similar to: How to use results of distribution fitting for further processing?

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

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, 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

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

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))

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)

Good-day Sir, I am R.Language users but am try to? estimate parameter of beta distribution particular dataset but give this error, which is not clear to me: (Initial value in "vmmin" is not finite) beta.fit <- fitdistr(data,densfun=dbeta,shape1=value , shape2=value) kindly assist. expecting your reply:

Please guide me through to resolve the error message that I get this is what i have done. >x1<- rnorm(100,2,1) >x1fitbeta<-fitdistr(x1,"beta") Error in fitdistr(x1, "beta") : 'start' must be a named list Yes, I do understand that sometime for the distribution to converge to the given set of data, it requires initial parameters of the distribution, to

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,

Hi, I have some data, that when plotted looks very close to a log-normal distribution. My goal is to build a regression model to test how this variable responds to several independent variables. To do this, I want to use the fitdistr tool from the MASS package to see how well my data fits the actual distribution, and also build a generalized linear model using the glm command. The summary

I have split my original dataframe to generate a list of dataframes each of which has 3 columns of factors and a 4th column of numeric data. I would like to use lapply to apply the fitdistr() function to only the 4th column (x$isi) of the dataframes in the list. Is there a way to do this or am I misusing lapply? As a second solution I tried splitting only the numeric data column to yield a

Dear R developers, I have encountered strange behaviour of fitdistr for gamma in recent R build i.e. 2.2.0. I have attached the code for data at the end of this mail so you can reproduce the problem. In short, I am able to run fitdistr under 2.1.0 without problems, while I get the following error under 2.2.0 (Version 2.2.0 Patched (2005-11-15 r36348)) > fitdistr(otm, "gamma") Error

I have split my original dataframe to generate a list of dataframes each of which has 3 columns of factors and a 4th column of numeric data. I would like to use lapply to apply the fitdistr() function to only the 4th column (x$isi) of the dataframes in the list. Is there a way to do this or am I misusing lapply? As a second solution I tried splitting only the numeric data column to yield a list

Hello. I am not certain even how to search the archives for this particular question, so if there is an obvious answer, please smack me with a large halibut and send me to the URLs. I have been experimenting with fitting curves by using both maximum likelihood and maximum spacing estimation techniques. Originally, I have been writing distribution-specific functions in 'R' which work

Hi all, I'm using the function "fitdistr" (library MASS) to fit a distribution to given data. What I have to do further, is getting the log-Likelihood-Value from this estimation. Is there any simple possibility to realize it? Regards, Carsten

Dear all I've got the dataset data:2743;4678;21427;6194;10286;1505;12811;2161;6853;2625;14542;694;11491; ?? ?? ?? ?? ?? 14924;28640;17097;2136;5308;3477;91301;11488;3860;64114;14334 I know from other testing that it should be possible to fit the data with the exponentialdistribution. I tried to get parameterestimates for the exponentialdistribution with R, but as the values of the parameter

Dear all, I have a set of data which seem to be distributed almost exponentially but only on [0;1]. I guess that the probability distribution in this case would look like \frac{lambda}{1-e^{-\lambda}) e^{-\lambda x} I would like to use fitdistr to estimate the value of \lambda. 1) Would it be correct first to find lambda for the exponential distribution and then to substitute it in the formula

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))} >

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

Dear R-users I've got the next problem: I've got this *function*: fitcond=function(x,densfun,pcorte,start,...){ myfn <- function(parm,x,pcorte,...) -sum(log(dens(parm,x,pcorte,...))) Call <- match.call(expand.dots = TRUE) if (missing(start)) start <- NULL dots <- names(list(...)) dots <- dots[!is.element(dots, c("upper",