From: milind.a.joshi@gmail.com   
      
   MrElbi wrote:   
   > mathlover wrote:   
   > > Ted Dunning wrote:   
   > > > mathlover wrote:   
   > > > > All the usual clustering methods I am aware of (such as k-means, Fuzzy   
   > > > > k-means, etc.) find all the clusters together. I mean, as you know,   
   > > > > they consider some cost function minimizing which (by an iterative   
   > > > > method for instance) finds all the clusters.   
   > > >   
   > > > As a trivial answer, you can just run any of the clustering methods   
   > > > that you are familiar with with a progressively larger number of   
   > > > clusters using the old clusters as approximate seeds for the next set.   
   > > > This will have some strange properties but will give approximately the   
   > > > behavior you seek.  Since clustering is generally O(m n), this is   
   > > > reasonably efficient.   
   > > >   
   > > > More interestingly, there are heirarchical clustering algorithms that   
   > > > divide the data set into progressively finer divisions.  Look for   
   > > > dendograms, for instance.   
   > > >   
   > > > This not really what you are asking for, however, since the neither of   
   > > > these really find a single cluster in the sense you are asking.   
   > > >   
   > > > If you know something about the distribution of the data that you are   
   > > > looking at, you might be able to define something better.  For   
   > > > instance, you can define a cost function such that it must describe a   
   > > > membership rule for a subset of your data and the distribution of   
   > > > instances in that subset.  It would be best fi the membership function   
   > > > is based on a distance formulation.  You would need to add cost for   
   > > > poor description of the distribution of the data in the subset and for   
   > > > the number of data instances outside the subset (or, inversely, use the   
   > > > average log of the probability of the members of the subset  plus the   
   > > > size of the subset).  There is probably an interesting mixture model   
   > > > that could be used for this, but the key thing is that can't penalize   
   > > > having a small subset too heavily, nor should you penalize a model for   
   > > > an inability to describe the distribution of the  excluded instances.   
   > > > You can then progressively add subsets.   
   > > >   
   > > > This is much closer to what you are looking for, but I have no idea if   
   > > > it will actually work well on your data.   
   > > >   
   > >   
   > > Dear Ted Dunning,   
   > >   
   > > I think I might have badly phrased my question. Indeed, in the problem   
   > > I am working on simple k-means clustering attains satisfying quality.   
   > > Of course, I know the total number of clusters (so I can use the   
   > > k-means clustering).   
   > >   
   > > However, because of the very large size of the problem it takes a lot   
   > > of time to find all the clusters (I mean using k-means). But, the point   
   > > is that I don't need all the clusters the k-means finds in the later   
   > > stages of my problem. Actually, I even know exactly how many of them I   
   > > need.   
   > >   
   > > Thus, I am looking for a method that gives in its output clusters near   
   > > those k-means gives, but it can find them one at a time (not like   
   > > k-means that finds all of them together) so that I can quit after   
   > > finding the exact number of clusters I need.   
   > >   
   > > Best regards,   
   > > mathlover.   
   > >   
   >   
   > Hi,   
   >   
   > You should search informations about the BFR algorithm. It is a   
   > partitional algorithm which is used to do clustering on very huge   
   > datasets: the datas are treated by pages, which seems to be close to   
   > what you are looking for.   
   >   
   > Regards,   
   >   
      
   What about running the clustering algorithm on a small section of your   
   data set partitioned randomly into 'n' pieces, where 'n' is a size   
   chosen based on your resource constraints?   
      
   Regards,   
   Milind Joshi   
      
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