similar to: kmeans: number of cluster centres must lie between 1 and nrow(x)

Dear R, I have a simple question concerning with a special case of spare matrix multiplications. Say A is a 200-by-10000 dense matrix. B is a 10000-by-10000 block- diagonal matrix, and each diagonal block B_i is 100-by-100. The usual way I did A%*%B will take about 30 seconds which is to time consuming because I have to do this thousands of times. I also tried to partition A into 100 small blocks

Dear all, Why does not the three-dot accept arguments from the parent environment? I am just confused with this error, can someone give me a hint? > rm(list=ls()) > testFun <- function(a, ...) + { + if(a){ + print(a) + }else + { + print(b) + } + } > > myTask <- function(a) + { + b <- 3 + testFun(a, b = b) + } >

Dear R, How can I disable using "enter" key to exit the browser() in debug mode? I would love to have this option because it is so annoying to jump out of the debugging mode unexpectedly when I don't want to. I guess some of us have encouraged at least one of these situations, 1, Accidentally pressed the enter key within the browser. 2, Copy and paste a piece of debugging code

Dear R, Is there an efficient way to make a list that each element is from the corresponding column of a matrix. For example, if I have a matrix "a" > a <- matrix(1:10, 5, 2) > a [,1] [,2] [1,] 1 6 [2,] 2 7 [3,] 3 8 [4,] 4 9 [5,] 5 10 I would like to have a list "b" like this > b <- list(a[, 1], a[, 2]) > b [[1]] [1] 1 2 3

Dear R, I have a silly question concerns with *apply. Say I have a list called A, A <- list(a = array(1:20, c(2, 2, 5)), b = array(1:30, c(2, 3, 5))) I wish to calculate the mean of A$a, and A$b w.r.t. their third dimension so I did lapply(A,apply,c(1,2),mean) Now if I still wish to do the above task but take away some burn-in, e.g. do not take A$a[,,1:2],and A$b[,,1:2] into account.

Dear R, Why rm(list<-ls()) gives an error but rm(list=ls()) not?. I remember the operator ‘<-’ can be used anywhere... Thanks! Feng -- Feng Li Department of Statistics Stockholm University 106 91 Stockholm, Sweden http://feng.li/ [[alternative HTML version deleted]]

Dear R, I am looking for R-package related to INARMA (Integer-valued ARMA). Can anyone give me some information? I did not get information from task view. Many thanks. Feng -- Feng Li Department of Statistics Stockholm University 106 91 Stockholm, Sweden http://feng.li/ [[alternative HTML version deleted]]

Dear R, I have a covariates matrix with 10 observations, e.g. > X <- matrix(rnorm(50), 10, 5) > X [,1] [,2] [,3] [,4] [,5] [1,] 0.24857135 0.30880745 -1.44118657 1.10229027 1.0526010 [2,] 1.24316806 0.36275370 -0.40096866 -0.24387888 -1.5324384 [3,] -0.33504014 0.42996246 0.03902479 -0.84778875 -2.4754644 [4,] 0.06710229 1.01950917

Dear R, I rare use the standard R-gui on Windows. Yesterday I tried the latest stable release on Windows 7 and XP and found one thing interesting. Assume currently I am running R code, say > example(plot) Then I click the "close window" button on the R main window. R asks me to save workspace image or not. Then I click "cancel". I suspect that my program

Dear R, I have two questions: 1, Why both R and Matlab give 0*Inf==NaN? To my knowledge, it should be zero mathematically. Am I right? 2, I need to calculate e.g. exp(a)/(exp(b)+c), where both a and b are very large numbers (>>1000, e.g a=1000, b=1007, and c=5). R gives me NaN when I use the following command: > exp(1000)/(exp(1007)+5) [1] NaN I am pretty sure this should be close to

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)

Dear R, If I have remembered correctly, a square matrix is singular if and only if its determinant is zero. I am a bit confused by the following code error. Can someone give me a hint? > a <- matrix(c(1e20,1e2,1e3,1e3),2) > det(a) [1] 1e+23 > solve(a) Error in solve.default(a) : system is computationally singular: reciprocal condition number = 1e-17 Thanks in advance! Feng --

Dear R, I have a vector, say a = c(1,2,4,5,6,8). Can I generate a vector or array (2-by-3-by-3) of this form c(1,2,1,2,1,2,4,5,4,5,4,5,6,8,6,8,6,8), in which every two elements in "a" have been repeated twice? I am to stupid today and could not figure this simple question out... Many many thanks! Feng -- Feng Li Department of Statistics Stockholm University 106 91 Stockholm,

Dear R, I have two small questions confused me recently. Now assume I have a matrix "a", like this, > a <- matrix(1:6, 2, 3) > a [,1] [,2] [,3] [1,] 1 3 5 [2,] 2 4 6 I sometimes need each row of "a" raised to a different exponent. So I do a trick like this, > a^c(2, 3) [,1] [,2] [,3] [1,] 1 9 25 [2,] 8 64 216 My first

Dear R, I found a very wired behavior for a 2-by-2 matrix, see this example > A <- matrix(1:4, 2) > idx4A <- matrix(1:4, 2) > A[idx4A] Error in A[idx4A] : subscript out of bounds But other matrices are fine, > B <- matrix(1:9, 3) > idx4B <- matrix(1:9, 3) > B[idx4B] [1] 1 2 3 4 5 6 7 8 9 I can reproduce this for both 32bit windows and 64bit linux with R 2.12.1

I''m sorry to posting this message in this list, but it''s the only way iI found. Who known a mailing list telling about the cisco''s router? -- ____________________________________________________________ PIII 500MHz - 96MB RAM - HD 8.2GB - Diamond Stealth III s540 USR Sportster 56K Int Voice - SoundBlaster AWE 64 -- CL 6 //LinuxMan by Friends - The power of a GNU

Hi, I want to know which distance is using in the function kmeans and if we can change this distance. Indeed, in the function pam, we can put a distance matrix in parameter (by the line "pam<-pam(dist(matrixdata),k=7)" ) but we can't do it in the function kmeans, we have to put the matrix of data directly ... Thanks in advance, Nicolas BOUGET

Dear R, I have two R instances running at the same time, say instance A and instance B. Is there a simpler way to pass the data in A to B? More precise, I have a stupid example: In instance A, I am running a function "test1" test1 <- function (x1) { x2 <- sin(x1) return(x2) } In instance B, another function "test2" test2 <- function (x2) { x3 <-

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.

Hello, here is a working reproducible example which crashes R using kmeans or gives empty clusters using the nstart option with R 15.2. library(cluster) kmeans(ruspini,4) kmeans(ruspini,4,nstart=2) kmeans(ruspini,4,nstart=4) kmeans(ruspini,4,nstart=10) ?kmeans either we got empty always clusters and or, after some further commands an segfault. regards, Detlef Groth ------------ [R] Empty