Slicing a list into n nearly-equal-length partitions

def partition(lst, n):
    division = len(lst) / float(n)
    return [ lst[int(round(division * i)): int(round(division * (i + 1)))] for i in xrange(n) ]

>>> partition([1,2,3,4,5],5)
[[1], [2], [3], [4], [5]]
>>> partition([1,2,3,4,5],2)
[[1, 2, 3], [4, 5]]
>>> partition([1,2,3,4,5],3)
[[1, 2], [3, 4], [5]]
>>> partition(range(105), 10)
[[0, 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, 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, 100, 101, 102, 103, 104]]

Python 3 version:

def partition(lst, n):
    division = len(lst) / n
    return [lst[round(division * i):round(division * (i + 1))] for i in range(n)]

Just a different take, that only works if [[1,3,5],[2,4]] is an acceptable partition, in your example.

def partition ( lst, n ):
    return [ lst[i::n] for i in xrange(n) ]

This satisfies the example mentioned in @Daniel Stutzbach's example:

partition(range(105),10)
# [[0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100],
# [1, 11, 21, 31, 41, 51, 61, 71, 81, 91, 101],
# [2, 12, 22, 32, 42, 52, 62, 72, 82, 92, 102],
# [3, 13, 23, 33, 43, 53, 63, 73, 83, 93, 103],
# [4, 14, 24, 34, 44, 54, 64, 74, 84, 94, 104],
# [5, 15, 25, 35, 45, 55, 65, 75, 85, 95],
# [6, 16, 26, 36, 46, 56, 66, 76, 86, 96],
# [7, 17, 27, 37, 47, 57, 67, 77, 87, 97],
# [8, 18, 28, 38, 48, 58, 68, 78, 88, 98],
# [9, 19, 29, 39, 49, 59, 69, 79, 89, 99]]

Here's a version that's similar to Daniel's: it divides as evenly as possible, but puts all the larger partitions at the start:

def partition(lst, n):
    q, r = divmod(len(lst), n)
    indices = [q*i + min(i, r) for i in xrange(n+1)]
    return [lst[indices[i]:indices[i+1]] for i in xrange(n)]

It also avoids the use of float arithmetic, since that always makes me uncomfortable. :)

Edit: an example, just to show the contrast with Daniel Stutzbach's solution

>>> print [len(x) for x in partition(range(105), 10)]
[11, 11, 11, 11, 11, 10, 10, 10, 10, 10]