similar to: How to look into the asterisked function?

Dear R-Helpers, I suspect I'm about to ask a FAQ, but I haven't been able to find an answer in the FAQ, AItR or an R Site Search. When I look at the methods of summary (below) it says, "Non-visible functions are asterisked". I looked at the help file for summary.princomp, which did not comment on it being non-visible. I ran its help file example, which printed visible output. I

Hi all, I've recently upgraded to R version 2.1.1 and when trying to inspect the contents of many packages in the library (for instance library\MASS\R) I've realized wordpad, or the notepad, won't open them since they have *.RDB and *.RDX extensions which these editors cannot recognize. However, libraries in previous versions of R did not have these extensions and I could inspect

Hello, I am a new R user. I am trying to use the arima command, but I have a question on intermediate lags. I want to run in R the equivalent Stata command of ARIMA d.yyy, AR(5) MA(5 7). This would tell the program I am interested in AR lag 5, MA lag 5, and MA lag 7, all while skipping the intermediate lags of AR 1-4, and MA 1-4, 6. Is there any way to do this in R? Thank you. -- View this

Dear List, This may be related to this email thread initiated by Ben Bolker last month: https://stat.ethz.ch/pipermail/r-devel/2011-July/061630.html In answering this Question on StackOverflow http://stackoverflow.com/q/7195628/429846 I noticed that `methods()` was not listing some S3 methods for `plot()` provided by the mgcv package. At the time I wanted to check the development version of R as

Hi, Although my doubt is pretty,as i m not from stats background i am not sure how to proceed on this. Currently i am doing a forecasting.I used ARIMA to forecast and time series was volatile i used garchFit for residuals. How to use the output of Garch to correct the forecasted values from ARIMA. Here is my code: ###delta is the data fit<-arima(delta,order=c(2,,0,1)) fit.res <-

Hello, I was wondering if anybody could help me with this? I have plotted an acf function for a time series and am very happy with it. Now I am interested in calculating for myself the two values for the confidence intervals that are plotted on the graph of the acf. The confidence intervals do not appear to be returned from the acf function (is this true?). So far I haven't managed to

Dear all, I would like to ask one question related to statistics, for specifically on defining dummy variables. As of now, I have come across 3 different kind of dummy variables (assuming I am working with Seasonal dummy, and number of season is 4): > dummy1 <- diag(4) > for(i in 1:3) dummy1 <- rbind(dummy1, diag(4)) > dummy1 <- dummy1[,-4] > > dummy2 <- dummy1 >

Good afternoon! I'm trying to model a time series on the following data, which represent a monthly consumption of juices: >x<-scan() 1: 2859 3613 3930 5193 4523 3226 4280 3436 3235 3379 3517 6022 13: 4465 4604 5441 6575 6092 6607 6390 6150 6488 5912 6228 10196 25: 7612 7270 8617 9535 8449 8520 9148 8077 7824 7991 7660 12130 37: 9135 9512 9631 12642

Dear R-helpers, I have a time series analysis problem in R: I want to analyse the output of my simulation model which is proportional cover of shrubs in a savanna plot for each of 500 successive years. I have run the model (which includes stochasticity, especially in the initial conditions) 17 times generating 17 time series of shrub cover. I am interested in a possible periodicity of shrub

Hola, Estoy visualizando una serie temporal para determinar sus órdenes ARIMA y no consigo lo siguiente: ¿Cómo puedo sacar la ACF de los retardos múltiplos del periodo? Es decir, sólo ver en el gráfico ACF los retardos 12, 24, 36... Gracias!! David [[alternative HTML version deleted]]

Please see that correlogram for a arbitrary time series : acf(zooreg(rnorm(39), start=as.yearmon("2008-01-01"), frequency=12)) What is the meaning of lag 0.2, 0.4, ........ in the plot? Those should not be integers? Or I am missing something? Thanks -- View this message in context: http://n4.nabble.com/Meaning-of-lag-0-2-0-4-tp1765093p1765093.html Sent from the R help mailing list

Hi, I've noted that not all `predict' methods are hidden in the namespace: > methods("predict") [1] predict.ar* predict.Arima* [3] predict.arima0* predict.glm [5] predict.HoltWinters* predict.lm [7] predict.loess* predict.mlm [9] predict.nls* predict.poly [11] predict.ppr* predict.prcomp* [13]

Below is my full implementation (tried to make it simple as for demonstration) Lapply_me = function(X = X, FUN = FUN, Apply_MC = FALSE, ...) { if (Apply_MC) { return(mclapply(X, FUN, ...)) } else { if (any(names(list(...)) == 'mc.cores')) { myList = list(...)[!names(list(...)) %in% 'mc.cores'] } return(lapply(X, FUN, myList)) } } Lapply_me(as.list(1:4), function(xx) { if (xx ==

Hi all, please see my code: > library(zoo) > a <- as.yearmon("March-2010", "%B-%Y") > b <- as.yearmon("May-2010", "%B-%Y") > > nn <- (b-a)*12 # number of months in between them > nn [1] 2 > as.integer(nn) [1] 1 What is the correct way to find the number of months between "a" and "b", still

The reason that it works for Apply_MC=TRUE is that in that case you call mclapply(X,FUN,...) and the mclapply() function strips off the mc.cores argument from the "..." list before calling FUN, so FUN is being called with zero arguments, exactly as it is declared. A quick workaround is to change the line Lapply_me(as.list(1:4), function(xx) { to Lapply_me(as.list(1:4),

My modified function looks below : Lapply_me = function(X = X, FUN = FUN, Apply_MC = FALSE, ...) { if (Apply_MC) { return(mclapply(X, FUN, ...)) } else { if (any(names(list(...)) == 'mc.cores')) { myList = list(...)[!names(list(...)) %in% 'mc.cores'] } return(lapply(X, FUN, myList)) } } Here, I am not passing ... anymore rather passing myList On Sun, Mar 4, 2018 at 10:37 PM,

@Eric - with this approach I am getting below error : Error in FUN(X[[i]], ...) : unused argument (list()) On Sun, Mar 4, 2018 at 10:18 PM, Eric Berger <ericjberger at gmail.com> wrote: > Hi Christofer, > You cannot assign to list(...). You can do the following > > myList <- list(...)[!names(list(...)) %in% 'mc.cores'] > > HTH, > Eric > > On Sun, Mar

Hello! I have two questions. First of all, I have a problem dealing with acf (Autocovariance function) and need help. First I defined a time series, x, which is a vector created by x <- ts(rnorm(200)). So I plugged the series directly into the acf function, acf(x) and an error message popped up as: Error in .C("acf", as.double(x), as.integer(sampleT), as.integer(nser), :

Good afternoon! I'm trying to model a ts but unfortunately i'.m very new to this kind of modeling so i 'll be very grateful if you have an advice. This was my syntax:

In one of my scripts I used plot.acf function, but I had to modify it a little to suit my needs. Although my modifications are minor, I believe some people might benefit from it. The problem: currently, in cross-correlation plots main title captions are constructed of 2 names separated by '&' symbol. Such captions occupy too much horizontal space and usually don't fit in plot. A