similar to: weibull in R

Hi, I posted a question (bellow) a few weeks ago and had a reply (thanks Christian) that partly solves the problem, but I still would like to be able to restrict some of the independent variables in a nls model to be always >0, (is there a way to do it)?? Thanks, Angel >From: "Christian Ritz" <ritz at dina.kvl.dk> >To: "Angel -" <angel_lul at

I was wondering if someone familiar with survival analysis can help me with the following. I would like to fit a Weibull curve, that may be dependent on a covariate, my dataframe "labdata" that has the fields "cov", "time", and "censor". Do I do the following? wieb<-survreg(Surv(labdata$time, labadata$censor)~labdata$cov,

After some futile searches, I decided to ask the list to see if any of the sages out there would have an answer: I have a function I wrote a few years ago in S, which calls glim numerous times. I'd like to port it to R, but glm works differently from glim, which takes as part of its input an X design matrix. I probably could write a function to convert glim to glm, but hope this

Dear Knowledgeable R Community Members, Please excuse my ignorance, I apologize in advance if this is an easy question, but I am a bit stumped and could use a little guidance. I have a finite mixture modeling problem -- for example, a 2-component Weibull mixture -- where the components have a large overlap, and I am trying to adapt the "mclust" package which concern to normal

GUYS, I NEED HELP WITH ERROR: library(MASS) > dados<-read.table("mediaRGinverno.txt",header=FALSE) > vento50<-fitdistr(dados[[1]],densfun="weibull") Erro em fitdistr(dados[[1]], densfun = "weibull") : Weibull values must be > 0 WHY RETURN THIS ERROR? WHAT CAN I DO? BEST REGARDS [[alternative HTML version deleted]]

Hi, I am trying to fit a sigmoid curve to some data with nls but I am getting into some trouble. Seems that the optimization method is getting down to some parameter estimates that make the equation unsolvable. This is an example: >growth<-data.frame(Time=c(5,7,9,11,13,15,17,19,21,23,25,27),BodyMass=c(45,85,125,210,300,485,570,700,830,940,1030,1120))

I'm getting errors when running what seems to be a simple Weibull distribution function: This works: x <- c(23,19,37,38,40,36,172,48,113,90,54,104,90,54,157,51,77,78,144,34,29,45,16,15,37,218,170,44,121) rate <- c(.01,.02,.04,.05,.1,.2,.3,.4,.5,.8,.9) year <- c(100,50,25,20,10,5,3.3,2.5,2,1.2,1.1) library(MASS) x <- sort(x) tryCatch( f<-fitdistr(x, 'weibull'), error

Hello, is there a quick way of estimating Weibull parameters for some data points that are assumed to be Weibull-distributed? I guess I'm just too lazy to set up a Maximum-Likelihood estimation... ...but maybe there is a simpler way? Thanks for any hint (and yes, I've read help(Weibull) ;) Kaspar Pflugshaupt -- Kaspar Pflugshaupt Geobotanical Institute ETH Zurich, Switzerland

Dear All, I have two questions regarding distribution fitting. I have several datasets, all left-truncated at x=1, that I am attempting to fit distributions to (lognormal, weibull and exponential). I had been using fitdistr in the MASS package as follows: fitdistr<-(x,"weibull") However, this does not take into consideration the truncation at x=1. I read another posting in this

Hi I have been asked to provide Weibull parameters from a paper using Kaplan Meir survival analysis. This is something I am not familiar with. The survival analysis in R works nicely and is the same as commercial software (only the graphs are superior in R). The Weibull does not and produces an error (see below). Any ideas why this error should occur? My approach may be spurious.

Hello, I'm quite new to R and want to make a Weibull-regression with the survival package. I know how to build my "Surv"-object and how to make a standard-weibull regression with "survreg". However, I want to fit a translated or 3-parametric weibull dist to account for a failure-free time. I think I would need a new object in survreg.distributions, but I don't know how

Hi I am using R to fit a survival function to my data (with a weibull distribution). Data: Survival of individuals in relation to 4 treatments ('a','b','c','g') syntax: ---- > survreg(Surv(date2)~males2, dist='weibull') But I have some problems interpreting the outcome and getting the parameters for each curve. --------- Value Std.

Hello, I would like to fit a mixture of two Weibull distributions to my data, estimate the model parameters, and compare the fit of the model to that of a single Weibull distribution. I have used the mix() function in the 'mixdist' package to fit the mixed distribution, and have got the parameter estimates, however, I have not been able to get the log-likelihood for the fit of this model

Hi All, I have conducted the following survival analysis which appears to be OK (thanks BRipley for solving my earlier problem). > surv.mod1 <- survreg( Surv(timep1, relall6)~randgrpc, data=Dataset, dist="weibull", scale = 1) > summary(surv.mod1) Call: survreg(formula = Surv(timep1, relall6) ~ randgrpc, data = Dataset, dist = "weibull", scale = 1)

Hi, I was wondering if someone can help me interpret the results of running weibreg. I run the following and get the following R output. > weibreg(Surv(time, censor)~covar) fit$fail = 0 Call: weibreg(formula = Surv(time, censor)~covar) Covariate Mean Coef Rel.Risk L-R p Wald p covar 319.880 -0.002 0.998 0.000 log(scale) 0.000 8.239

Hi R-Community, I have a vector of Weibull distributed observations and I would like to estimate the parameters "shape" and "scale" of the Weibull distribution. Is there a way to do this in R? Much thanks in advance, Hagen Schm?ller -- ----------------------------------------------------------------------- Dipl.-Ing. Hagen K. Schm?ller Institut f?r Elektrische Anlagen und

Dear R experts, How should lambda and gamma (with std.errors) be calculated for a weibull model with age as an independent predictor? I have assumed that this can be done with survreg with e. g. (summary(survreg(Surv(time, status) ~ age, dist = 'weibull')) ) and predict.survreg with e.g. (predict(model, se.fit = T, newdata = data.frame(age = seq(50, 80, 5)) but unfortunately I'm

Hi all, In my continuous transition of GLIM to R I try to make a glm with binomial errors. The data file have 3 vectors: h -> the factor that is ajusted (have 3 levels) d -> number of animais alive (the response) n -> total number of animals To test proportion of alive, make d/n. In GLIM: $yvar d$ $error binomial n$ $fit +h$ scale deviance = 25.730 (change = -9.138) at cycle 4

Hello, How can I do a test of two weibull distributions? I have two weibull distribution sets from two wind datasets in order to check whether they are same. I thought 2 sample t-test would be applicable but I couldn't find any ways to do that on the Internet. Does anyone know what type of test is applicable to my purpose? and what R function can you recommend? Plus, if it turned out that

Dear R-users I have some datasets, all left-censoring, and I would like to fit distributions to (weibull,exponential, etc..). I read one solution using the function survreg in the survival package. i.e survreg(Surv(...)~1, dist="weibull") but it returns only the scale parameter. Does anyone know how to successfully fit the exponential, weibull etc... distributions to left-censoring