最具代表性的集群实例

Question

在对我的数据集(名为data.matrix的数据帧)执行聚类分析之后,我在末尾添加了一个名为cluster的新列(第27列),其中包含每个实例所属的群集名称.

我现在想要的是来自每个集群的代表性实例.我试图从群集的质心中找到与欧氏距离最小的实例(并为我的每个群集重复该过程)

这就是我做的.你能想到其他 - 或许更优雅的方式吗？(假设没有空值的数字列).

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

如果距离公式好的话,我现在不会.我认为应该有sqrt(sum((x-cent)^2))或sum(abs(x-cent)).我先假设.第二个想法是,只是打印解决方案并不是一个好主意.所以我先计算,然后打印.第三 - 我建议使用plyr,但我同时给予(有和没有plyr)解决方案.

# 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)
}))