similar to: survreg asking for parameter "special"

Hello! I have used read.csv to read in a data frame, and there are a few variables in it, however, when I tried is.list(data$V1) >FALSE In fact, I have tried, they are not vectors either. I'm wondering: 1. What objects are these "lists" of data? 2. How could I find out about the type/inheritence of an object in general? 3. The reason I want it to be a list or vector, is that I

Hi, I hope somebody can help me with this issue: I am doing population pyramids using the barplot command, so in the left side I have male age structure and in the right side the female age structure. To plot the male age structure I put the data in negative numbers. Now, I want to change the sign in the bar plot in such way that I have no-sign numbers, both in left and right side of the graph. I

I have a matrix containing means and CIs (lower and upper in two columns, so three columns for every data point) for several points. I have to build a graph of these means accompained by the CIs (as wiskers). No problems with making the graph of means, but I don't know how to introduce CIs. Can anybody advise? -- View this message in context:

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,

Hello, I'm trying to do cor(x1,x2) and I get the following error: Error in cor.default(x1, x2) : missing observations in cov/cor A few things: 1. I've used cor() many times and have never encountered this error. 2. length(x1) = length(x2) 3. is.numeric(x1) = is.numeric(x2) = TRUE 4. which(is.na(x1)) = which(is.na(x2)) = integer(0) {the same goes for is.nan()} 5. I also try

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-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

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

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 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

Hi all, Sorry if this has been answered already, but I couldn't find it in the archives or general internet. Is it possible to implement the gompertz distribution as survreg.distribution to use with survreg of the survival library? I haven't found anything and recent attempts from my side weren't succefull so far. I know that other packages like 'eha' and

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)

The myoglobin sequence, with reference number NM_005368 in Gen bank, has the following frequencies of DNA nucleotides: A C G T 237 278 309 242 Do these data provide sufficient evidence, at the 1% level of significance, that the DNA nucleotides have an unequal distribution, that is the DNA nucleotides are not evenly utilised? Clearly state your hypothesis, test statistic and conclusion.

Hi, the standard errors of the coefficients in two regressions that I computed by hand and using lm() differ by about 1%. Can somebody help me to identify the source of this difference? The coefficient estimates are the same, but the standard errors differ. ####Simulate data happiness=0 income=0 gender=(rep(c(0,1,1,0),25)) for(i in 1:100){ happiness[i]=1000+i+rnorm(1,0,40)

I am trying to fit a lognormal distribution on interval-censored data. Some of my intervals have a lower bound of zero. Unfortunately, it seems like survreg() cannot deal with lower bounds of zero, despite the fact that plnorm(0)==0 and pnorm(-Inf)==0 are well defined. Below is a short example to reproduce the problem. Does anyone know why survreg() must behave that way? Is there an alternate

Full_Name: Tim Cohn Version: 1.6.1 OS: Macintosh OS X Submission from: (NULL) (130.11.34.250) The Mac version of survreg does not handle left-censored data correctly (at least the results are not what I get doing it other ways, and they are not the same as I get running R 1.6.1 in Windows 98se; the Windows 98 results are correct). On the windows version of R 1.6.1. >

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 <-

Hi So my particular problem is this: I have a row vector of length 5200 elements - specifically created by x<-rbinom(5200,1,0.5) y<-matrix(x,nrow=1,ncol=5200) y now, each element is either a 0 or a 1 - e.g. it could be (0,1,1,1,1,0,0,0,1,1,1) e.t.c. when the element is a 1, i need to multiply a number (say 1000) by 1.005, and if it is 1 again, multiply it _again_ by 1.005. so for

Hi, I'm estimating a loglogistic aft (accelerated failure time) model, just a simple plain vanilla one (without time dependent covariates), I'm comparing the results that I obtain between aftreg (eha package) and survreg(surv package). If I don't use any covariate the results are identical , if I add covariates all the coefficients are the same until a precision of 10^4 or 10^-5 except

Dear R help list, I am modeling some survival data with coxph and survreg (dist='weibull') using package survival. I have 2 problems: 1) I do not understand how to interpret the regression coefficients in the survreg output and it is not clear, for me, from ?survreg.objects how to. Here is an example of the codes that points out my problem: - data is stc1 - the factor is dichotomous