similar to: algorithm used in k-mean clustering

Hi! I am new to R just 3 days in it and i apologize if my questions seem very trivial and consumed your valuable time. I am coding in perl and i stumbled upon R regarding plotting good statistical graphs. I tried the kmean clustering for a large matrix ,say > 150 * 150 . I tried the example code given in the tutorial to perform 2d plot # i ranges from 2 to 10 cl <- kmeans(x, i, 20)

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

Your message doesn't help us very much. You haven't said what kind of calculation it is you want to do, and that certainly matters. For example, for some kinds of computations the solution you started below would work fine: > M <- matrix(1:16, 4, 4) > is.na(diag(M)) <- TRUE > M [,1] [,2] [,3] [,4] [1,] NA 5 9 13 [2,] 2 NA 10 14 [3,] 3 7 NA

Dear R users I have the matrix of the centres of some clusters, e.g. 20 clusters each with 100 dimentions, so this matrix contains 20 rows * 100 columns numeric values. I have collected new data (each with 100 numeric values) and would like to keep the above 20 centres fixed/'unmoved' whilst just see how my new data fit in this grouping system, e.g. if the data is close to cluster 1

I was wondering if there was an R equivalent to the two phased approach that MATLAB uses in performing the Kmeans algorithm. If not is there away that I can determine if the kmeans in R and the kmeans in MATLAB are essentially giving me the same clustering information within a small amount of error? -- View this message in context:

Hello, I'm new using R. I'm trying to develop a K-means Clustering with R for some data I have, however each time I use that instruction with the same data my cluster means, clustering vector and within cluster sum of square change and I don't understand why because I use the same parameters and the same data. Can anybody explain me why does it happen? Thank you Act. Calef

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

Hello, I am hoping you can help me with a question concerning kmeans clustering in R. I am working with the following data-set (abbreviated): BMW Ford Infiniti Jeep Lexus Chrysler Mercedes Saab Porsche Volvo [1,] 6 8 2 8 4 5 4 4 7 7 [2,] 8 7 4 6 4 1 6 7 8 5 [3,] 8 2 4

Hello. Is it possible to choose the distance in the kmeans algorithm? I have m vectors of n components and I want to cluster them using kmeans algorithm but I want to use the Mahalanobis distance or another distance. How can I do it in R? If I use kmeans, I have no option to choose the distance. Thanks in advance, Arnau.

Hi All, Can someone please tell me* R function for Bisecting K-means algorithm*. I have used *kmeans() *function but not getting good results. Please help. -- Thanks and Regards, Vivek Kumar Singh Research Assistant, School of Computing, National University of Singapore Mobile:(0065) 82721535 [[alternative HTML version deleted]]

Hello, I have used the functions agnes and cutree to cluster my data (4977 objects x 22 variables) into 8 clusters. I would like to refine the solution using a k-means or similar algorithm, setting the initial cluster centres as the group means from agnes. However my data matrix has NA's in it and the function kmeans does not appear to accept this? > dim(centres) [1] 8 22 > dim(data)

Buen día a todos. mi pregunta es si alguien sabe si el algoritmo de k medias siempre da los mismos resultados con los mismos datos de entrada. o si al correrlo dos veces con los mismos datos de entrada se pueden obtener grupos distintos. [[alternative HTML version deleted]]

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

Hello, I've been working on the KMeans clustering algorithm recently and since the past week, I have been stuck on a problem which I'm not able to find a solution to. Since we are representing documents as Tf-idf vectors, they are really sparse vectors (a usual corpus can have around 5000 terms). So it gets really difficult to represent these sparse vectors in a way that would be

Hi all, I am wandering.. is it possible to cluster data which is in a single column ? for example.. I have some data as follows: 4013 7362 7585 9304 11879 14785 21795 30500 30669 30924 33988 36975 40422 42911 50501 51593 53729 54338 55497 57337 61993 62601 66229 69815 69933 70760 71340 75921 83972 90134 91061 . . . is it possible to cluster this data since it is in a single column ? I have

Hi, all I would like to use R to perform k-means clustering on my data which included 33 samples measured with ~1000 variables. I have already used kmeans package for this analysis, and showed that there are 4 clusters in my data. However, it's really difficult to plot this cluster in 2-D format since the "huge" number of variables. One possible way is to project the

*Apologies if this is not the right way to ask a question, I'm a first timer posting here. Does anyone have a solution to this? I'm having trouble figuring out how to use weighting with K Means Clustering. So say if my dataset is: Column 1 = x coords Column 2 = y coords Column 3 = frequency each coordinate occurs So I'm basically trying to weight the points more heavily if

Hi the devel list, I am using K means with a non standard distance. As far as I see, the function kmeans is able to deal with 4 differents algorithm, but not with a user define distance. In addition, kmeans is not able to deal with missing value whereas there is several solution that k-means can use to deal with them ; one is using a distance that takes the missing value in account, like a

Folks! Netweb Group, Inc. fully supports connectivity to any Asterisk PBX systems you have and can provide A-Z termination with immediate effect. Any volume is good enough for us, even an amount as small as $1.00 a day will do for us. We will provide connectivity from our Softswitch IP 216.162.116.46. If anyone is interested, add this to your Asterisk IPBX and then email me for setting up a

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