search for: ruimaximo

Hello, This is very basic and very frustrating. Suppose this: >A=5 >B=5 >C=10 > ls() "A" "B" "C" I would like this >xpto() 5 5 10 How can I do xpto()? Thanks Rui [[alternative HTML version deleted]]

Hello, I am trying to run a small example with foreach, but I am having some problems. Here is the code: *library(doMC) registerDoMC() zappa = list() frank = list() foreach (i = 1:4) %dopar% { zappa[[i]] = kmeans (iris[-5],4) frank[[i]] = warnings() }* The code runs without error. However the zappa and frank will be empty lists. If I use regular *for *instead, the list will be filled up

Hi, I downloaded a dataset from UCI repositories named Bag of Words: http://archive.ics.uci.edu/ml/machine-learning-databases/bag-of-words/readme.txt The dataset is in a text file with the following structure: --- docID1 wordID1 count docID1 wordID2 count docID1 wordID3 count docID1 wordID4 count ... docID2 wordID2 count docID2 wordID5 count docID2 wordID6 count --- Where docIDx is an integer

Hello, I have a matrix with values, with columns c1..cn. I need the values to be normalized between 0 and 1 by column. Therefor, the 0 should correspond to the minimum value in the column c1 and 1 should correspond to the maximum value in the column c1. The remaining columns should be organized in the same way. Does a function in R exists for this purpose? Thanks, Rui [[alternative HTML

Hello, I am newbie to R and I want to do this: for(i in 1:6) { ds[i] <- list(df=data.frame(oilDF[,1],oilDF[,i+2]), df2=data.frame(oilDF2[,1],oilDF2[,i+2])) } #oilDF and oilDF2 are 2 data frames with several columns. They have different number of rows #I want to have for example ds[1]$df, ds[1]$df2 with the respective data.frames. #How can I instantiate a list of data.frames pairs

Hi, This question seems very basic but I cannot find an answer on google. I have a R session on a linux command line. I called a function that is taking ages. I want to cancel the function but without killing the R session. What is the shortcut? Thanks, Rui [[alternative HTML version deleted]]

Hello, I need a Fuzzy C Means algorithm. I found some documentation about cmeans {e1071} at http://rss.acs.unt.edu/Rdoc/library/e1071/html/cmeans.html Does someone knows where I can find it? Thank you Rui [[alternative HTML version deleted]]

Hello. I have this matrix: cy [,1] [,2] [,3] [,4] [,5] [,6] [1,] 1.0000000 1.0000000 -0.5164570 -0.5164087 -0.4004139 -0.4003719 [2,] 1.0000000 1.0000000 -0.5164570 -0.5164087 -0.4004139 -0.4003719 [3,] -0.5164570 -0.5164570 1.0000000 1.0000000 -0.1186691 -0.1186972 [4,] -0.5164087 -0.5164087 1.0000000 1.0000000 -0.1186995 -0.1187276 [5,]

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

Hi. Is there any constant that represents the maximum value of an integer? If I need to setup by myself what is the maximum value? Best, Rui

Dear all, I need to know in which number of iterations the kmeans converge each time I run it. Any idea how to do it? Thank you for your attention, Rui

Hello, I have a trained Back Propagation Neural Network model in weka. I would like to re-evaluate the NN using R with a given input. How can I do this? I could not find an example of RWeca that applies to NN Thanks, Rui

Dear all, I just installed the package hydroPSO and when I try to run the example of the lhoat() function I got this error: nparam <- 5 lhoat( fn=sphere, lower=rep(-100,nparam), upper=rep(100,nparam), control=list(N=10, f=0.1, write2disk=TRUE) ) *Error in fn.name == "hydromod" : * * comparison (1) is possible only for atomic and list types* Any

Dear all. I am searching for KMeans ++ for R. I cannot find it. Do you know any package with it? Best regards, Rui

Hello, I have 2 functions (a and b) a = function(n) { matrix (runif(n*2,0.0,1), n) } > > > b = function (m, matrix) { > n=nrow (matrix) > p=ceiling (n/m) > lapply (1:p, function (l,n,m) { > inf = ((l-1)*m)+1 > if (l<p) sup=((l-1)*m)+m > else sup=n >