similar to: ARIMA - Skipping intermediate lags

Hi friends, there is methods() function to see the all available methods for a particular function, for example: > head(methods("print")) [1] "print.acf" "print.anova" "print.aov" "print.aovlist" "print.ar" "print.Arima" In this list, there are some functions which are asterisked like print.acf().

Dear all, The standard call to ARIMA in the base package such as arima(y,c(5,0,0),include.mean=FALSE) gives a full 5th order lag polynomial model with for example coeffs Coefficients: ar1 ar2 ar3 ar4 ar5 0.4715 0.067 -0.1772 0.0256 -0.2550 s.e. 0.1421 0.158 0.1569 0.1602 0.1469 Is it possible (I doubt it but am

Dear all R users, I am really struggling to determine the most appropriate lag order of ARIMA model. My understanding is that, as for MA [q] model the auto correlation coeff vanishes after q lag, it says the MA order of a ARIMA model, and for a AR[p] model partial autocorrelation vanishes after p lags it helps to determine the AR lag. And most appropriate model choosed by this argument gives

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 >

Hi, I am trying to forecast a model using predict.Arima I found arima model for a data set: x={x1,x2,x3,...,x(t)} arima_model = arima(x,order=c(1,0,1)) I am forecasting the next N lags using predict: arima_pred = predict(arima_model,n.ahead = N, se.fit=T) If I have one more point in my series, let's say x(t+1). I do not want to recalibrate themodel, I just want to forecast the next N-1

Hi I am fitting an arima model to some time series X. When I was comparing the fitted values of the model to the true time series I realized that the true time series lags one time step behind the fitted values of the arima model. And this is the case for any model. When I did a simple linear regression using lm to check, I also find the same results, that the true series lags behind the

I am trying to use auto.arima to fit a univariate time series and do forecast. This is an imaginary data on monthly outcomes of 2 years and I want to forecast the outcome for next 12 months of next year. data Data1; input RR; datalines; 12 14 17 15 13 15 15 14 15 14 16 15 15 18 16 16 15 14 15 16 16 14 13 12 ; run; I successfully took this data into R and used the auto.arima codes but am getting

Hi, I am trying to estimate an ARIMA model in the case where I have some specific knowledge about the coefficients that should be included in the model. Take a classical ARIMA (or even ARMA) model: P(B) X(t) = Q(B) epsilon(t), where X(t) is the data, epsilon is a white noise, B is the backward operator and P and Q are some polynoms. Additionally, assume that you know in advance how P and Q

Hello, The "arma" function in the "tseries" package allows estimation of models with specific "ar" and "ma" lags with its "lag" argument. For example: y[t] = a[0] + a[1]y[t-3] +b[1]e[t-2] + e[t] can be estimated with the following specification : arma(y, lag=list(ar=3,ma=2)). Is this possible with the "arima" function in the

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

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 <-

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

Dear all, let say I have following list: Dat <- vector("list", length = 26) names(Dat) <- LETTERS My_Function <- function(x) return(rnorm(5)) Dat1 <- lapply(Dat, My_Function) However I want to apply my function 'My_Function' for all elements of 'Dat' except the elements having 'names(Dat) == "P"'. Here I have specified the name

Hi again, I am struggling to extract the number part from below string : "\"cm_ffm\":\"563.77\"" Basically, I need to extract 563.77 from above. The underlying number can be a whole number, and there could be comma separator as well. So far I tried below : > library(stringr) > str_extract("\"cm_ffm\":\"563.77\"",

Dear all, when I start R, I want that the console window should be in the Maximized size automatically. Can somebody help me how to achieve that? Thanks and regards,

Hello again, I want to remove "$" sign and replace with nothing in my text. Therefore I used following code: > gsub("$|,", "", "$232,685.35436") [1] "$232685.35436" However I could not remove '$' sign. Can somebody help me why is it so? Thanks and regards

Hi Joshua, thanks for your prompt reply. However as I said, sum() function I used here just for demonstrating the problem, I have other custom function to implement, not necessarily sum() I am looking for a generic solution for above problem. Any better idea? Thanks, On Fri, Aug 11, 2017 at 12:04 AM, Joshua Ulrich <josh.m.ulrich at gmail.com> wrote: > Use a `width` of integer index