similar to: survreg 3-way interaction

Hi, I'm a bit puzzled by the gamlss fitting of exponential and lognormal data. Gamlss seems to think that exponentially distributed data fits better with a lognormal distribution, and vice versa. For example, X <- rexp(1000) Gexp <- gamlss(X~1,family=EXP) # X~1 is X tilde 1 GAMLSS-RS iteration 1: Global Deviance = 2037.825 GAMLSS-RS iteration 2: Global Deviance = 2037.825 Glno

I have hit a problem with the design of the mcmc package I can't figure out, possibly because I don't really understand the R function call mechanism. The function metrop in the mcmc package has a ... argument that it passes to one or two user-supplied functions, which are other arguments to metrop. When the two functions don't have the same arguments, this doesn't work.

Hello list, I'm trying to make sure that I'm performing a path analysis correctly using the sem package. the figure at http://flame.cs.dal.ca/~sstewart/regressDiag.png has a detailing of the model. The challenge I'm having is that reuse is an indicator (0/1) variable. Here's the code I'm using: corr =

Dear R helpers, I have following loss data and I need to fit LEFT truncated Log Normal distribution to this data which is Truncated at 1000000. dat = c(1333834,5710254,9987567,7809469,6940935,3473671,1270209,1102523,1124002, 5830159,4302300,3925242,2638409,2324421,7238436,9088709,7439250,4976551,4864319, 8741334,1863770,7098310,4942288,4971829,4986372) library(gamlss.tr) gen.trun(5, LOGNO)

I have the following questions about the variance of the random effects in the survreg() function in the survival package: 1) How can I extract the variance of the random effects after fitting a model? For example: set.seed(1007) x <- runif(100) m <- rnorm(10, mean = 1, sd =2) mu <- rep(m, rep(10,10)) test1 <- data.frame(Time = qsurvreg(x, mean = mu, scale= 0.5, distribution =

Dear R-users, I have noticed small discrepencies in the reported estimate of the variance of the frailty by the print method for survreg() and the 'theta' component included in the object fit: # Examples in R-2.6.2 for Windows library(survival) # version 2.34-1 (2008-03-31) # discrepancy fit1 <- survreg(Surv(time, status) ~ rx + frailty(litter), rats) fit1 fit1$history[[1]]$theta

I am trying to model survival data with a Weibull distribution using survreg. Units are clustered two apiece, sometimes receiving the same treatment and sometimes opposing treatment.

Does someone know the rules by which 'coxph' returns 'frail', the predicted frailty terms? In my test function: ----------------------------------------------- fr <- function(){ #testing(frailty terms in 'survival' require(survival) dat <- data.frame(exit = 1:6, event = rep(1, 6), x = rep(c(0, 1), 3),

Hi all, I am trying to run Weibull PH model in R. Assume in the data set I have x1 a continuous variable and x2 a categorical variable with two classes (0= sick and 1= healthy). I fit the model in the following way. Test=survreg(Surv(time,cens)~ x1+x2,dist="weibull") My questions are 1. Is it Weibull PH model or Weibull AFT model? Call: survreg(formula = Surv(time, delta) ~ x1

I am using psm to model some parametric survival data, the data is for length of stay in an emergency department. There are several ways a patient's stay in the emergency department can end (discharge, admit, etc..) so I am looking at modeling the effects of several covariates on the various outcomes. Initially I am trying to fit a survival model for each type of outcome using the psm

Henrik wrote: I think what people are also thinking about is that the policy for publishing a package on CRAN is that it have to pass R CMD check with no errors, warnings *or* notes. So, in that sense notes are no different from warnings. --------------------------------- Getting rid of these notes would be very hard in the survival package. The population survival routines (survexp, pyears)

Hi, I'm using survreg() from the survival package for parametric survival regression (modelling inter-arrival times of patients to a waiting list as exponentially distributed, with various regressors such as queue size and season). Does anyone know which algorithm survreg() uses for this? Thanks, Gad -- Gad Abraham Department of Mathematics and Statistics The University of Melbourne

Dear all, I am using gamlss (Package gamlss version 4.0-0, R version 2.10.1, Windows XP Service Pack 3 on a HP EliteBook) to relate bird counts to habit variables. However, most models fail because “the global deviance is increasing” and I am not sure what causes this behaviour. The dataset consists of counts of birds (duck) and 5 habit variables measured in the field (n= 182). The dependent

Dear all I am trying to build a package in Windows. I use the following command (which it used to work with previous versions ) and I am getting the following error #-------------------------------------------------------------------------------------------------------------- C:\PROGRA~1\R\rw2000\bin>Rcmd build --binary --use-zip C:\PROGRA~1\R\rw2000\src\library\gamlss * checking for

Dear all I am trying to create a package in R 1.9.0 and I a getting an error message which I do not understand. (I am using R in Windows XP and 2000) For example the following works well in 1.8.1 C:\Program Files\R\rw1081\src\gnuwin32>make pkg-gamlss ---------- Making package gamlss ------------ adding build stamp to DESCRIPTION installing inst files installing indices not zipping

Hola a tod en s, tengo una duda que la comunidad R me puede ayudar. Estoy trabajando con gamlss, porque tengo una variable respuesta con valores entre 0 y 1 e incluidos estos. La distribución que utilizo com gamlss para este caso es "beta inflated" (Stasinopulos and Rigby 2007. Journal of Statistical Software 23(7)). El modelo que intento correr es: m1<-gamlss(Teleosteos ~

Dear all and Mikis I have the opportunity to compare fits with the 'classical' glm and gamlss and no smoother of any kind just the same model formula (both with the binomial family). I get exactly the same coefficients but very different residuals, gamlss giving residuals which are extremely close to 'normal' and glm very far... How can this be ? Thanks in advance for

Dear List, How do I extract the approximate Wald test for the frailty (in the following example 17.89 value)? What about the P-values, other Chisq, DF, se(coef) and se2? How can they be extracted? ######################################################> kfitm1 Call: coxph(formula = Surv(time, status) ~ age + sex + disease + frailty(id, dist = "gauss"), data = kidney)

Hello all: Please help me with this "can't find object" issue. I'm trying to get leave-one-out predicted values for Beta-binomial regression. It may be the gamlss issue because the code seems to work when %do% is used. I have searched for similar issues, but haven't managed to figure it out. This is on Windows 10 platform. Thanks in advance, Nik #

Dear R users: I'm generating the following survival data: set.seed(123) n=200 #sample size x=rbinom(n,size=1,prob=.5) #binomial treatment v=rgamma(n,shape=1,scale=1) #gamma frailty w=rweibull(n,shape=1,scale=1) #Weibull deviates b=-log(2) #treatment's slope t=exp( -x*b -log(v) + log(w) ) #failure times c=rep(1,n) #uncensored indicator id=seq(1:n) #individual frailty indicator