9
$\begingroup$

For example having data

data = {16, 4, 17, 10, 15, 4, 4, 6, 7, 14, 9, 17, 27, 6, 1, 9, 0, 12, 20, 8, 0, 3, 4, 0, 3, 4}

I want to cluster them int 4 partitions

FindClusters[data, 4]

So that each partition is as close to 55 = (220/4) as possible.

The sequence order should not be changed.


I was also considering a knapsack based solution:

GetSample[n_,mydata_] := #[[Flatten@
     Position[LinearProgramming[-#, -{#}, -{Total[data]/n},
       {0, 1} & /@ #, Integers], 1], 1]] &@ mydata
GetSample[4, data]

after what i create a copy from where i remove sample elements and repeat n times.

$\endgroup$
  • $\begingroup$ @PlatoManiac I imagine, that each partition would have their elements in the same order as they appear in the full list and, perhaps, the partitions themselves would be ordered in such a way, that their first elements are also in the same order. But Mean[data] returns 110/13, not 55. I guess, that's just the wording of the title slightly failing. $\endgroup$ – LLlAMnYP Jul 3 '15 at 12:19
  • $\begingroup$ Ya that might be the case @LLlAMnYP $\endgroup$ – PlatoManiac Jul 3 '15 at 12:21
  • $\begingroup$ I take it to mean that the first cluster can only be $\sum_i^n x_{i}$ $\endgroup$ – Feyre Jul 3 '15 at 12:22
  • $\begingroup$ wolframalpha.com/input/… $\endgroup$ – Margus Jul 3 '15 at 12:22
  • $\begingroup$ Should the total deviation Total(Abs[x1-55],Abs[x2-55],Abs[x3-55],Abs[x4-55]) be as small as possible, or what metric do you want to use? $\endgroup$ – Feyre Jul 3 '15 at 12:35
5
$\begingroup$

For lists of the size in the example, brute-forcing s/b fine:

Module[{d = #1, m = #2, l = Length@#1, dt = (Tr@#1)/#2, dtp, parted},
   parted = 
    Internal`PartitionRagged[d, #] & /@ 
     Join @@ Permutations /@ IntegerPartitions[l, {m}];
   dtp = Total[Abs[Total[parted, {3}] - dt], {2}];
   Pick[parted, dtp, Min@dtp]] &[data, 4]

(*
{{{16, 4, 17, 10}, {15, 4, 4, 6, 7, 14, 9}, {17, 27, 6, 1, 9}, {0, 12, 20, 8, 0, 3, 4, 0, 3, 4}}, 
{{16, 4, 17, 10}, {15, 4, 4, 6, 7, 14, 9}, {17, 27, 6, 1, 9, 0}, {12, 20, 8, 0, 3, 4, 0, 3, 4}}}
*)

I chose minimization of sum of absolute differences from goal as the metric, you can change it (dtp=...) to whatever floats your boat.

$\endgroup$
2
$\begingroup$

Here is a simple approach that splits the data into 4 sequences, with each sequence being as close to 55 as possible

drop[list_, m_] := 
 Drop[list, (Position[#, Sequence @@ Nearest[#, m]] &@ Accumulate@list)[[1, 1]]]
take[list_, m_] := 
 Take[list, (Position[#, Sequence @@ Nearest[#, m]] &@ Accumulate@list)[[1, 1]]]

NestList[{take[#[[2]], 55], drop[#[[2]], 55]} &, {{}, data}, 4][[2 ;;, 1]]
{{16, 4, 17, 10, 15}, {4, 4, 6, 7, 14, 9, 17}, {27, 6, 1, 9, 0, 12}, 
 {20, 8, 0, 3, 4, 0, 3, 4}}
Total /@ %

{62, 61, 55, 42}

One could use

pos[list_, m_] := (Position[#, Sequence @@ Nearest[#, m]] &@ Accumulate@list)[[1, 1]]

NestList[{pos[#[[2]], 55], drop[#[[2]], 55]} &, {{}, data}, 4][[2 ;;, 1]]

{5, 7, 6, 8}

to get some starting values, that could be used for some code that splits the data into sequences using a global measure for the difference to 55.

$\endgroup$
  • $\begingroup$ Minimizing the maximum difference of each sublist sum to 55 results in the following split positions: {{4, 6, 5, 11}, {4, 7, 4, 11}, {4, 7, 5, 10}, {4, 7, 6, 9}}. All resulting in having 8 as the maximum absolute difference. $\endgroup$ – Karsten 7. Jul 3 '15 at 19:20
2
$\begingroup$

Here's an overengineered approach using the method of simulated annealing. I apologize for the poor style of coding, this is something I had lying around found somewhere online and modified just now for this task.

data = {16, 4, 17, 10, 15, 4, 4, 6, 7, 14, 9, 17, 27, 6, 1, 9, 0, 12, 20, 8, 0, 3, 4, 0, 3, 4};

Solution = Partition[Range@26, 7, 7, 1, {}];
Optimum = Solution;

Fitness[list_] := Norm@(Total /@ (data[[#]] & /@ list) - 55);

Iterate[Sol_, temp_] :=
 Module[{nSol = Sol, fromto = RandomSample[Range@4, 2], index, val},
  While[Length@nSol[[First@fromto]] == 0, 
   fromto = RandomSample[Range@4, 2]];
  index = RandomInteger[{1, Length@nSol[[First@fromto]]}];
  val = nSol[[First@fromto, index]];
  nSol = Delete[nSol, {First@fromto, index}];
  nSol = Insert[nSol, val, {Last@fromto, 1}];

  If[Fitness[nSol] < Fitness[Optimum], Optimum = nSol];

  Solution = 
   RandomChoice[{Exp[-Fitness[nSol]/temp], 
      Exp[-Fitness[Sol]/temp]} -> {nSol, Sol}]]

ListPlot@(tempSched = (.9 + Tanh[(10000 - #)/15000]) & /@ 
    Range[30000])

ListPlot@(Fitness /@ FoldList[Iterate, Solution, tempSched])

The current solution is a partition of the list of indices. With each step the program randomly takes a number from one partition and sticks it into another. It takes then a random choice between the newly proposed solution and the old one based on the current "temperature" and the fitness of either solution (how close they are to summing to 55). Also, if it finds a better solution than any previous one it saves it for future use.

After running the above, let's check the results.

Solution
data[[#]] & /@ Solution
Total /@ %
{{21, 24, 17, 7, 2, 16, 4, 18, 5}, {14, 9, 12, 13}, {15, 22, 6, 26, 8,
   23, 11, 20, 1}, {25, 10, 3, 19}}
{{0, 0, 0, 4, 4, 9, 10, 12, 15}, {6, 7, 17, 27}, {1, 3, 4, 4, 6, 4, 9,
   8, 16}, {3, 14, 17, 20}}
{54, 57, 55, 54}

Quite good, but we can do better.

Optimum
data[[#]] & /@ Optimum
Total /@ %
{{17, 8, 14, 9, 11, 18, 5}, {25, 20, 12, 13}, {15, 21, 4, 26, 6, 24, 
  7, 22, 23, 16, 1}, {2, 10, 3, 19}}
{{0, 6, 6, 7, 9, 12, 15}, {3, 8, 17, 27}, {1, 0, 10, 4, 4, 0, 4, 3, 4,
   9, 16}, {4, 14, 17, 20}}
{55, 55, 55, 55}

Exactly as desired. Finally OP requests to maintain the order of the data, so

data[[#]] & /@ SortBy[Sort /@ Optimum, First]
{{16, 10, 4, 4, 1, 9, 0, 3, 4, 0, 4}, {4, 17, 14, 20}, {15, 6, 7, 9,  6, 0, 12}, {17, 27, 8, 3}}
$\endgroup$
1
$\begingroup$

Since the size of the list is relatively small, you could use

clusters = ReplaceList[data, {w__, x__, y__, z__} -> {{w}, {x}, {y}, {z}}]

to split the original list data into 4 sublists while preserving the order. There are about 2300 clusters. Then sort these clusters of four sublists by the difference of their sums from 55.

SortBy[clusters, Total[Abs[Total[#, {2}] - 55]] &]

The first cluster in the sorted list has totals of {47,59,60,54}. This cluster is

(* {{16, 4, 17, 10},
    {15, 4, 4, 6, 7, 14, 9},
    {17, 27, 6, 1, 9},
    {0, 12, 20, 8, 0, 3, 4, 0, 3, 4}}  *)
$\endgroup$
0
$\begingroup$
data = {16, 4, 17, 10, 15, 4, 4, 6, 7, 14, 9, 17, 27, 6, 1, 9, 0, 12, 
   20, 8, 0, 3, 4, 0, 3, 4};
data1 = {};
i = 1;
While[i < Length[data] && 
  Total[data1] < 55, {data1 = 
   Append[data1, x = Total[{Nearest[data, 55 - Total[data1]][[1]]}]], 
  data = Delete[data, FirstPosition[data, x]]}; i++]
data2 = {};
i = 1;
While[i < Length[data] && 
  Total[data2] < 55, {data2 = 
   Append[data2, x = Total[{Nearest[data, 55 - Total[data2]][[1]]}]], 
  data = Delete[data, FirstPosition[data, x]]}; i++]
data3 = {};
i = 1;
While[i < Length[data] && 
  Total[data3] < 55, {data3 = 
   Append[data3, x = Total[{Nearest[data, 55 - Total[data3]][[1]]}]], 
  data = Delete[data, FirstPosition[data, x]]}; i++]
enddata={data1,data2,data3,data}

This yields exactly four times 55. In:

Total[data1]
Total[data2]
Total[data3]
Total[data]

Out: 55 55 55 55

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.