similar to: survreg.distributions() error

sorry for cross-posting Dear all, I have tow (several) bivariate distributions with a known mean and variance-covariance structure (hence a known density function) that I would like to compare in order to get an intersect that tells me something about "how different" these distributions are (as t-statistics for univariate distributions). In order to visualize what I mean hear a little

Hi, I'm using the package quadprog to solve the following quadratic programming problem. I want to minimize the function (b_1-b_2)^2+(b_3-b_4)^2 by the following constraints b_i, i=1,...,4: b_1+b_3=1 b_2+b_4=1 0.1<=b_1<=0.2 0.2<=b_2<=0.4 0.8<=b_3<=0.9 0.6<=b_4<=0.8 In my opinion the solution should be b_1=b_2=0.2 und b_3=b_4=0.8. Unfortunately R doesn't find

Dear R-user: I am using survreg(Surv()) for fitting a Tobit model of left-censored longitudinal data. For logarithmic transformation of y data, I am trying use survreg.distributions in the following way: tfit=survreg(Surv(y, y>=-5, type="left")~x + cluster(id), dist="gaussian", data=y.data, scale=0, weights=w) my.gaussian<-survreg.distributions$gaussian

Dear R Helpers, I am trying to carry out survival analysis in the presence of long term survivors (immunes/cureds). This involves using a split population model (some call it a mixture model) where the assumption of eventual failure is relaxed. I am following closely the formulation by Maller and Zhou (1996). Parametric modelling in this situation involves the introduction of a new

Dear R-Helpers: I have tried everything I can think of and hope not to appear too foolish when my error is pointed out to me. I have some real data (18 points) that look linear on a log-log plot so I used them for a comparison of lm() and survreg. There are no suspensions. survreg.df <- data.frame(Cycles=c(2009000, 577000, 145000, 376000, 37000, 979000, 17420000, 71065000, 46397000,

Dear R Listers- I am attempting to analyse the survival of seeds in cages (exclosures) that differ in their permeability to rainforest mammals. Because I did not observe the moment of seed disappearance, my data is interval censored. This limits my options for analysis (as I understand it) to survreg, in the survival package. Because I repeated the experiment in 8 sites, I have a random

Can anybody give me a neat example of interval censored data analysis codes in R? Given that suvreg(Surv(c(1,1,NA,3),c(2,NA,2,3),type="interval2")~1) works why does survreg(Surv(data[,1],data[,2],type="interval2")~1) not work where data is : T.1 T.2 Status 1 0.0000000 0.62873036 1 2 0.0000000 2.07039068 1 3 0.0000000

Dear R-users, I know that a few people have asked whether survreg handles left-truncation data and the reply that i have seen so far is that it does. However, when I try to use survreg on left-truncated data, I got the following error message. > survcs3<-survreg(Surv(start,end,status)~AG, data=DPONEcs3, dist="exponential") Error in survreg(Surv(start, end, status) ~ AG, data =

Hi, I would like to fit parametric survival models to time-to-event data that are left truncated. I have checked the help page for survreg and looked in the R-help archive, and it appears that the R function survreg from the survival library (version 2.16) should allow me to take account of left truncation. However, when I try the command

---------------------------- Original Message ---------------------------- Subject: Different results from survreg with version 2.6.1 and 2.10.1 From: nathalcs at ulrik.uio.no Date: Fri, March 19, 2010 16:00 To: r-help at r-project.org -------------------------------------------------------------------------- Dear all I'm using survreg command in package survival.

Dear R help, I've encountered a problem with survreg's anova(). I am currently writing general code to fit a variety of models using different fitting functions. Here's a simple example of what I'm trying to do: ---begin code--- # general function to analyse data analyse.data <- function(formula, FUN, data, ...) { fit <- FUN(formula, data=data, ...) anova(fit)

I am trying to use R to do loglogistic hazard estimation. My plan is to generate a loglogistic hazard sample data and then use survreg to estimate it. If everything is correct, survreg should return the parameters I have used to generate the sample data. I have written the following code to do a time invariant hazard estimation. The output of summary(modloglog) shows the factor loading of

Dear R-list I have been trying to make survreg fit a normal regression model with left truncated data, but unfortunately I am not able to figure out how to do it. The following survreg-call seems to work just fine when the observations are right censored: library(survival) n<-100000 #censored observations x<-rnorm(n) y<-rnorm(n,mean=x) d<-data.frame(x,y) d$ym<-pmin(y,0.5)

Hello all, I want to extract the survival function estimate from a model fitted by survreg(). Using predict.survreg(..., type="quantile", p=seq(0,1,0.001)), gives the quantiles, which I managed to turn around into a survival function estimate (Prob{T > t} as function of t). Is there a more straightforward way of doing this? I have had difficulties using pweibull() with the

Hey all, I am trying to use the survreg function in R to estimate the mean and standard deviation to come up with the MLE of alpha and lambda for the weibull distribution. I am doing the following: times<-c(10,13,18,19,23,30,36,38,54,56,59,75,93,97,104,107,107,107) censor<-c(1,0,0,1,0,1,1,0,0,0,1,1,1,1,0,1,0,0) survreg(Surv(times,censor),dist='weibull') and I get the following

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

Hello, I'm working with some left censored survival data using accelerated failure time models. I am interested in fitting different distributions to the data but seem to be getting the same results from the model fit using survreg regardless of the assumed distribution. These two codes seem to provide the same results: aft.gaussian <-

On Wed, 13 Nov 2002, Jan de Leeuw wrote: > > No problemo. And, in fact, I get the same results in > the R-1.6.0 Carbon version. I don't. Could there be a G3/G4 issue? -thomas > --- Jan > > On Wednesday, November 13, 2002, at 02:05 PM, tim@timcohn.com wrote: > > > Full_Name: Tim Cohn > > Version: 1.6.1 > > OS: Macintosh OS X > > Submission

1. The weibull is the only distribution that can be written in both a proportional hazazrds for and an accelerated failure time form. Survreg uses the latter. In an ACF model, we model the time to failure. Positive coefficients are good (longer time to death). In a PH model, we model the death rate. Positive coefficients are bad (higher death rate). You are not the first to be confused

R users, I am making model selection with an accelerated failure time model using the command survreg within the library survival. As I want to compare models with different probability distributions I need to have the full likelihood. How can I find out what survreg generates: the full likelihood or a likelihood with "unnecessary" constants dropped? Example I want to