similar to: Clustering Large Applications..sort of

Hola, Daniel: Quizá deberías ser más explícito porque de la información que suministras yo solo te puedo decir que no veo la relación entre los 3 tipos de algoritmos que nombras: - un análisis de componentes principales puede ser una fase previa de los otros dos - hacer un cluster es un tipo de aprendizaje no supervisado, mientras que un clasificador normalmente es utilizado en aprendizaje

Hi, I would like to know if someone knows an extended implementation of k-means in R to find appropriate number of clusters for a given k-dimensional data. Also, I am working on clustering for forecasting, if someone is interested or has knowledge on implementational details please mail me, I would appreciate it. Regards Skanda Kallur "Cogito, ergo sum" (I think, therefore I

Dear R user, I am a biochemist/bioinformatician, at the moment working on protein clusterings by conformation similarity. I only started seriously working with R about a couple of months ago. I have been able so far to read my way through tutorials and set-up my hierarchical clusterings. My problem is that I cannot find a way to obtain information on the rooting of specific nodes, i.e. of

Hi, I am using code below to get a plot that will show me on the X axis the number of clusters and on the Y axis the cluster average widths. However I am getting this error: Error in summary(silhouette(cutree(d, x), dist(iris[, -5])))$si.summary : $ operator is invalid for atomic vectors the code I am using is: avgs<-sapply(1:20,function(x) summary(silhouette(cutree(d,x),

CREO QUE ESTE TIPO DE CONSULTA, EXCEDE EL PROPÓSITO DE ESTE FORO. El miércoles, 26 de noviembre de 2014, Daniel Carrillo Zapata < daniel.carrillo2 en um.es> escribió: > Hola Isidro, > > mira, te explico mejor: tengo una base de datos con información de > 10 conductores en un recorrido de 30 minutos en coche. Para cada > conductor, se le midió parámetros

I have a pretty big similarity matrix (2870x2870). I will produce even bigger ones soon. I am using PAM to generate clusters. The desired number of output clusters is a PAM input parameter. I do not know a-priopri what is the best clusters layout . I resorted to the silhouette test. It takes forever as I have to run PAM with all possible numbers of clusters. I wonder whether there is some faster

Hello everyone. I just started learning R for clustering analysis in my research project. I tried k-means method and PAM method, both of which were properly processed with my data. I have some questions about PAM graphical output. Suppose to do the commands shown below; pm<-pam(D, 6) plot(pm) I got two charts after prompted. In the first chart, 6 oval clusters are drawn together

Hi all, I am running a series of experiments where after manipulating my data I run several clustering algorithms (agnes, diana and a clustering method of my own) on the data. I wanted to determine which clustering method did the best job, so therefore I had defined my own quality measure using two criteria: compactness of the data within the clusters themselves and the amount of seperation

Hola. Yo en vez de utilizar análisis cluster que impliquen distancias, probaría con un kmedias o con un pam (partition around medoids) pero utilizando muestras, la función clara de la librería cluster puede ayudarte. Pego el details de la ayuda de 'clara' Details clara is fully described in chapter 3 of Kaufman and Rousseeuw (1990). Compared to other partitioning methods such as pam,

Hola de nuevo a todos, me gustaría pediros disculpas por los correos que he enviado. La razón de enviarlos es que pensaba que era también un foro en el que podía plantear cuestiones estadísticas, no solo sobre R en concreto. Siempre es importante aprender algo de todo lo que haces, así que lo que me llevo es el conocimiento de que aquí solo puedo plantear cuestiones de implementación en R, y

Hi list, When I was using 'silhouette' from the 'cluster' package to calculate clustering performances, R crashed. I traced the problem to the fact that my clustering labels only have 2's and 3's. when I replaced them with 1's and 2's, the problem was solved. Is the function purposely written in this way so when I have clustering labels, "2" and

Hi, I am using fanny and I have estrange results. I am wondering if someone out there can help me understand why this happens. First of all in most of my tries, it gives me a result in which each object has equal membership in all clusters. I have read that that means "the clustering is entirely fuzzy". Looking at the graphics it is really difficult to understand how objects with so

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

Dear all, I am trying (without much success) to use the fuction 'silhouette'. Would anyone encountered that before (or would know where I am wrong ?) Please find below the R ouput. Thanks in advance, L. > s <- silhouette(ct, as.dist(metric)) Error in "[<-"(*tmp*, iC, "sil_width", value = s.i) : number of items to replace is not a multiple of replacement

Hello all, This might have been fixed in later versions (I am using R1.7.0), r-help archive contains messages reporting similar problems but no reports of codes fixes. I have encountered a couple of problems using the silhouette function. one occurs when the clustering contains clusters composed of 1 element (Martin Maechler posted code few months ago that fixes a similar problem that occurs

Hello everyone, What is the simplest, most RELIABLE way to import data from MS EXCEL (.xls) format to R? In the past I have used the read.xls() function from the xlsReadWrite package, however, I have been wrestling with it all afternoon long with no success. I continue to receive the following error message: > {widge<-read.xls("F:\\Classes\\Z1.Data\\stat.3010\\WidgeOne.xls", +

Hi Suppose you have the code below. The result I get from the cat function is from the avgs object. Now, I have 30 diferent objects like this and I wish to make a summary table, something like: Avgs1 Avgs2 Avgs3 i= 2 average= 0.515983i i= 2 average= 0.746983 i= 2 average= 0.2665983 i= 3 average=

Hello all. I wish to run k-means with "manhattan" distance. Since this is not supported by the function "kmeans", I turned to the "pam" function in the "fpc" package. Yet, when I tried to have the algorithm run with different starting points, I found that pam ignores and keep on starting the algorithm from the same starting-points (medoids). For my

Hello, I was using version 2.13.2 and I have just downloaded the latest version 2.14.1. However, I''m trying to install the packages I was using and when I look for them in the packages list, I can´t find many in the CRAN binaries (e.g. "vegan"). I do find them in the CRAN sources but the installation fails. I tried downloading the version 2.14.0 and I had the same problem. I

Hello, I am a new user of R, and I'd be grateful if someone could help me with the following: I would like to compute the mean of variable "trust" in dataframe "foo", but separately for each level of variable V2. That is, I'd like to compute the mean of trust at each level of V2. I have done this: > tmp <- by(foo, foo$V2, function(x) mean(foo$trust, na.rm=T))