Question

所以如果我有数字[1,2,2,3]并且我想要k = 2个分区我会[1] [2,2,3]，[1,2] [2,3] ，[2,2] [1,3]，[2] [1,2,3]，[3] [1,2,2]等。

Answer 1

在Code Review处查看Python中的答案。

Answer 2

user3569在Code Review的解决方案为下面的测试用例生成五个2元组，而不是仅仅为3元组。但是，删除对返回的元组的frozenset()调用会导致代码仅返回3元组。修订后的代码如下：

from itertools import chain, combinations

def subsets(arr):
    """ Note this only returns non empty subsets of arr"""
    return chain(*[combinations(arr,i + 1) for i,a in enumerate(arr)])

def k_subset(arr, k):
    s_arr = sorted(arr)
    return set([i for i in combinations(subsets(arr),k) 
               if sorted(chain(*i)) == s_arr])

s = k_subset([2,2,2,2,3,3,5],3)

for ss in sorted(s):
    print(len(ss)," - ",ss)

用户3569说“它运行得很慢，但相当简洁”。

（编辑：见下面的Knuth解决方案）

输出结果为：

3  -  ((2,), (2,), (2, 2, 3, 3, 5))
3  -  ((2,), (2, 2), (2, 3, 3, 5))
3  -  ((2,), (2, 2, 2), (3, 3, 5))
3  -  ((2,), (2, 2, 3), (2, 3, 5))
3  -  ((2,), (2, 2, 5), (2, 3, 3))
3  -  ((2,), (2, 3), (2, 2, 3, 5))
3  -  ((2,), (2, 3, 3), (2, 2, 5))
3  -  ((2,), (2, 3, 5), (2, 2, 3))
3  -  ((2,), (2, 5), (2, 2, 3, 3))
3  -  ((2,), (3,), (2, 2, 2, 3, 5))
3  -  ((2,), (3, 3), (2, 2, 2, 5))
3  -  ((2,), (3, 5), (2, 2, 2, 3))
3  -  ((2,), (5,), (2, 2, 2, 3, 3))
3  -  ((2, 2), (2, 2), (3, 3, 5))
3  -  ((2, 2), (2, 3), (2, 3, 5))
3  -  ((2, 2), (2, 5), (2, 3, 3))
3  -  ((2, 2), (3, 3), (2, 2, 5))
3  -  ((2, 2), (3, 5), (2, 2, 3))
3  -  ((2, 3), (2, 2), (2, 3, 5))
3  -  ((2, 3), (2, 3), (2, 2, 5))
3  -  ((2, 3), (2, 5), (2, 2, 3))
3  -  ((2, 3), (3, 5), (2, 2, 2))
3  -  ((2, 5), (2, 2), (2, 3, 3))
3  -  ((2, 5), (2, 3), (2, 2, 3))
3  -  ((2, 5), (3, 3), (2, 2, 2))
3  -  ((3,), (2, 2), (2, 2, 3, 5))
3  -  ((3,), (2, 2, 2), (2, 3, 5))
3  -  ((3,), (2, 2, 3), (2, 2, 5))
3  -  ((3,), (2, 2, 5), (2, 2, 3))
3  -  ((3,), (2, 3), (2, 2, 2, 5))
3  -  ((3,), (2, 3, 5), (2, 2, 2))
3  -  ((3,), (2, 5), (2, 2, 2, 3))
3  -  ((3,), (3,), (2, 2, 2, 2, 5))
3  -  ((3,), (3, 5), (2, 2, 2, 2))
3  -  ((3,), (5,), (2, 2, 2, 2, 3))
3  -  ((5,), (2, 2), (2, 2, 3, 3))
3  -  ((5,), (2, 2, 2), (2, 3, 3))
3  -  ((5,), (2, 2, 3), (2, 2, 3))
3  -  ((5,), (2, 3), (2, 2, 2, 3))
3  -  ((5,), (2, 3, 3), (2, 2, 2))
3  -  ((5,), (3, 3), (2, 2, 2, 2))

如果不需要重复，可以如下调用由Adeel Zafar Soomro在同一Code Review页面上实现的Knuth解决方案：

s = algorithm_u([2,2,2,2,3,3,5],3)
ss = set(tuple(sorted(tuple(tuple(y) for y in x) for x in s)))

我还没有计时，但Knuth的解决方案明显更快，即使对于这个测试案例也是如此。

然而，它返回63个元组而不是user3569的解决方案返回的41个元组。我还没有仔细检查输出，以确定哪个输出是正确的。

Answer 3

这是Haskell中的一个版本：

import Data.List (nub, sort, permutations)

parts 0 = []
parts n = nub $ map sort $ [n] : [x:xs | x <- [1..n`div`2], xs <- parts(n - x)]

partition [] ys result = sort $ map sort result
partition (x:xs) ys result = 
  partition xs (drop x ys) (result ++ [take x ys])

partitions xs k = 
  let variations = filter (\x -> length x == k) $ parts (length xs)
  in nub $ concat $ map (\x -> mapVariation x (nub $ permutations xs)) variations
    where mapVariation variation = map (\x -> partition variation x [])


OUTPUT:
*Main> partitions [1,2,2,3] 2
[[[1],[2,2,3]],[[1,2,3],[2]],[[1,2,2],[3]],[[1,2],[2,3]],[[1,3],[2,2]]]

Answer 4

Python解决方案：

pip install PartitionSets

然后：

import partitionsets.partition
filter(lambda x: len(x) == k, partitionsets.partition.Partition(arr))

PartitionSets实现似乎非常快，但遗憾的是，您无法将多个分区作为参数传递，因此您需要从所有子集分区中过滤k-set分区。

您可能还想看一下： similar topic on researchgate

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