similar to: error in dweibull (PR#405)

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

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.

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))

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],

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))} >

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)

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

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)

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]]

my problem actually arised with fitting the data to the weibulldistribution, where it is hard to see, if the proposed parameterestimates make sense. data1:2743;4678;21427;6194;10286;1505;12811;2161;6853;2625;14542;694;11491; ?? ?? ?? ?? ?? 14924;28640;17097;2136;5308;3477;91301;11488;3860;64114;14334 how am I supposed to know what starting values i have to take? i get different

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

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 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

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

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 R-experts: I am using survreg() to estimate the parameters of a Weibull density having right-censored observations. Some observations are weighted. To do that I regress the weighed observations against a column of ones. When I enter the data as 37 weighted observations, the parameter estimates are exactly the same as when I enter the data as the corresponding 70 unweighted observations.

Hello, I am a new user of R and found the function dtnorm() in the package msm. My problem now is, that it is not possible for me to get the mean and sd out of a sample when I want a left-truncated normal distribution starting at "0". fitdistr(x,dtnorm, start=list(mean=0, sd=1)) returns the error message "Fehler in "[<-"(`*tmp*`, x >= lower & x <= upper,

I'm having trouble figuring out how to create a function using "body<-" (). The help file for body() says that the argument should be a list of R expressions. However if I try that I get an error: > tmpfun <- function(a, b=2){} > body(tmpfun) <- list(expression(z <- a + b),expression(z^2)) Error in as.function.default(c(formals(f), value), envir) :

Dear all, I need to plot an cumulative distribution plot of a variable and then to fit a distribution to that, probably a weibull or lognormal. I have plotted the ecdf as > plot(ecdf(x)) but I haven't managed to fit the distribution. I have as well attached the data. I would appreciate if you could help me on that. Thank you. Kind regards Maria -------------- next part --------------