similar to: Fitting a Weibull/NaNs

I've been using R to do logistic regresssion, and that's working well, but there are two things I haven't figured out how to do. (1) Is there some pre-existing function that will let you compute the odds ratios and confidence intervals for them for a specific fit. I know how to do this manually or even write a function that I can call with the coefficients and se, but

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

I've noticed that glm and lrm give extremely different results if you attempt to fit a saturated model to a dataset with zero cells. Consider, for instance the data from, Agresti's Death Penalty example [0]. The crosstab table is: , , PENALTY = NO VIC DEF BLACK WHITE BLACK 97 52 WHITE 9 132 , , PENALTY = YES VIC DEF BLACK WHITE BLACK 6 11

Is there some package out there that implements functions for reliability analysis, especially for software reliability? In particular, I'm looking for: * Laplace factor (Cox & Lewis 1978) * Goel-Okumoto fitting Thanks in advance, -Ekr -- [Eric Rescorla ekr at rtfm.com]

Full_Name: G?ran Brostr?m Version: 2.3.1 OS: Linux, ubuntu Submission from: (NULL) (85.11.40.53) > dweibull(0, 0.5, 1) [1] NaN Warning message: NaNs produced in: dweibull(x, shape, scale, log) should give Inf (and no Warning). Compare with > dgamma(0, 0.5, 1) [1] Inf This happens when 'shape' < 1.

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],

Frank E Harrell Jr wrote: > Eric Rescorla <ekr at rtfm.com> wrote: > >> (2) I'd like to compute goodness-of-fit statistics for my fit >> (Hosmer-Lemeshow, Pearson, etc.). I didn't see a package that >> did this. Have I missed one? > > Hosmer-Lemeshow has low power and relies on arbitrary binning of > predicted probabilities. The Hosmer-Le Cessie

Full_Name: R. Woodrow Setzer, Jr. Version: 0.90.1 OS: linux - Redhat 6.1 Submission from: (NULL) (165.247.155.206) dweibull(0,1,1) evaluates to 0; it should be 1. Note that dweibull(.Machine$double.eps) evaluates to 1. > dweibull(.01,1,1) [1] 0.9900498 > dweibull(.00001,1,1) [1] 0.99999 > dweibull(.Machine$double.eps,1,1) [1] 1 > dweibull(0,1,1) [1] 0

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))

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]]

R 2.3.0 Linux, SuSE 10.0 Hi I have two problems with mle - probably I am using it the wrong way so please let me know. I want to fit different distributions to an observed count of seeds and in the next step use AIC or BIC to identify the best distribution. But when I run the script below (which is part of my original script), I get one error message for the first call of mle: Error in

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

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

hi all, I need to generate a function inside a loop: tmp is an array for (i in 1:10) { func<- func * function(beta1) dweibull(tmp[i],beta1,eta) } because then i need to integrate this function on beta. I could have written this : func<-function(beta1) prod(dweibull(tmp,beta1,eta)) (with eta and beta1 set) but it is unplottable and no integrable... i could make it a bit different but

Dear R-users, Using Maximum-likelihood Fitting (fitdistr function) I've got the next error: > fitdistr(datos,"weibull",lower=0) Error in optim(x = c(1.4625e-06, 0.257854, 0.0001217545, 0.11421005, 0.028721576, : L-BFGS-B *needs finite values of 'fn' * where "datos" is a vector of length=1000 between 1.4625e-06 and 0.8867114 I add the lower argument in

This is probably simple, but I have a hard time finding the solution. Any help greatly appreciated.   I would like to use the results of fitdistr(z,densfun=dweibull,start=list(scale=1,shape=1)) for further processing.  How do I assign the values of scale and shape to b and a without manually entering the numbers?   TIA __________________________________________________________________

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

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

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

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)