Pythonic split list into n random chunks of roughly equal size As part of my implementation of cross-validation, I find myself needing to split a list into chunks of roughly equal size.

import random

def chunk(xs, n):
    ys = list(xs)
    random.shuffle(ys)
    ylen = len(ys)
    size = int(ylen / n)
    chunks = [ys[0+size*i : size*(i+1)] for i in xrange(n)]
    leftover = ylen - size*n
    edge = size*n
    for i in xrange(leftover):
            chunks[i%n].append(ys[edge+i])
    return chunks


This works as intended

&gt;&gt;&gt; chunk(range(10), 3)
[[4, 1, 2, 7], [5, 3, 6], [9, 8, 0]]


But it seems rather long and boring. Is there a library function that could perform this operation? Are there pythonic improvements that can be made to my code?
import random

def chunk(xs, n):
    ys = list(xs)


Copies of lists are usually taken using xs[:]

    random.shuffle(ys)
    ylen = len(ys)


I don't think storing the length in a variable actually helps your code much

    size = int(ylen / n)


Use size = ylen // n // is the integer division operator

    chunks = [ys[0+size*i : size*(i+1)] for i in xrange(n)]


Why the 0+? 

    leftover = ylen - size*n


Actually, you can find size and leftover using size, leftover = divmod(ylen, n)

    edge = size*n
    for i in xrange(leftover):
            chunks[i%n].append(ys[edge+i])


You can't have len(leftovers) &gt;= n. So you can do:

   for chunk, value in zip(chunks, leftover):
       chunk.append(value)


    return chunks


Some more improvement could be had if you used numpy. If this is part of a number crunching code you should look into it.