similar to: why is generating the same graph???

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]]

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

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 __________________________________________________________________

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'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

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],

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:

This is no doubt trivial but after searching the help files and the web, I cannot seem to find it. 1) How do I convert 'hgt' into 'HGT' in R? 2) How should I have used the help facilities to find this? At the end of the day, all I want to do is case insensitive string matching... i.e. 'if ("HGT" == 'hgt') print('this should be true')' I tried

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))

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

Hi, there! I'm having kind this same problem https://stat.ethz.ch/pipermail/r-help/2008-October/178221.html but I want to display another plot of my data, which is a point with two arrows indicating confidence interval, in the same graph that I've just plotted another, but the "add=T" is not functioning, I'm getting the same error Warning messages: 1: In plot.window(...)

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))} >

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 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

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 list

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

How can I configure oracle to run under wine in linux. I tried copy the configuration as I had in windows, but did not work. Someone can help me ? thanks. piu.

This may be a begining question. If so, please bear with me. If I have some data that based on the historgram and other plots it "looks" like a beta distribution. Is there a function or functions within R to help me determine the model parameters for such a distirbution? Similarily for other "common" distirbutions, Poisson(lambda), Chi-Square(degrees of freedom, chi-square

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,