similar to: kmeans cluster stability

Regarding a previous question concerning the kmeans function I've tried the same example and I also get a strange result (at least according to what is said in the help of the function kmeans). Apparently, the function is disregarding the initial cluster centers one gives it. According to the help of the function: centers: Either the number of clusters or a set of initial cluster

Hello All, I'm having some trouble figuring out what the clearest way to plot my k-means clustering result on an my existing MDS. First I performed MDS on my distance matrix (note: I performed k-means on the MDS coordinates because applying a euclidean distance measure to my raw data would have been inappropriate) canto.MDS<-cmdscale(canto) I then figured out what would be my optimum

Hi, I am using kmeans to cluster a dataset. I test this example: > data<-matrix(scan("data100.txt"),100,37,byrow=T) (my dataset is 100 rows and 37 columns--clustering rows) > c1<-kmeans(data,3,20) > c1 $cluster [1] 1 1 1 1 1 1 1 3 3 3 1 3 1 3 3 1 1 1 1 3 1 3 3 1 1 1 3 3 1 1 3 1 1 1 1 3 3 [38] 3 1 1 1 3 1 1 1 1 3 3 3 1 1 1 1 1 1 3 1 3 1 1 3 1 1 1 1 3 1 1 1 1 1 1 3

Hi Team, I am trying to run kmeans in R, and I need to save the different clusters into different folders. How can I achieve this? # this is how my data looks. $ *cat 1.tsv | head* userid bookid rating bookTotalRatings bookAvgRating userTotalRatings userAvgRating 1 100 0 24 2.7916666666666665 291 2.6735395189003435 2 200 7 24 2.9583333333333335 6 7.0

Dear helpers I'm sorry to insist but I still think there is something wrong with the function kmeans. For instance, let's try the same small example: > dados<-matrix(c(-1,0,2,2.5,7,9,0,3,0,6,1,4),6,2) I will choose observations 3 and 4 for initial centers and just one iteration. The results are > A<-kmeans(dados,dados[c(3,4),],1) > A $cluster [1] 1 1 1 1 2 2 $centers

Hi all, I am trying to run the k-means cluster analysis using the function kmeans in the package cluster. The data are: x = c(-0.26, -0.23, -0.05, -0.20, 0.30, -0.84, -0.10, -0.12, 0.10, -0.31, -0.19, 0.18, -0.26, -0.23, -0.37, -0.23) I've got two different solutions when I ran this function over a few times: kmeans(x, centers=2) The first solution gives the following: $cluster [1]

Dear members. I am having problems to understand the kmeans- results in R. I am applying kmeans-algorithms to my big data file, and it is producing the results of the clusters. Q1) Does anybody knows how to find out in which cluster (I have fixed numberofclusters = 5 ) which data have been used? COMMAND (kmeans.results <- kmeans(mydata,centers =5, iter.max= 1000, nstart =10000)) Q2) When I

Sorry, to solve your question I had tried: data(faithful) kmeans(faithful[c(1:20),1],10) Error: empty cluster: try a better set of initial centers But when I run this a second time it will be ok. It seems, that kmeans has problems to initialize good starting points, because of the random choose of these starting initial points. With kmeans(data,k,centers=c(...) the problem can be solved.

Dear helpers I was working with kmeans from package mva and found some strange situations. When I run several times the kmeans algorithm with the same dataset I get the same partition. I simulated a little example with 6 observations and run kmeans giving the centers and making just one iteration. I expected that the algorithm just allocated the observations to the nearest center but think this

Hello, I am running cluster analysis, and am attempting to create a graph of my clusters. I keep on getting an error that says that my figure margins are too large. d <- file.choose() d <- read.csv(d,header=TRUE) mydataS <- scale(d, center = TRUE, scale=TRUE) #Converts mydataS from a matrix to a data frame mydataS2 <- as.data.frame(mydataS) #removes "coden"

Hello, how is, for example, the Schwarz criterion is defined for kmeans? It should be something like: k <- 2 vars <- 4 nobs <- 100 dat <- rbind(matrix(rnorm(nobs, sd = 0.3), ncol = vars), matrix(rnorm(nobs, mean = 1, sd = 0.3), ncol = vars)) colnames(dat) <- paste("var",1:4) (cl <- kmeans(dat, k)) schwarz <- sum(cl$withinss)+ vars*k*log(nobs) Thanks

I am trying to generate kmean of 10 clusters for a 165 x 165 matrix. i do not see any errors known to me. But I get this error on running the script Error: empty cluster: try a better set of initial centers the commands are M <-matrix(scan("R_mutual",n = 165 * 165),165,165,byrow = T) cl <- kmeans(M,centers=10,20) len = dim(M)[1] .... .... I ran the same script last night and

Hi All, I was using the following command for performing kmeans for Iris dataset. Kmeans_model<-kmeans(dataFrame[,c(1,2,3,4)],centers=3) This was giving proper results for me. But, in my application we generate the R commands dynamically and there was a requirement that the column names will be sent instead of column indices to the R commands.Hence, to incorporate this, i tried using the R

Dear R users, When I run the following code, R crashes: require(cclust) x <- matrix(c(0,0,0,1.5,1,-1), ncol=2, byrow=TRUE) cclust(x, centers=x[2:3,], dist="manhattan", method="kmeans") While this works: cclust(x, centers=x[2:3,], dist="euclidean", method="kmeans") I'm posting this here because I am not sure if it is a bug. I've been searching

Hello, I have been getting the following intermittent error from kmeans: >str(cavint.p.r) num [1:1967, 1:13] 0.691 0.123 0.388 0.268 0.485 ... - attr(*, "dimnames")=List of 2 ..$ : chr [1:1967] "6" "49" "87" "102" ... ..$ : chr [1:13] "HYD" "NEG" "POS" "OXY" ... > set.seed(34) >

I am using R 1.3.0 on Windows 2000. For an experiment, I am wanting to find the most diverse 400 items to study in a possible 3200 items. Diversity here is based on a few hundred attributes. For this, I would like to do a clustering analysis and find 400 clusters (i.e. different from each other in some way hopefully). From each of these 400 clusters, I will pick a representative. I expect

Hi. I am trying to modify kmeans function. It seems that is failing something obvious with the workspace. I am a newbie and here is my code: myk = function (x, centers, iter.max = 10, nstart = 1, algorithm = c("Hartigan-Wong", + "Lloyd", "Forgy", "MacQueen")) + { + do_one <- function(nmeth) { + Z <- switch(nmeth, { + Z

Buen día, no sé si estoy utilizando bien la lista, es la primera vez. Si lo hago mal me corrigen por favor. Sobre tu comentario Pedro, muchas gracias. Lo qeu entiendo con tu sugerencia de set.seed es qeu de esa forma fijas los resultados, pero no estoy seguro si otra agrupación funcione mejor. Es decir me interesa un método de agrupación que genere la "mejor" agrupación y como los

Sir, Would like to know what sort of input matrix are taken by the kmeans function defined in mva library of R application. As per the documentation for the Kmeans it takes the following 2 data sets: 1) data 2) centers The commands to be executed in R are as follows: library(mva) data <- read.table('file1',header=TRUE,sep="\t") centers <-

Hi All, For the function "bclust"(e1071), the argument "base.method" is explained as "must be the name of a partitioning cluster function returning a list with the same components as the return value of 'kmeans'. In my understanding, there are three partitioning cluster functions in R, which are "clara, pam, fanny". Then I check each of them to