(range(1,10))]*3)]
Out[9]: [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

Or, if you want to include the left over elements that would be truncated by zip, use a slice syntax:

In [16]: l=range(14)

In [17]: [l[i:i+3] for i in range(0,len(l),3)]
Out[17]: [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10, 11], [12, 13]]

Answer 3

You can use numpy.reshape here as well:

import numpy as np

x = np.array([1,2,3,4,5,6,7,8,9])

new_x = np.reshape(x, (3,3))

Result:

>>> new_x
array([[1, 2, 3],
       [4, 5, 6],
       [7, 8, 9]])

Answer 4

>>> map(None,*[iter(s)]*3)
[(1, 2, 3), (4, 5, 6), (7, 8, 9)]

Answer 5

Here's a much less "clever" way of doing it with recursion:

from itertools import chain

def groupsof(n, xs):
    if len(xs) < n:
        return [xs]
    else:
        return chain([xs[0:n]], groupsof(n, xs[n:]))

print list(groupsof(3, [1,2,3,4,5,6,7,8,9,10,11,12,13]))