similar to: Lasso with Categorical Variables

Hello list. For some reason, the makers of glmnet do not accept a dataframe as input. They expect the input to be a matrix, where the dummies are already precoded. Now I have created a sample dataset with . 11 factor columns with two levels . 4 factor columns with three levels . 135 continuous columns (from a standard normal) . 100 observations (rows) Say this dataframe is in dfrPredictors. What

Hi all, I am stuck at ploting multiple graphs on one page. I don't why it doesn't work. All the 6 plots are either exactly the same, or they simply don't plot at all. I made sure that in each iteration the "datasub" and "factorsub" are different ... Could you please help me? Thanks a lot! I did my homework and followed the following advice:

Hi all, I am whitening my data: # code begins N <- 300 M <- 2 x <- matrix(data=rnorm(N*M, 0, 3)-10, ncol=M, nrow=N) y <- matrix(c(1,-2,-2,1), ncol=M, nrow=M) z <- data.frame(x %*% y) colnames(z) <- c('x','y') par(mfrow=c(1,3)) plot(z, pch=5, col="blue") whiten <- function(x) { (x-mean(x))/sd(x) } zz <- transform(z, x=whiten(x), y=whiten(y))

Hi everyone, I'm using the following code to go over every element of a data frame (row wise). The problem I am facing is that the outer 'x' variable is not incrementing itself, thus, only one row of values is obtained, and the program does not proceed to the next row. This is the code: while(x<=coln) { while(y<=rown) { n<-as.numeric(df[[y]][x]);

Hello All, I am new to R and am having a problem. I have the following 29 x 5 array called temp with a mode "list". Sincere Exciting Competent Sophisticated Rugged 3.5 2.5 3.8 3.5 3.0 2.0 2.5 3.5 2.5 2.3 2.8 3.0 2.3 1.5 1.8 2.0 4.0 2.8 1.5 4.5 2.5 1.5 1.8 1.3 1.3

Full_Name: Raimar Falke Version: R version 2.10.0 (2009-10-26) OS: Linux 2.6.27-16-generic #1 SMP Tue Dec 1 19:26:23 UTC 2009 x86_64 GNU/Linux Submission from: (NULL) (134.102.222.56) The construction of a data frame in the way shown below requires much more memory than expected. If we assume a cell value takes 8 bytes the total amount of the data is 128mb. However the process takes about

Dear List, First off, my deepest gratitude to the Sweave developers: this tool has improved my quality greatly. A question in my work I use \Sexpr{} statements scalar values and the xtable package for all manner of tables. What I'd like to do is to use a vector inline, rather than a whole separate table. Something like: %%%%%%%%%%%%%%%% begin code % Latex junk % Sweave block:

Excelente! Ahora corre muy rápido. No conocía ese método, creo que me va a resultar muy útil. Muchas gracias y saludos. Fernando Macedo El 24/02/15 a las 10:51, Jorge I Velez escribió: Fernando, Podrias intentar R> a <- rep('a', 5) R> b <- rep('b', 5) R> a [1] "a" "a" "a" "a" "a" R> b [1] "b"

DeaR useRs: I have a list with 500 elements, in each other there are data.frames and I want to take the first row and the first column of each elements of my list since the first to the 500-th. Thanks and excuse my bad English. --- [[alternative HTML version deleted]]

In response to those who asked for a better explanation of what the Vilno software does, here's a simple example that gives some idea of what it does. LABRESULTS is a dataset with multiple rows per patient , with lab sodium measurements. It has columns: PATIENT_ID, VISIT_NUM, and SODIUM. DEMO is a dataset with one row per patient, with demographic data. It has columns: PATIENT_ID, GENDER.

Gracias, Carlos. Habia pensado en algo similar usando sapply(): sapply(seq(1, ncol(vtmp), by = 2), function(i) c(rbind(as.character(vtmp[, i]), as.character(vtmp[, i+1])))) Dependiendo de la dimension de los datos, quizas mapply() sea mas eficiente que sapply(). Saludos cordiales, Jorge.- 2015-02-25 1:01 GMT+11:00 Carlos Ortega <cof en qualityexcellence.es>: > Hola, > > Este

Hi All, I need to parse data from a file, example shown below. The first two lines can be skipped, the third line contains the column names. The next 13 lines can be skipped. The next line "1991" is a year value, with the following 13 values data for that year. The file then repeats this format with (year, 13 lines of data for that year). I would ideally like to end up with an

On Sat, Nov 30, 2019 at 06:49:29PM +0800, zxs11289 wrote: > I face a problem on KVM using virt-df。 I have two computer,they > have installed tools for KVM. The computer（ip 15）has no virtual > machine and computer（ip 33）has two virtual machines。on the > computer(ip33)， i run the command “virt-df” if on the computer(ip15)， > i run the command “virt-df --connect

Dear R-users I'm trying to use lasso in lars package for subset regression, I have a large matrix of size 1000x100 and my aim is to select a subset k of the 100 variables. Is there any way in lars to fix the number k (i.e. to select the best 10 variables) library(lars) aa=lars(X,Y,type="lasso",max.steps=200) plot(aa,plottype="Cp") aa$RSS which.min(aa$RSS)

Hi, I am trying to do a lasso regression using the lars package with the following data (see attached): FastestTime WinPercentage PlacePercentage ShowPercentage BreakAverage FinishAverage Time7Average Time3Average Finish 116.90 0.14 0.14 0.29 4.43 3.29 117.56 117.77 5.00 116.23 0.29 0.43 0.14 6.14 2.14 116.84 116.80 2.00 116.41 0.00 0.14 0.29 5.71 3.71 117.24

Hi, all, I am using the lars package for lasso estimate. So I get a lasso fit first: lassofit = lars(x,y,type ="lasso",normalize=T, intercept=T) Then I want to get coefficient with respect to a certain value of \lambda (the tuning parameter), I know lars has three mode options c("step", "fraction", "norm"), but can I use the \lambda value instead

Hi all, I have a data set containing variables LOSS, GDP, HPI and UE. (I have attached it in case it is required). Having renamed the variables as l,g,h and u, I wish to run a Lasso Regression with l as the dependent variable and all the other 3 as the independent variables. data=read.table("data.txt", header=T) l=data$LOSS h=data$HPI u=data$UE g=data$GDP matrix=data.frame(l,g,h,u)

I have always assumed that having a package in the 'Depends' field would automatically also?import?the namespace. However, it seems that in R 2.15, dependencies do not become available until the package is actually?attached?to the searchpath. Is this intended behavior? The problem appears as follows: Suppose there is a package 'Child' which?Depends, but does not explicitly

Hello, I have a question about the lars package. I am using this package to get the coefficients at a specific LASSO parameter s. data(diabetes) attach(diabetes) object <- lars(x,y,type="lasso") cvres<-cv.lars(x,y,K=10,fraction = seq(from = 0, to = 1, length = 100)) fits <- predict.lars(object, type="coefficients", s=0.1, mode="fraction") Can I assign

I'm using the package 'lars' in R with the following code: > library(lars) > set.seed(3) > n <- 1000 > x1 <- rnorm(n) > x2 <- x1+rnorm(n)*0.5 > x3 <- rnorm(n) > x4 <- rnorm(n) > x5 <- rexp(n) > y <- 5*x1 + 4*x2 + 2*x3 + 7*x4 + rnorm(n) > x <- cbind(x1,x2,x3,x4,x5) > cor(cbind(y,x)) y x1 x2