Iterator over all partitions into k groups?

This works, although it is probably super inneficient (I sort them all to avoid double-counting):

def clusters(l, K):
    if l:
        prev = None
        for t in clusters(l[1:], K):
            tup = sorted(t)
            if tup != prev:
                prev = tup
                for i in xrange(K):
                    yield tup[:i] + [[l[0]] + tup[i],] + tup[i+1:]
    else:
        yield [[] for _ in xrange(K)]

It also returns empty clusters, so you would probably want to wrap this in order to get only the non-empty ones:

def neclusters(l, K):
    for c in clusters(l, K):
        if all(x for x in c): yield c

Counting just to check:

def kamongn(n, k):
    res = 1
    for x in xrange(n-k, n):
        res *= x + 1
    for x in xrange(k):
        res /= x + 1
    return res

def Stirling(n, k):
    res = 0
    for j in xrange(k + 1):
        res += (-1)**(k-j) * kamongn(k, j) * j ** n
    for x in xrange(k):
        res /= x + 1
    return res

>>> sum(1 for _ in neclusters([2,3,5,7,11,13], K=3)) == Stirling(len([2,3,5,7,11,13]), k=3)
True

It works !

The output:

>>> clust = neclusters([2,3,5,7,11,13], K=3)
>>> [clust.next() for _ in xrange(5)]
[[[2, 3, 5, 7], [11], [13]], [[3, 5, 7], [2, 11], [13]], [[3, 5, 7], [11], [2, 13]], [[2, 3, 11], [5, 7], [13]], [[3, 11], [2, 5, 7], [13]]]

A simple alternative view of this problem is the assignment of one of the three cluster labels to each element.

import itertools
def neclusters(l, k):
    for labels in itertools.product(range(k), repeat=len(l)):
        partition = [[] for i in range(k)]
        for i, label in enumerate(labels):
            partition[label].append(l[i])
        yield partition

as with @val's answer, this can be wrapped to remove partitions with empty clusters.