similar to: HELP ERROR Weibull values must be > 0

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

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

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

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:

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

In english: Buddies, The Eucalyptus is a system to build a cluster and provide a service such as Amazon EC2 on its own structure. It is very promising, flexible, stable and well documented. The team of Eucalyptus just created an Internet election to choose which distributions will be officialy supported by it. I would ask everyone for voting in Debian! Let me explain why. I have

In english: Buddies, The Eucalyptus is a system to build a cluster and provide a service such as Amazon EC2 on its own structure. It is very promising, flexible, stable and well documented. The team of Eucalyptus just created an Internet election to choose which distributions will be officialy supported by it. I would ask everyone for voting in Debian! Let me explain why. I have

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

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

Hi R-Community, I have a vector of Weibull distributed observations and I would like to estimate the parameters "shape" and "scale" of the Weibull distribution. Is there a way to do this in R? Much thanks in advance, Hagen Schm?ller -- ----------------------------------------------------------------------- Dipl.-Ing. Hagen K. Schm?ller Institut f?r Elektrische Anlagen und

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

Buenos días para cada una/o. Acabo de suscribirme. Tengo una inquietud que vi tratada por partes (y usando distintos paquetes) en las consultas anteriores. ¿Cuál es la mejor forma de ajustar distribuciones datos a distribuciones de probabilidad teóricas? ¿Cuáles son los paquetes para hacerlo en forma ágil? Gracias por su atención. Saludos. César Escalante C.

I am planning to schedule a monthly XCP meeting for the community and am struggling with when to host the call. As we are a global community, there is no single optimal time to host the meeting. In an effort to support the most likely attendees, please send me your time zone if you plan to participate in these calls. I will track the most common time zones in an effort to maximize attendance. All

I am planning to schedule a monthly XCP meeting for the community and am struggling with when to host the call. As we are a global community, there is no single optimal time to host the meeting. In an effort to support the most likely attendees, please send me your time zone if you plan to participate in these calls. I will track the most common time zones in an effort to maximize attendance. All

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

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 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