similar to: specify seed for Random Number Generator

I have a dataset with transactions and a timestamp at which they occoured during a day. The time stamp is in the format YYYY/MM/DD hh:mm:ss. I would like to plot a timeseries of the transactions to see if there is a particular time in the day when there is a spike in transactions. Ofcourse the YYYY/MM/DD can be dropped since I am monitoring activity for the day and the actual date is

Hello: Are there any libraries that will do a subset selection for glm's? I looked through leaps, but seems like it is specifically for linear regressions. Thank you. -Dhiren

Hello: I have a matrix 'dat' with 2 columns. I have the following code: for (i in 1:nrows(dat)) { if (dat[i,1] < dat[i,2]) { dat[i,2]<-0 } else { dat[i,2]<-1 } Is there a way to accomplish this without the for loop? Thank you. -Dhiren

Hi: I am extremely new to linux so bare with the questions. I am trying to install the SJava package for R on linux (debian). I issue the R CMD INSTALL SJava_0.68-0.tar.gz and get the following error Cannot find Java. Please set your path to include the directory in which the java executable resides, or set the environment variable JAVA_HOME before this configure script is run. I have

Hi all, I'm trying to make a little script to determine an "unknown" rate for a number of known exponential trials. My Code: #Set Trials and generate number trials=100 rand<-runif(1,0,1) vector=0 #Generate vector of 100 random exponentials and sum them for (i in 1:100) { vector<-rexp(trials,rate=rand) } sumvect=sum(vector) #Create the log likelihood function

Does anyone know if R have any problems with the exponential random number generation (function rexp)? I comment it because I executed data<-sort(rexp(100)) plot(data,dexp(data)/(1-pexp(data)),type="l") and the graphic isn't constant. (Note: exponential distribution have a constant hazard failure rate). Thank you, Juan

I am using rexp to generate several exponential distributions. I am passing rexp a vector of rates , r. I am wanting to simulate a sample of size 200 for each rate so the code looks like: rexp(n=200*length(r),rate=r) this gives me a vector of the random exponential variables, but they are all disjointed b/c rexp goes through and simulates an exponential variable for each rate and it does that 200

dear list I use runif to generate a ramdom number between min and max runif(n, min=0, max=1) however , the syntaxe of rexp does not allow that rexp(n, rate = 1) and it generate a number with the corresponding rate. The question is: how to generate a number between min and max using rexp(). Regards -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000

Hi everybody, while performing ks.test for a standard exponential distribution on samples of dimension 2500, generated everytime as new, i had this strange behaviour: >data<-rexp(2500,0.4) >ks.test(data,"pexp",0.4) One-sample Kolmogorov-Smirnov test data: data D = 0.0147, p-value = 0.6549 alternative hypothesis: two.sided >data<-rexp(2500,0.4)

Hi, Wondering if anyone could help me out with this error.Im trying to fill a matrix with random numbers taken from an exponential distribution using a loop: x.3<-matrix(rep(0,3000),nrow=1000,byrow=T)for(i in 1:1000){x[i,]<-rexp(3,rate=2/3)} I get the error message: Error in x[i, ] <- rexp(3, rate = 2/3) : incorrect number of subscripts on matrix Any ideas??? Appreciate any thoughts.

l want to create a column of 1 and 2 randomly what command should l use eg treatment strata 1 1 2 0 1 1 2 1 2 0 2 1 2 0 1 0 these should be created randomly secondly if l have

Hello, I have written a function that demonstrates the CLT by generating samples following the exponential distribution, calculating the means, plotting the histogram, and drawing the limiting normal curve as an overlay. I have the title of each histogram state the sample size and rate (1/theta) for the exponential (the output is actually 4 histograms), but I can't get the greek letter theta

I have some data measured with a coarsely-quantized clock. Let's say the real data are q<- sort(rexp(100,.5)) The quantized form is floor(q), so a simple quantile plot of one against the other can be calculated using: plot(q,type="l"); points(floor(q),col="red") which of course shows the characteristic stair-step. I would like to smooth the quantized

Dear Mr. for writing program about Gibbs sampling, i have a question. if i want to generate data from Exponential distribution but range of X is restricted, how can i do? regards, A.Rashidi   [[alternative HTML version deleted]]

Any one know is there any package or function to generate bivariate exponential distribution? I gusee there should be three parameters, two rate parameters and one correlation parameter. I just did not find any function available on R. Any suggestion is appreciated. -- View this message in context:

I am trying to generate survival data using R .Im trying to randomly generate a column of 1s and 0 and another column randomly generated using an exponential distribution but l cant seem to get the random function. how do l go about it thanks in advance rt chiruka --------------------------------- [[alternative HTML version deleted]]

Hi R community, I have some data set and construct the likelihood as follows likelihood <- function(alpha,beta){ lh<-1 d<-0 p<-0 k<-NULL data<-read.table("epidemic.txt",header = TRUE) attach(data, warn.conflicts = F) k <-which(inftime==1) d <- (sqrt((x-x[k])^2+(y-y[k])^2))^(-beta) p<-1 - exp(-alpha*d) for(i in 1:100){

I have a string '982323.1' and would like to replace everything after the '.' with a '41'. So the string should look like '982323.41'. The code I use to do this is sub('\.$','41',982323.1) This works fine as long as there is only 1 digit after the decimal. If I have '982323.10', then the result of the code is '982323.141'

Hi, At each iteration in my program,I need to generate tree vectors,X1,X2,X3, from exponential distribution with parameters a1,a2,a3. Can you help me please how can I do it such that it take a little time? thank you khazaei

Hello, I want to estimate the exponential parameter by using?optim?with the following input, where t contains 40% of the data and q contains 60% of the data within an interval. In implementing the code command for optim i want it to contain both the t and q data so i can obtain the correct estimate. Is there any suggestion as to how this can be done. I have tried h<-c(t,q) but it is not working