similar to: Merge list to list - as list

Dear all, I have dataset x <- list(matrix(1:20, 5, 4),matrix(1:20, 5, 4),matrix(1:20, 5, 4)) y <- list(matrix(110:114, 5, 1),matrix(110:114, 5, 1),matrix(110:114, 5, 1)) I need merge x and y as list (y put in last column). The result is something like [[1]] [,1] [,2] [,3] [,4] [,5] [1,] 1 6 11 16 110 [2,] 2 7 12 17 111 [3,] 3 8 13 18 112 [4,] 4

Dear all, I apologize if my question is quite simple. I have a dataset (20 columns & 1000 rows) which some of columns have the same value and the others have different values. Here are some piece of my dataset: obj <- cbind(c(1,1,1,4,0,0,1,4,-1), c(0,1,1,4,1,0,1,4,-1), c(1,1,1,4,2,0,1,4,-1), c(1,1,1,4,3,0,1,4,-1), c(1,1,1,4,6,0,1,5,-1),

Thank you very much for your useful suggestions. These are exactly what I was looking for. foo <- list(foo1, foo2, foo3) lapply(foo, function(x) matrix(unlist(x), nrow = length(x), byrow = TRUE)) or lapply(foo, function(x) do.call('rbind', x)) Best, Muhammad Subianto On 4/11/06, Muhammad Subianto <msubianto at gmail.com> wrote: > Dear all, > I have a result my experiment

Dear all, I have a dataset train <- cbind(c(0,2,2,1,0), c(8,9,4,0,2), 6:10, c(-1, 1, 1, -1, 1)) test <- cbind(1:5, c(0,1,5,1,3), c(1,1,2,0,3) ,c(1, 1, -1, 1, 1)) I want to find which rows of train and test it different in its last column (column 4). The solution must be something like train [,1] [,2] [,3] [,4] [1,] 0 8 6 -1 [3,] 2 4 8 1 [4,] 1 0 9 -1

Dear all, Are there any functions to convert decimals to fractions in R? I have the result: > summary(as.factor(complete.ID)) 0 0.0133333333333333 0.04 2256 488 230 0.0666666666666667 0.0933333333333333 0.106666666666667 2342 310 726 0.133333333333333

I have a question, which very easy to solve, but I can't find a solution. I want to convert a data frame to matrix. Here my toy example: > L3 <- c(1:3) > L10 <- c(1:6) > d <- data.frame(cbind(x=c(10,20), y=L10), fac=sample(L3, + 6, repl=TRUE)) > d x y fac 1 10 1 1 2 20 2 1 3 10 3 1 4 20 4 3 5 10 5 2 6 20 6 2 > is.data.frame(d) [1] TRUE > sapply(d,

Dear all, I have a result my experiment like this below (here my toy example): foo1 <- list() foo1[[1]] <- c(10, 20, 30) foo1[[2]] <- c(11, 21, 31) foo2 <- list() foo2[[1]] <- c(100, 200, 300) foo2[[2]] <- c(110, 210, 310) foo3 <- list() foo3[[1]] <- c(1000, 2000, 3000) foo3[[2]] <- c(1100, 2100, 3100) list(foo1,foo2,foo3) The result: > list(foo1,foo2,foo3) [[1]]

Dear all R-users, (My apologies if this subject is wrong) I have dataset: mydat <- as.data.frame( matrix(c(14,0,1,0,1,1, 25,1,1,0,1,1, 5,0,0,1,1,0, 31,1,1,1,1,1, 10,0,0,0,0,1), nrow=5,ncol=6,byrow=TRUE)) dimnames(mydat)[[2]]

Dear all, I tried to install rgl package on my linux machine fc5, I got an error. Here I run R as user, $ R > options(repos=c(CRAN="http://cran.at.r-project.org/")) > install.packages("rgl", lib="/home/subianto/local/lib/R/library/site-packages", dependencies=TRUE) trying URL 'http://cran.at.r-project.org/src/contrib/rgl_0.67-2.tar.gz' Content type

Hi, Excuse me for this simple question. How to convert as.data.frame to as.character? ?data.frame > L3 <- LETTERS[1:3] > L10 <- LETTERS[1:10] > d <- data.frame(cbind(x=c("XYZ"), y=L10), fac=sample(L3, 10, repl=TRUE)) > d x y fac 1 XYZ A A 2 XYZ B A 3 XYZ C A 4 XYZ D A 5 XYZ E B 6 XYZ F C 7 XYZ G A 8 XYZ H C 9 XYZ I B 10 XYZ

Dear all, I have a dataset. I want to make barplot from this data. Zero1 <- " V1 V2 V3 V4 V5 V6 V7 V8 V9 1 1 0 0 0 1 0 0 0 Positive 2 0 0 1 0 1 0 1 1 Negative 3 0 0 1 0 0 0 1 1 Positive 4 0 1 0 1 1 1 0 1 Negative 5 0 0 1 0 1 1 0 0 Positive 6 0 1 0 0 1 1 1 1 Negative 7 1 0 1 1 1 1 1 1 Negative 8 0 0 0 0 1 0 0 1

Dear R users, My situation: (1) I have limited workspace for my work harddisk (about 10 GiB). (2) I have a lot of data files in R workspace (*.RData) which most of them > 200 MiB. For some reason I zip some of them, for instance "filename.RData (250 MiB)" to "filename.zip (3MiB)". In this work I have a lot of more space of my harddisk. Normally, If I want to use

Dear all, I was wondering if someone can help me. I am learning SVM for classification in my research with kernlab package. I want to know about classification performance using Area Under Curve (AUC). I know ROCR package can do this job but I found all example in ROCR package have include prediction, for example, ROCR.hiv {ROCR}. My problem is how to produce prediction in SVM and to find

Dear R-helpers, First I apologize if my question is quite simple I have a large datasets which more 100 variables. For a research I need to change all name of variables with add one or more letters on each variables. For example, > data(Pima.tr) > Pima.tr[1:5,] npreg glu bp skin bmi ped age type 1 5 86 68 28 30.2 0.364 24 No 2 7 195 70 33 25.1 0.163 55 Yes 3 5

Dear R users, I have a problem, which I can not find a solution. Probably someone could help me? I have a result from my classification, like this > credit.toy [[1]] age married ownhouse income gender class 1 20-30 no no low male good 2 40-50 no yes medium female good [[2]] age married ownhouse income gender class 1 20-30 yes yes high male

Dear R-helper, Is there possible to make this array: > a <- array(1:12, c(4, 3)) > a [,1] [,2] [,3] [1,] 1 5 9 [2,] 2 6 10 [3,] 3 7 11 [4,] 4 8 12 > like: c(1,5,9) c(2,6,10) c(3,7,11) c(4,8,12) Thank you very much in advance. Regards, Muhammad Subianto

Dear R-helpers, I have dataset (data.frame) like below, x1 x2 x3 x4 x5 x6 x7 x8 x9 ... x1200 0 0 0 1 1 0 0 1 1 1 0 0 1 1 0 0 1 1 0 1 0 1 1 0 0 1 1 1 1 0 1 1 0 0 1 1 ... How can I change automatically 0=no and 1=yes. Thank you very much in advance. Kindly regards, Muhammad

Dear useR, I have a dataset like this below, > prevRND.dat <- read.table("C:\\workdir\\prevRND.txt", + header=FALSE, # No header. + col.names = c("X","Y","Z"), + sep = ",") > prevRND.dat X Y Z 1 A A 0.950933 2 A B 0.143600 3 A C 0.956133 4 B A

Dear R-help, First I apologize if my question is quite simple. I need add some of data in the first place my dataset, how can I do that. I have tried with rbind, but I did not succes. 0.1 3.6 0.4 0.9 rose 4.1 4.0 1.2 1.2 rose 4.4 3.2 1.9 0.5 rose 4.6 1.1 1.1 0.2 rose For example,

Dear R-helper, I am new learning R. Now, I have a data set like: 24,2,3,3,1,1,2,3,0,1 45,1,3,10,1,1,3,4,0,1 43,2,3,7,1,1,3,4,0,1 42,3,2,9,1,1,3,3,0,1 36,3,3,8,1,1,3,2,0,1 19,4,4,0,1,1,3,3,0,1 38,2,3,6,1,1,3,2,0,1 21,3,3,1,1,0,3,2,0,1 27,2,3,3,1,1,3,4,0,1 45,1,1,8,1,1,2,2,1,1 ... with 3730 rows I want to remove comma (,) in data set. The result like: 24 2 3 3 1 1 2 3 0 1 45 1 3 10 1 1 3 4 0 1