similar to: Plotting K-means clustering results on an MDS

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, 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"

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]

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

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

I'm doing kmeans partitioning on a small (n=26) dataset that has 5 variables. I noticed that if I repeatedly run the same command, the cluster centers change and the cluster membership changes. Using RW1022 under Windows NT & Windows 2000 >kmeans(pottery[,1:5], 4, 20) [...snip] $size [1] 7 3 9 7 [...snip] $size [1] 7 10 4 5 [...snip] $size [1] 6 10 5 5 yields a different

H i, all! First of all, I'd like to apologize for my poor English. It's for years I don't use it. This is a R-version of a function I wrote a long ago for my HP48 calculator. It works with the binary expression of the power and just need to duplicate the mem used by X. Hope this helps. mtx.exp<-function(X,n) #Function to calculate the n-th power of a matrix X; { phi <-

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

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

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

Dear all, I wrote a simple script in order to put together some functions and method to be executed on various files I am trying to have to possibility to call the script changing few parameters in order to use the different files. I succeeded partly using the function method. However in my script I call the sink() function in order to output the results to a .txt file. When using the function the

Hi, in the past I set up a master/master replication through TCP using dovecot 2.2.10 on both servers and everything is running fine, I filter the users to replicate using the iterate_query on file dovecot-sql.con.ext; now I want to set up a new replication between two new servers running dovecot 2.2.32 but the replicator service tries to replicate every user on my database even when the

doveconf -n # 2.2.32 (dfbe293d4): /etc/dovecot/dovecot.conf # OS: Linux 2.6.32-696.1.1.el6.x86_64 x86_64 CentOS release 6.9 (Final) ext3 auth_socket_path = /var/run/dovecot/auth-userdb doveadm_password = # hidden, use -P to show it doveadm_port = 61800 first_valid_uid = 150 last_valid_uid = 150 mail_location = maildir:/var/vmail/%d/%n mail_plugins = notify replication mail_privileged_group =

Dear all, I have a population with three age-classes, at time t=0 the population is: n.zero <- c(1,0,0) I have a transition matrix A which denotes "fertility" and "survival": A <- matrix(c(0,1,5, 0.3,0,0, 0,0.5,0), ncol=3, byrow=TRUE) To obtain the population at t=1, I calculate: A %*% n.zero To obtain the population t=2, I calculate: A %*% (A %*% n.zero) ... and so

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 aporte?excelente!! Atte. Ricardo Alva Valiente De: Jose Luis Cañadas Reche [mailto:canadasreche en gmail.com] Enviado el: miércoles, 29 de abril de 2015 12:51 PM Para: Alva Valiente, Ricardo (RIAV); 'javier.ruben.marcuzzi en gmail.com'; R-help-es en r-project.org Asunto: Re: [R-es] cantidad de datos Podrías hacer varios kmedias con diferente número de clusters y comprobar como

El inconveniente con un K-medias, es que se tiene que se tiene que pre definir el número de segmentos, pero eso es algo con lo q no cuento. La solución de Javier me parece q sería la única opción. Atte. Ricardo Alva Valiente -----Mensaje original----- De: R-help-es [mailto:r-help-es-bounces en r-project.org] En nombre de javier.ruben.marcuzzi en gmail.com Enviado el: miércoles, 29 de abril de

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

I have mostly succeeded in getting master and slave DNS servers operational. Mostly, because the zone file is not written when a zone is updated on the master server when the notify and transfer process happens. The slave DNS server gets the changes to the modified zone, but the slave zone file remains as before. I've found a few tutorials and lots of discussions, many of which talk about

Sorry I didn't answer before. Martin, thanks very much for your notes and for tidying the function up! About matPower, I didn't realise that where it says "pot <- pot %*% p ot", consumption of memory was higer than it was on the original algorithm, And about the other function in this thread, matExp, I can remember I us look to m >Message: 27