我有一组文档,我试图根据他们的词汇表进行聚类(即首先使用DocumentTermMatrix
命令创建一个语料库然后是一个稀疏矩阵,依此类推)。为了改进群集并更好地理解哪些特征/单词使特定文档落入特定群集,我想知道每个群集的最显着特征是什么。
Lantz的 Machine Learning with R 一书中有一个例子,如果你碰巧知道它 - 他根据他们所盯住的兴趣聚集青少年社交媒体档案,并最终得到像这样的表格显示“每个集群......具有最能将其与其他集群区分开来的特征”:
cluster 1 | cluster 2 | cluster 3 ....
swimming | band | sports ...
dance | music | kissed ....
现在,我的功能并不是那么有用,但是我仍然希望能够构建类似的东西。
但是,这本书没有解释表格是如何构建的。我已经尽力去创造性地谷歌了,也许答案是对集群意味着一些明显的计算,但作为R的新手以及统计数据,我无法弄明白。非常感谢任何帮助,包括我之前可能错过的先前问题或其他资源的链接!
感谢。
答案 0 :(得分:2)
我前段时间遇到过类似的问题..
这是我做的:
require("tm")
require("skmeans")
require("slam")
# clus: a skmeans object
# dtm: a Document Term Matrix
# first: eg. 10 most frequent words per cluster
# unique: if FALSE all words of the DTM will be used
# if TRUE only cluster specific words will be used
# result: List with words and frequency of words
# If unique = TRUE, only cluster specific words will be considered.
# Words which occur in more than one cluster will be ignored.
mfrq_words_per_cluster <- function(clus, dtm, first = 10, unique = TRUE){
if(!any(class(clus) == "skmeans")) return("clus must be an skmeans object")
dtm <- as.simple_triplet_matrix(dtm)
indM <- table(names(clus$cluster), clus$cluster) == 1 # generate bool matrix
hfun <- function(ind, dtm){ # help function, summing up words
if(is.null(dtm[ind, ])) dtm[ind, ] else col_sums(dtm[ind, ])
}
frqM <- apply(indM, 2, hfun, dtm = dtm)
if(unique){
# eliminate word which occur in several clusters
frqM <- frqM[rowSums(frqM > 0) == 1, ]
}
# export to list, order and take first x elements
res <- lapply(1:ncol(frqM), function(i, mat, first)
head(sort(mat[, i], decreasing = TRUE), first),
mat = frqM, first = first)
names(res) <- paste0("CLUSTER_", 1:ncol(frqM))
return(res)
}
一个小例子:
data("crude")
dtm <- DocumentTermMatrix(crude, control =
list(removePunctuation = TRUE,
removeNumbers = TRUE,
stopwords = TRUE))
rownames(dtm) <- paste0("Doc_", 1:20)
clus <- skmeans(dtm, 3)
mfrq_words_per_cluster(clus, dtm)
mfrq_words_per_cluster(clus, dtm, unique = FALSE)
HTH