Most representative instance of a cluster

Question

After performing a cluster analysis to my dataset (a dataframe named data.matrix), I added a new column, named cluster, at the end (col 27) containing the cluster name that each instance belongs to.

What I want now, is a representative instance from each cluster. I tried to find the instance having the smallest euclidean distance from the cluster's centroid (and repeat the procedure for each one of my clusters)

This is what I did. Can you think of other -perhaps more elegant- ways? (assume numeric columns with no nulls).

clusters <- levels(data.matrix$cluster)
cluster_col = c(27)

for (j in 1:length(clusters)) {
    # get the subset for cluster j
    data = data.matrix[data.matrix$cluster == clusters[j],]

    # remove the cluster column
    data <- data[,-cluster_col]

    # calculate the centroid
    cent <- mean(data)

    # copy data to data.matrix_cl, attaching a distance column at the end
    data.matrix_cl <- cbind(data, dist = apply(data, 1, function(x) {sqrt(sum((x - cent)^2))}))

    # get instances with min distance
    candidates <- data.matrix_cl[data.matrix_cl$dist == min(data.matrix_cl$dist),]

    # print their rownames
    print(paste("Candidates for cluster ",j))
    print(rownames(candidates))
}

Answer 1

Is the technique you're interested in 'k-means clustering'? If so, here's how the centroids are calculated at each iteration:

1. choose a k value (an integer that specifies the number of clusters to divide your data set);

2. random select k rows from your data set, those are the centroids for the 1st iteration;

3. calculate the distance that each data point is from each centroid;

4. each data point has a 'closest centroid', that determines its 'group';

5. calculate the mean for each group--those are the new centroids;

6. back to step 3 (stopping criterion is usually based on comparison with the respective centroid values in successive loops, i.e., if they values change not more than 0.01%, then quit).

Those steps in code:

# toy data set
mx = matrix(runif60, 10, 99), nrow=12, ncol=5, byrow=F)
cndx = sample(nrow(mx), 2)
# the two centroids at iteration 1
cn1 = mx[cndx[1],]
cn2 = mx[cndx[2],]
# to calculate Pearson similarity
fnx1 = function(a){sqrt((cn1[1] - a[1])^2 + (cn1[2] - a[2])^2)}
fnx2 = function(a){sqrt((cn2[1] - a[1])^2 + (cn2[2] - a[2])^2)}
# calculate distance matrix
dx1 = apply(mx, 1, fnx1)
dx2 = apply(mx, 1, fnx2)
dx = matrix(c(dx1, dx2), nrow=2, ncol=12)
# index for extracting the new groups from the data set
ndx = apply(dx, 1, which.min)
group1 = mx[ndx==1,]
group2 = mx[ndx==2,]
# calculate the new centroids for the next iteration
new_cnt1 = apply(group1, 2, mean)
new_cnt2 = apply(group2, 2, mean)

Answer 2

At first I don't now if you distance formula is alright. I think there should be sqrt(sum((x-cent)^2)) or sum(abs(x-cent)). I assumed first. Second thought is that just printing solution is not good idea. So I first compute, then print. Third - I recommend using plyr but I give both (with and without plyr) solutions.

# Simulated data:
n <- 100
data.matrix <- cbind(
  data.frame(matrix(runif(26*n), n, 26)),
  cluster=sample(letters[1:6], n, replace=TRUE)
)
cluster_col <- which(names(data.matrix)=="cluster")

# With plyr:
require(plyr)
candidates <- dlply(data.matrix, "cluster", function(data) {
  dists <- colSums(laply(data[, -cluster_col], function(x) (x-mean(x))^2))
  rownames(data)[dists==min(dists)]
})

l_ply(names(candidates), function(c_name, c_list=candidates[[c_name]]) {
    print(paste("Candidates for cluster ",c_name))
    print(c_list)
})

# without plyr
candidates <- tapply(
  1:nrow(data.matrix),
  data.matrix$cluster,
  function(id, data=data.matrix[id, ]) {
    dists <- rowSums(sapply(data[, -cluster_col], function(x) (x-mean(x))^2))
    rownames(data)[dists==min(dists)]
  }
)

invisible(lapply(names(candidates), function(c_name, c_list=candidates[[c_name]]) {
    print(paste("Candidates for cluster ",c_name))
    print(c_list)
}))