CODEWITHSUNDEEP
rcluster-analysis
Shreyes
Shreyes

Reputation: 3751

How to obtain cluster label in HierarchicalSparseCluster() in R

While performing sparse Hierarchical clustering in R using sparcl package in R, I could not obtain the cluster labeling for the data. In the help documentation, they have the following code:

# Generate 2-class data
set.seed(1)
x <- matrix(rnorm(100*50),ncol=50)
y <- c(rep(1,50),rep(2,50))
x[y==1,1:25] <- x[y==1,1:25]+2

# Do tuning parameter selection for sparse hierarchical clustering
perm.out <- HierarchicalSparseCluster.permute(x, wbounds=c(1.5,2:6), nperms = 5)

# Perform sparse hierarchical clustering
sparsehc <- HierarchicalSparseCluster(dists=perm.out$dists,
wbound=perm.out$bestw, method="complete")

Now, how do I get the cluster label from object sparsehc is my question?

For Kmeans, we have a simple attribute "cs" created. Eg. 

## Choosing tuning parameters
km.perm <- KMeansSparseCluster.permute(data_mat, K = 10, wbounds= seq(3,7, len =
20),     nperms=5)

## Performing kmean sparce clustring 
sparse_data_clus <- KMeansSparseCluster(data_mat, K = 10, wbounds= km.perm$bestw)
clusterlabel <- sparse_data_clus[[1]]$Cs

How can I get a similar label in HierarchicalSparseCluster()?

Thanks!

Upvotes: 0

Views: 518

Answers (2)

a o
a o

Reputation: 1

It's a little late to respond to this, but I had the same problem. Here's what worked for me:

set.seed(1)
x <- matrix(rnorm(100*50),ncol=50)
y <- c(rep(1,50),rep(2,50))
x[y==1,1:25] <- x[y==1,1:25]+2
data_mat <- x

Do the permutation on the matrix you created

hier.perm <- HierarchicalSparseCluster.permute(data_mat, 
                                               wbounds= seq(3,7, len = 20),
                                               nperms=5)

Run HierarchicalSparse on the results

hier.sparse <- HierarchicalSparseCluster(dists=hier.perm$dists, 
                                         wbound=hier.perm$bestw,
                                         method='complete')

Run hclust on the value $u from previous line, then use cutree to divvy it up as you'd like

cluster = hclust(dist(hier.sparse$u))    
cutree(cluster,3)

Upvotes: 0

Has QUIT--Anony-Mousse
Has QUIT--Anony-Mousse

Reputation: 77454

Hierarchical clustering will usually return a dendrogram (i.e. a hierarchy of clusters, with single elements at the bottom and the complete data set at the top), not a strict partitioning.

If you want a strict partitioning (such as produced by regular k-means), you will have to extract such a partitioning from this hierarchy. There are many methods available to do so, the simplest is to use a threshold level.

As I don't use R a lot (too slow), I cannot give you details here though. Have a look at ?cutree.

Upvotes: 1

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