search for: withinss

...sure to my raw data would have been inappropriate) canto.MDS<-cmdscale(canto) I then figured out what would be my optimum k-value by plotting the within sums of squares for K1-K15 > wss <- (nrow(canto.MDS)-1)*sum(apply(canto.MDS,2,var)) > wss[2] <- sum(kmeans(canto.MDS,centers=2)$withinss) > wss[3] <- sum(kmeans(canto.MDS,centers=3)$withinss) > wss[4] <- sum(kmeans(canto.MDS,centers=4)$withinss) > wss[5] <- sum(kmeans(canto.MDS,centers=5)$withinss) > wss[6] <- sum(kmeans(canto.MDS,centers=6)$withinss) > wss[7] <- sum(kmeans(canto.MDS,centers=7)$wit...

...cluster centers... Now a small dataset: > data<-matrix(c(-1,0,2,2.5,7,9,0,3,0,6,1,4),6,2) If I use rows 3 and 4 as cluster centers and a single iteration of kmeans I get: > kmeans(data,data[c(3,4),],1) $cluster [1] 1 1 1 1 2 2 $centers [,1] [,2] 1 0.875 2.25 2 8.000 2.50 $withinss [1] 32.9375 6.5000 $size [1] 4 2 If I now use rows 1 and 6 as cluster centers I get exactly the same solution after the first iteration: > kmeans(data,data[c(1,6),],1) $cluster [1] 1 1 1 1 2 2 $centers [,1] [,2] 1 0.875 2.25 2 8.000 2.50 $withinss [1] 32.9375 6.5000 $size [1] 4 2 So...

...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] 2 2 1 2 1 2 2 2 1 2 2 1 2 2 2 2 $centers [,1] 1 0.1325000 2 -0.2783333 $withinss [1] 0.0646750 0.4033667 $size [1] 4 12 The second solution gives the following: $cluster [1] 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 $centers [,1] 1 -0.1313333 2 -0.8400000 $withinss [1] 0.5035733 0.0000000 $size [1] 15 1 I don't understand why this is happening, and how do I choose betwee...

...4 4 2 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 Within cluster sum of squares by cluster: [1] 2.731015e+26 8.785281e+22 4.726557e+26 3.513411e+22 5.092071e+25 (between_SS / total_SS = 98.9 %) Available components: [1] "cluster" "centers" "totss" "withinss" "tot.withinss" [6] "betweenss" "size" *Now how to I save these 5 clusters into 5 separate folders? * Please advise, Thanks. [[alternative HTML version deleted]]

...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 1 [75] 1 3 1 3 1 1 1 1 3 1 1 1 1 1 3 1 1 3 1 1 3 3 1 2 1 1 $withinss [1] 1037.5987 0.0000 666.9701 $size [1] 68 1 31 > c4<-kmeans(data,3,20) $withinss [1] 0.0000 865.7628 851.1214 $size [1] 1 54 45 Does any one tell me why the results are very different with the same dataset and parameters when I run some times this command 'kmeans(data,3,20)...

...bership 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 answer every time a run it. Sometimes the answer is different only in the order of withinss (and the ordering of the numbers of cases assigned to each group). Other times there are completely different centers, withinss and completely different cluster configurations. This variability doesn't happen in either S-Plus 2000 or S-Plus 6.0 (Beta 2). I can see from the help that the R...

...1 5 5 5 1 [39] 3 1 5 5 3 1 1 1 1 5 5 1 4 1 3 5 5 5 5 5 5 1 Within cluster sum of squares by cluster: [1] 0.6702803 0.0000000 0.2453294 0.1860180 1.3535263 (between_SS / total_SS = 76.8 %) Available components: [1] "cluster" "centers" "totss" "withinss" "tot.withinss" [6] "betweenss" "size" > Q3)I would like to understand which raw data are in which cluster ? Does somebody knows how to access the table of raw data which are in the same cluster ? Thanks for help DZU -- View this message in...

...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 [,1] [,2] 1 0.875 2.75 2 8.000 2.50 $withinss [1] 38.9375 6.5000 $size [1] 4 2 If I do it by hand, after one iteration, the results are $cluster [1] 1 2 1 2 1 2 So I think that something is wrong with the function kmeans; probably the initial centers given by the user are not being taken into account.

...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 for your help, Serguei ________________________________________ Austrian Institute of Economic Research (WIFO) P.O.Box 91 Tel.: +43-1-7982601-231 1103 Vienna, Austria Fax: +43-1-7989386 Mail: Serguei.Kaniovski@wifo.ac.at http://www.wifo.a...

..."")) file <- read.table(paste("R_out/", filename, ".txt", sep="")) obj <- kmenas(file, 5) sink(paste("R_out/res", filename, ".txt", sep="")) "Kluster centers" obj$centers "Within size for clusters" obj$withinss "Cluster size" obj$size sink() } [[alternative HTML version deleted]]

...centers) <- list(1L:k, dimnames(x)[[2L]]) + cluster <- Z$c1 + if (!is.null(rn <- rownames(x))) + names(cluster) <- rn + totss <- sum(scale(x, scale = FALSE)^2) + print(Z$iter) + structure(list(cluster = cluster, centers = centers, totss = totss, + withinss = Z$wss, tot.withinss = best, betweenss = totss - + best, size = Z$nc, iter = Z$iter), class = "kmeans") + } > <environment: namespace:stats> Error: unexpected '<' in "<"

...ame(mydataS) #removes "coden" variable myvars <- names(mydataS2) %in% c("coden") mydataSNc <- mydataS2[!myvars] #Determine number of clusters wss <- (nrow(mydataSNc)-1)*sum(apply(mydataSNc,2,var)) for (i in 2:15) wss[i] <- sum(kmeans(mydataSNc, centers=i)$withinss) plot(1:15, wss, type="b", xlab="Number of Clusters",ylab="Within groups sum of squares") #perform Kmeans with 2 clusters kmSNc2 <- kmeans(mydataSNc,2) #print conponents of kmSNc2 print(kmSNc2) #plot 2 clusters plot(mydataSNc, col = kmSNc2$cluster) #plot...

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

...-1.0 0 [2,] 0.0 3 [3,] 2.0 0 [4,] 2.5 6 [5,] 7.0 1 [6,] 9.0 4 > plot(dados) > dados<-matrix(c(-1,0,2,2.5,7,9,0,5,0,6,1,4),6,2) > plot(dados) > A<-kmeans(dados,dados[c(3,4),],1) > A $cluster [1] 1 1 1 1 2 2 $centers [,1] [,2] 1 0.875 2.75 2 8.000 2.50 $withinss [1] 38.9375 6.5000 $size [1] 4 2 Any hints? Thanks a lot Luis Silva

...1018 [29] 0.9034660 1.2406042 0.9529861 3.3889001 0.8462411 0.8338748 1.8540691 [36] 1.3624104 6.9509700 > kmeans(univ,ncl) $cluster [1] 5 3 5 5 4 5 3 3 3 3 1 3 5 3 4 1 2 5 1 3 3 5 3 1 5 5 5 5 3 3 3 1 3 3 2 3 4 $centers [,1] 1 3.5806021 2 2.0872314 3 0.9808016 4 5.9624282 5 0.5968146 $withinss [1] 0.4622251 0.1087293 0.3637454 1.7637280 0.1768656 $size [1] 5 2 16 3 11 Thanks a lot, Unung Istopo On Tue, 2004-05-11 at 17:19, TEMPL Matthias wrote: > Hello, > When clustering with kmeans, your data should have more than one > variable. Matthias

...all" "silinfo" "data" But "kmeans" has: K-means clustering with 2 clusters of sizes 50, 50 Cluster means: Clustering vector: Within cluster sum of squares by cluster: Available components: [1] "cluster" "centers" "withinss" "size" [[alternative HTML version deleted]]

...<- subset(grid40km.data,grid40km.cluster==i, select=c(X1, X2, X3, X4, X5)) SSgroup <- (nrow(data_group-1)*sum(apply(data_group,2,var))) # SS for all variables for a given cluster SStot_grid40km=append(SStot_grid40km, SSgroup,after=length(SStot_grid40km)) } ssw_grid40km = sum(SStot_grid40km) #withinSS (??) as the sum of SS for all clusters ssbetween_grid40km = SSTot-ssw_grid40km RSQ_grid40km2 = ssbetween_grid40km/SSTot # R-square Am I right? Does this correspond to SAS's R2? Many thanks, Yan Ressources Naturelles Canada Service Canadien des Forêts - Centre de Foresterie des Laurenti...

I have checked that section already. Sorry, I should have mentioned that. Memory limit increase does not work. Installtion of msvcrt.dll does not work either. Thank you. -----Original Message----- From: ripley at stats.ox.ac.uk [mailto:ripley at stats.ox.ac.uk] Sent: Wednesday, March 05, 2003 2:44 PM To: Igor Oleinik Cc: r-help at stat.math.ethz.ch Subject: Re: [R] problem with cclust[er]

...hacer varios kmedias con diferente número de clusters y comprobar como varía la suma de cuadrados entre cluster para "elegir" el número óptimo. # Determine number of clusters wss <- (nrow(mydata)-1)*sum(apply(mydata,2,var)) for (i in 2:15) wss[i] <- sum(kmeans(mydata, centers=i)$withinss) plot(1:15, wss, type="b", xlab="Number of Clusters", ylab="Within groups sum of squares") El 29/04/15 a las 19:42, Alva Valiente, Ricardo (RIAV) escribió: El inconveniente con un K-medias, es que se tiene que se tiene que pre definir el número de segmentos, pero es...

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