similar to: Bootstrapped standard errors

Dear Friends, I'm trying to generate a sequence of 100 observations with either a 1 or -1. In other words the sequence should look something like this. y = 1 1 -1 1 -1 -1 -1 1 1 ...... Can somebody please give me some direction on how I can do this in R. Thanks Anup ____________________________________________________________________________________ Don't get soaked. Take a

Dear Friends, I have data set with around 220,000 rows and 17 columns. One of the columns is an id variable which is grouped from 1000 through 9000. I need to perform the following operations. 1) Remove all the observations with id's between 6000 and 6999 I tried using this method. remdat1 <- subset(data, ID<6000) remdat2 <- subset(data, ID>=7000) donedat <- rbind(remdat1,

Dear friends, I have a basic question with R. I'm generating a set of random variables and then combining them using the cbind statement. The code for that is given below. for (i in 1:100) { y <- rpois(i,lambda=10) X0 <- seq(1,1,length=i) X1 <- rnorm(i,mean=5,sd=10) X2 <- rnorm(i,mean=17,sd=12) X3 <- rnorm(i,mean=3, sd=24) ind <- rep(1:5,20) }

Dear Friends, Suppose I have a vector as follows RI Value 1 10 2 11 3 8 4 4 6 12 I would like a function which returns the"row index number" for the minimum value of VALUE. Can somebody please give me some direction on how I can do this. In effect I'm trying to find a comparable function for the GAUSS command "minindc" >From the GAUSS Manual

Friends, I'm trying to pass an equation as an argument to a function. The idea is as follows. Let us say i write an independent function Ideal Situation: ifunc <- function(x) { return((x*x)-2) } mainfunc <- function(a,b) { evala <- ifunc(a) evalb <- ifunc(b) if (evala>evalb){return(evala)} else return(evalb) } Now I want to try and write this entire program in a single

Dear friends, I have a data set with 23 columns and 38000 rows. It is a panel running from the years 1991 through 2005. I want to aggregate the data and get the medians of each of the 23 columns for each of the years. In other words my output should be like this Year Median 1991 123 1992 145 1993 132 etc. The sample lines of code to do this operation is set1 <-

Dear Friends, I'm trying to find if there is a way to automate creation of the design matrix. Suppose we are interested in say running an autoregressive model. The user inputs the following data myfunAR <- function(y, order) {..... ...... } now here y is the data series and order represents the level of the process. In other words if order=2 then we have an AR (2) process. Now it is easy

Hello everyone, I have a dataset which includes the first three variables from the demo data below (year, id and var). I need to create the new variable ans as follows If var=1, then for each year (where var=1), i need to create a new dummy ans which takes the value of 1 for all corresponding id's where an instance of one was recorded. Sample data with the output is shown below. year

Dear Friends, I have some data with three columns named ID, Year and Measure X. I need to create a column which gives me a lag for each ID (note not a continous lag), but a lag conditional on the id and the given year. Please find below a sample of the data Input file sample ID Year X AB12 2000 100 AB12 2001 120 AB12 2002 140 AB12 2003 80 BL14 2000 180 BL14

Hi all, Is there an R function to generate random integers? Thanks in advance. Sincerely Anup --------------------------------- Now that's room service! Choose from over 150,000 hotels [[alternative HTML version deleted]]

Dear R-helpers, I have a basic question on using loops. I have a panel data set with different variables measured for "n" firms over "t" time periods. A snapshot of the data is given below id    t    X1    X2 1    1     4    3 1    2    9    2 1    3    7    3 1    4    6    6 2    1    6    4 2    2    5    3 2    3    1    1 3    1    9    6 3    2    5    5 thus total

Dear friends, I'm writing a function with three arguments myfun <- function(theta, X, values) { .... .... } in this function, I'm trying to write consistency checks. In order to compute the statistic of interest I only need theta and values. The idea of having X in there is that, if values is not provided by the user, then values is computed from X. my problem is I'm trying to

Dear Friends, I'm trying to see if there is some possibility that I can do the following computations without a loop. I have attached a toy example below. A <- c(1,2,3,4,5) B <- rnorm(100) store <- matrix(0,5,2) for (i in 1:5) { store[i,1] <- mean(pnorm(A[i]*B)) store[i,2] <- var(pnorm(A[i]*B)) } store Thanks in advance for your help. Kind Regards Anup [[alternative

Dear Friends, My problem is related to how to measure probabilities from a probit model by changing one independent variable keeping the others constant. A simple toy example is like this Range for my variables is defined as follows y=0 or 1, x1 = -10 to 10, x2=-40 to 100, x3 = -5 to 5 Model output <- glim(y ~ x1+x2+x3 -1, family=binomial(link="probit")) outcoef <-

Dear Friends, I have a matrix of size 5000 X 20. The first two columns are indicator variables taking the value of either 0 or 1. Let us call the first two columns Y1 and Y2. I need to randomly sample 1000 rows with all the associated columns, in other words my new matrix should be of size 1000 X 20. I realize that using this command newmat <- mainmat[sample(1000,replace=F),] achieves

Dear Friends, My objective is to do element wise multiplication of two vectors. For example suppose I have a <- (1,1,1) b <- (2,4) My output should be (2,4,2,4,2,4). I managed to write it down with loops as follows r <- c(1,1,1) l <- c(2,4) x <- 1 for (j in 1:3) { for (i in 1:2) { new[x,] <- r[j]*l[i] x <- x+1 } } Is there a simpler solution to

Dear friends, Please find below the code that I have employed for a rejection sampler to draw from asymmetric laplace distributions. I was wondering if this code can be written more efficiently? Are there more efficient ways of drawing random numbers from asymmetric laplace distributions?? Thanks in advance for your help and have a great weekend. Regards Anup

 Dear R-helpers, My question is related to how to impose constraints when when sampling from a distribution. For example, suppose I'm sampling a vector from a multivariate normal distribution vbeta <- 100*diag(2) mbeta <- c(1,1) ans <- beta <- c(rmvnorm(1,mbeta,vbeta)) ans will thus be a vector with two elements. My question is how do I place a restriction on one of the

I have an outlier that I would still like to display, but would prefer to shorten the axis. For example, display 0% - 40%, and 90% - 100%. Is this possible? I am using an xyplot. Thanks Murray -- Murray Pung Statistician, Datapharm Australia Pty Ltd 0404 273 283 [[alternative HTML version deleted]]

Dear Friends, I'm using the optim command to maximize a likelihood function. My optim command is as follows estim.out <- optim(beta, loglike, X=Xmain, Y=Y, hessian=T, method="BFGS", control=c(fnscale=-1, trace=1, REPORT=1)) Setting the report=1, gives me the likelihood function value (if i'm correct) at each step. The output from running this is as follows initial value