similar to: Weibull Likelihod function.

Displaying 20 results from an estimated 2000 matches similar to: "Weibull Likelihod function."

2012 Feb 21
3
HELP ERROR Weibull values must be > 0
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]]
2009 Apr 30
1
finite mixture model (2-component Weibull): plotting Weibull components?
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
2001 Aug 28
2
Estimating Weibull Distribution Parameters - very basic question
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
2003 Oct 20
1
Fitting a Weibull/NaNs
I'm trying to fit a Weibull distribution to some data via maximum likelihood estimation. I'm following the procedure described by Doug Bates in his "Using Open Source Software to Teach Mathematical Statistics" but I keep getting warnings about NaNs being converted to maximum positive value: > llfunc <- function (x) { -sum(dweibull(AM,shape=x[1],scale=x[2], log=TRUE))} >
2008 Apr 15
1
Weibull
Dear R users, This is a basic question. I want to fit a Weibull distribution. fitdistr(data, "weibull") works and it is a maximum likelihood fitting. Is it a good method ? Or is it better to write a function for the log-likelihood and the gradient and to use a numerical routine ? Fitdistr works for uncensored data, but what can I use for censored (and uncensored) data ? Thank you
2002 Jan 17
1
weibull in R
Hi all I try to make a weibull survival analysis on R. I know make this on GLIM, and now I try to make the GLIM exercice GLEX8 on R to learning and compare the test. The variables are: time censor group bodymass In GLIM I make: $calc %s=1 $ to fit weibull rather than exponential $input %pcl weibull $ $macro model group*bodymass $endmac$ $use weibull t w %s $ Then, GLIM estimate an alpha for the
2011 Oct 28
1
weibull fitdistr problem: optimization failed
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
2004 Jan 14
1
estimation of lambda and gamma with std errors for a weibull model
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
2011 Apr 27
3
MASS fitdistr with plyr or data.table?
I am trying to extract the shape and scale parameters of a wind speed distribution for different sites. I can do this in a clunky way, but I was hoping to find a way using data.table or plyr. However, when I try I am met with the following: set.seed(144) weib.dist<-rweibull(10000,shape=3,scale=8) weib.test<-data.table(cbind(1:10,weib.dist))
2008 Oct 07
3
Fitting weibull, exponential and lognormal distributions to left-truncated data.
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
2001 Jul 02
1
nls newbie: help approximating Weibull distribution
Hi folks, I tried to retain the Weibull distribution using the `nls' function and proceeding along the lines of the example provided in the `SSweibull' help (at least I thought so): t <- (1:200)/100 v <- pweibull(t, shape=3, scale=1) df <- data.frame(Time=t, Value=v) Asym <- 1.0; Drop <- 1.0; lrc <- 0; pwr <- 1 df.estimate <- nls(Value ~ SSweibull(Time,
2012 Feb 21
5
help error: In dweibull(x, shape, scale, log) : NaNs produzidos
Guys, I'm having an error when I use the command: library(MASS)> dados<-read.table("inverno.txt",header=FALSE)> vento50<-fitdistr(dados[[1]],densfun="weibull")Mensagens de aviso perdidas:1: In dweibull(x, shape, scale, log) : NaNs produzidos2: In dweibull(x, shape, scale, log) : NaNs produzidos3: In dweibull(x, shape, scale, log) : NaNs produzidos4: In
2002 Apr 05
2
weighted 2 or 3 parameter weibull estimation?
I've figured out how to use optim (barely) to estimate 2 parameter = weibull distributions. I can't get over how easy this is. What I need to = do is use a weight in the observations..... For example,=20 the tree diameters and weights are are=20 4.70 , 100 6.00, 98 7.10, 75.0 8.10, 86.3 8.60, 80.456 8.90, 20.5 9.50, 16.6 11.40, 12.657 11.80, 12.47 14.50,
2012 Jan 29
1
r-help; weibull parameter estimate
Hello, If i write a function as below using log of weibull distribution i do not get the required results in estimating the parameters what do i do, please a/b * (t/b)^a-1 * exp(-t/b)^a n=500 x<-rweibull(n,2,2) z<-function(p) {(-n*log(p[1])+n*log(p[2])- (p[1]-1)*sum(log(x))+(p[1]-1)*log(p[2])+(sum(x/p[2])^(p[1]))  )} zz<-optim(c(0.5,0.5),z) zz [[alternative HTML version deleted]]
2011 Oct 20
1
R code Error : Hybrid Censored Weibull Distribution
Dear Sir/madam, I'm getting a problem with a R-code which calculate Fisher Information Matrix for Hybrid Censored Weibull Distribution. My problem is that: when I take weibull(scale=1,shape=2) { i.e shape>1} I got my desired result but when I take weibull(scale=1,shape=0.5) { i.e shape<1} it gives error : Error in integrate(int2, lower = 0, upper = t) : the integral is probably
2005 Jul 22
1
multiplicate 2 functions
Thks for your answer, here is an exemple of what i do with the errors in french... > tmp [1] 200 150 245 125 134 345 320 450 678 > beta18 Erreur : Objet "beta18" not found //NORMAL just to show it > eta [1] 500 > func1<-function(beta18) dweibull(tmp[1],beta18,eta) > func1<-func1(beta18) * function(beta18) dweibull(tmp[2],beta18,eta) Erreur dans dweibull(tmp[1],
2004 Nov 23
6
Weibull survival regression
Dear R users, Please can you help me with a relatively straightforward problem that I am struggling with? I am simply trying to plot a baseline survivor and hazard function for a simple data set of lung cancer survival where `futime' is follow up time in months and status is 1=dead and 0=alive. Using the survival package: lung.wbs <- survreg( Surv(futime, status)~ 1, data=lung,
2011 Nov 03
1
Fit continuous distribution to truncated empirical values
Hi all, I am trying to fit a distribution to some data about survival times. I am interested only in a specific interval, e.g., while the data lies in the interval (0,...., 600), I want the best for the interval (0,..., 24). I have tried both fitdistr (MASS package) and fitdist (from the fitdistrplus package), but I could not get them working, e.g. fitdistr(left, "weibull", upper=24)
2012 Feb 23
3
why is generating the same graph???
Hi, why my script iss always generating the same graph?when I change the parameters and the name of text file? library(MASS) dados<-read.table("inverno.txt",header=FALSE) vento50<-fitdistr(dados[[1]],densfun="weibull") png(filename="invernoRG.png",width=800,height=600) hist(dados[[1]], seq(0, 18, 0.5), prob=TRUE, xlab="Velocidade
1999 Dec 09
1
nlm() problem or MLE problem?
I am trying to do a MLE fit of the weibull to some data, which I attach. fitweibull<-function() { rt<-scan("r/rt/data2/triam1.dat") rt<-sort(rt) plot(rt,ppoints(rt)) a<-9 b<-.27 fn<-function(p) -sum( log(dweibull(rt,p[1],p[2])) ) cat("starting -log like=",fn(c(a,b)),"\n") out<-nlm(fn,p=c(a,b), hessian=TRUE)