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I have a 1000 by 3 data matrix in which for each person we have 3 scores. I want to cluster those persons into 8 cluster using Agglomerative method. I did the clusetring by:

 class=FindClusters[data, 8, DistanceFunction -> ManhattanDistance, 
 Method -> {"Agglomerate", "Linkage" -> "Complete"}]

I want to find out how and which persons are classified into those 8 clusters.In other words, I want to have a vector of class membership for each person (similar to what we can get using ClusteringComponents[data,8,1]).

As an example, suppose we have this data set:

data={{3, 2, 4}, {1, 1, 0}, {5, 2, 4}, {2, 3, 2}, {4, 4, 4}, {3, 2, 4}, {3,
   2, 3}, {2, 1, 3}}

result of FindClusters is:

class = FindClusters[data, 3, DistanceFunction -> ManhattanDistance, 
  Method -> {"Agglomerate", "Linkage" -> "Complete"}]

{{{3, 2, 4}, {2, 3, 2}, {3, 2, 4}, {3, 2, 3}, {2, 1, 3}}, {{1, 1, 
   0}}, {{5, 2, 4}, {4, 4, 4}}}

through the output it can be seen that persons number 1,4,6,7&8 are classified in first cluster; person number 2 in second cluster and persons number 3&5 are in third cluster. I want to have a vector like:

{1,2,3,1,3,1,1,1}

then I can determine which person is located in which cluster and how many persons are in each cluster.

I decided to use FindClusters because I didn't get reasonable results from Agglomerate function for hierarchical clustering. My main aim is doing hierarchical clustering with different linkage functions.

Any help and additional ideas are appreciated.

Amin.

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    $\begingroup$ Could you include maybe a smaller sample of data, and the results you expect from a proper clustering? $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Jun 11, 2013 at 3:38
  • $\begingroup$ @0x4A4D♦ I did. $\endgroup$
    – Amin
    Commented Jun 11, 2013 at 3:56

1 Answer 1

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You can supply an optional label/argument to FindClusters. In this case, I've just labeled them with a list of integers

class = FindClusters[data -> Range[Length[data]], 3, 
  DistanceFunction -> ManhattanDistance, 
  Method -> {"Agglomerate", "Linkage" -> "Complete"}]

{{1, 4, 6, 7, 8}, {2}, {3, 5}}

which is the clustering labels you expected. Or just use 

 ClusteringComponents[data, 8, 1, DistanceFunction -> ManhattanDistance, 
      Method -> "Agglomerate"]

to get the alternative form 

{1, 2, 3, 4, 5, 1, 6, 7}

Note that FindCLusters and ClusteringComponents have pretty much the same options available.

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  • $\begingroup$ this is close to what I want but I need to have that final vector which includes cluster membership. I can obtain that vector by using ClusteringComponents[d, 3, 1] but not from FindClusters. $\endgroup$
    – Amin
    Commented Jun 11, 2013 at 4:06
  • $\begingroup$ What advantage is there to using FindClusters over ClusteringComponents? $\endgroup$
    – bill s
    Commented Jun 11, 2013 at 4:26
  • $\begingroup$ I want to use hierarchical cluster analysis with different linkage functions so I thought that FindClusters is more appropriate.Do you have any idea? $\endgroup$
    – Amin
    Commented Jun 11, 2013 at 4:30
  • $\begingroup$ The Agglomerate function in HierarchicalClustering package didn't work for me (which uses different linkage functions) then I moved to FindClusters $\endgroup$
    – Amin
    Commented Jun 11, 2013 at 4:36
  • $\begingroup$ I used your ClusteringComponents[data, 8, 1, DistanceFunction -> ManhattanDistance, Method -> "Agglomerate"] but it didn't work for me.It gave me some error messages. $\endgroup$
    – Amin
    Commented Jun 11, 2013 at 4:38

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