Hi, yes that error name is indeed kind of weird. But I think its thrown due to the missing robustness of the estimation since every weight is one and hence the fit is likely to be influenced by outliers in the provided data which should be just an example.
But do you have an idea to extract the single components of the fit? I guess there must be a possibility to predict those stl models. Cheers, Konrad _____ Von: Dennis Murphy [mailto:djmu...@gmail.com] Gesendet: Sonntag, 7. Februar 2010 16:30 An: Konrad Hoppe Betreff: Re: [R] predicting with stl() decomposition Hi: When I ran your code, I got the following message in the first loess call: > llrSaison <- loess(seriesTs~time , span=decomp$win[1] , + degree=decomp$deg[1]) Warning messages: 1: Chernobyl! trL<k 1 2: Chernobyl! trL<k 1 Somebody has a sense of humor in their code writing, but I'm pretty sure the message trL < k 1 has some meaning, probably telling you the fit is unstable. I looked through the loess function code but couldn't find anything in it that would be immediately helpful. It calls a function simpleLoess(), but that function is evidently non-visible. You probably need expert guidance here. Dennis On Sun, Feb 7, 2010 at 5:48 AM, Konrad Hoppe <konradho...@hotmail.de> wrote: Hi mailinglist members, Im actually working on a time series prediction and my current approach is to decompose the series first into a trend, a seasonal component and a remainder. Therefore Im using the stl() function. But Im wondering how to get the single components in order to predict the particular fitted series. This code snippet illustrates my problem: series <- vector(length=300) noise <- rnorm(300,0,2) time <- 1:300 series[1] <- noise[1] for(i in 3:300){ series[i] <- 0.5*series[i-1]+ noise[i] + 0.01*time[i] } seriesTs <- ts(series, start=c(1980,1), frequency=12) decomp <- stl(seriesTs ,"periodic") plot(decomp) llrSaison <- loess(seriesTs~time , span=decomp$win[1] , degree=decomp$deg[1]) llrTrend <- loess(seriesTs~time, span=decomp$win[2] , degree=decomp$deg[2]) plot(llrSaison$fitted) The last plot differs much from the seasonal plot in the plot(decomp) call. This is why the llr estimator doesnt extract the seasonal component, but how can I predict the single components at last? Or is there a function which can predict the values of the stl-object. Predict() doesnt work, Ive already tried it. All the best, Konrad Hoppe [[alternative HTML version deleted]] ______________________________________________ R-help@r-project.org mailing list PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]]
______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
