将所有分区迭代为k个组？

这可行，尽管它可能超级无效（我将它们全部排序以避免重复计算）：

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)]

它还返回空簇，因此您可能希望将其包装起来以便仅获取非空簇：

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

计数只是为了检查：

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 resdef 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

有用 ！

输出：

>>> 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]]]