将所有分区迭代为k个组？ [英] Iterator over all partitions into k groups?

问题描述

说我有一个列表L。如何在K个组的所有分区上获得迭代器？

Say I have a list L. How can I get an iterator over all partitions of K groups?

示例：L = [2,3,5,7 ，11，13]，K = 3

Example: L = [ 2,3,5,7,11, 13], K = 3

3组所有可能分区的列表：

List of all possible partitions of 3 groups:

[ [ 2 ], [ 3, 5], [ 7,11,13] ]
[ [ 2,3,5 ], [ 7, 11], [ 13] ]
[ [ 3, 11 ], [ 5, 7], [ 2, 13] ]
[ [ 3 ], [ 11 ], [ 5, 7, 2, 13] ]
etc...

===更新===

我正在研究一个似乎可行的解决方案，所以我只复制粘贴它

I was working on a solution which seems to be working, so I will just copy paste it

# -*- coding: utf-8 -*-

import itertools 

# return ( list1 - list0 )
def l1_sub_l0( l1, l0 ) :
    """Substract two lists"""
    #
    copy_l1 = list( l1 )
    copy_l0 = list( l0 )

    #
    for xx in l0 :
        #
        if copy_l1.count( xx ) > 0 :
            #
            copy_l1.remove( xx )
            copy_l0.remove( xx )

    #
    return [ copy_l1, copy_l0 ]


#
def gen_group_len( n, k ) :
    """Generate all possible group sizes"""

    # avoid doubles
    stop_list = []
    #
    for t in itertools.combinations_with_replacement( xrange( 1, n - 1 ), k - 1 ) :
        #
        last_n = n - sum( t )

        # valid group size
        if last_n  >= 1 :
            res = tuple( sorted( t + ( last_n, ) ) )
            #
            if res not in stop_list :
                yield res
                stop_list.append( res )


# group_len = (1, 1, 3)

def gen( group_len, my_list ) :
    """Generate all possible partitions of all possible group sizes"""

    #
    if len( group_len ) == 1 :
        yield ( tuple( my_list ), )

    #
    else :

        # need for a stop list if 2 groups of same size
        stop_list = []

        #
        for t in itertools.combinations( my_list, group_len[ 0 ] ) :
            #
            reduced_list = l1_sub_l0( my_list, t )[ 0 ]

            #
            for t2 in gen( group_len[ 1: ], reduced_list ) :
                #
                tmp = set( ( t, t2[ 0 ] ) )
                #
                if tmp not in stop_list :
                    yield ( t, ) + t2
                    # avoid doing same thing twice
                    if group_len[ 1 ] == group_len[ 0 ] :
                        stop_list.append( tmp )


#
my_list = [ 3,5,7,11,13 ]
n = len( my_list )
k = 3

#
group_len_list = list( gen_group_len( n, k ) )
print "for %i elements, %i configurations of group sizes" % ( n, len(  group_len_list ) )
print group_len_list

#
for group_len in group_len_list :
    #
    print "group sizes", group_len
    #
    for x in gen( group_len, my_list ) :
        print x
    #
    print "==="

输出：

for 5 elements, 2 configurations of group sizes
[(1, 1, 3), (1, 2, 2)]
group sizes (1, 1, 3)
((3,), (5,), (7, 11, 13))
((3,), (7,), (5, 11, 13))
((3,), (11,), (5, 7, 13))
((3,), (13,), (5, 7, 11))
((5,), (7,), (3, 11, 13))
((5,), (11,), (3, 7, 13))
((5,), (13,), (3, 7, 11))
((7,), (11,), (3, 5, 13))
((7,), (13,), (3, 5, 11))
((11,), (13,), (3, 5, 7))
===
group sizes (1, 2, 2)
((3,), (5, 7), (11, 13))
((3,), (5, 11), (7, 13))
((3,), (5, 13), (7, 11))
((5,), (3, 7), (11, 13))
((5,), (3, 11), (7, 13))
((5,), (3, 13), (7, 11))
((7,), (3, 5), (11, 13))
((7,), (3, 11), (5, 13))
((7,), (3, 13), (5, 11))
((11,), (3, 5), (7, 13))
((11,), (3, 7), (5, 13))
((11,), (3, 13), (5, 7))
((13,), (3, 5), (7, 11))
((13,), (3, 7), (5, 11))
((13,), (3, 11), (5, 7))
===

推荐答案

这很有用，尽管它可能超级无效（我将它们全部排序以避免重复计算） ）：

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)]

它还会返回空簇，因此您可能想要将其包装以仅获取非空值：

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

仅用于检查：

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 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

它有效！

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]]]

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