search for: newxreg

...xreg = list.indep, include.mean = TRUE) and would like to use predict() to forecast: chn.forecast <- rep(0,times=num.record) chn.forecast[1] <- y[1] for (j in 2:num.record){ indep <- c(aa=chn.forecast[j-1], list.indep[j,2:num.indep]) # this is the newxreg in the forecast. chn.forecast[j] <- predict(fit, newxreg=indep, n.ahead = 1) } However, I got the error message as 'xreg' and 'newxreg' have different numbers of columns". So I debug into predict.Arima (as shown in (*)). (*): debugging in: predict.Arima(...

...al setting of errors,etc.)? I run the following commands and get the original code of the "predict" command, but I can't read it. Can anybody explain it to me? Thanks! saji from Shanghai the code: > getS3method("predict","Arima") function (object, n.ahead = 1, newxreg = NULL, se.fit = TRUE, ...) { myNCOL <- function(x) if (is.null(x)) 0 else NCOL(x) rsd <- object$residuals xr <- object$call$xreg xreg <- if (!is.null(xr)) eval.parent(xr) else NULL ncxreg <- myNCOL(xreg) if (myNCOL(newxreg) !=...

...us variables that have only 1 lag Question 1: ----------- Suppose I use arima to fit the model: df.y<-arima(yvec,order=c(L,0,0),xreg=xmat[,(1:K)],n.cond=maximum.lag) Now suppose I want to do a 1-period ahead prediction based on the results of this regression, using predict: predict(df.y,newxreg=newx,n.ahead=1) I'm expecting newx to be 1X3. After all, I just want to predict 1 value of y, so in my mind I should just need 1 time period's observation of x (i.e. # rows=n.ahead). I'm sort of expecting predict to grab the last two values of yvec to use as y(t),y(t-1) in prediction....

...ated of class type arimax.According to the documentation this is the same type as arima.So now I make a prediction,say 15 time steps ahead: >forecast=predict(air.m1, n.ahead=15) The following error is produced: Error in predict.Arima(air.m1, n.ahead = 15) : 'xreg' and 'newxreg' have different numbers of columns -------------------------------------------------------------------------------------------------------------------- Question is how to to get a prediction correctly using predict.(I've seen DSE package but that seems overkill to make just a simple predic...

...s not a multiple of vector length (arg 1) I think this is because I'm not running predict.Arima in the same environment that I did the fit, so the data object used in the fit is no longer present. Looking at the predict.Arima source, predict.Arima <- function (object, n.ahead = 1, newxreg = NULL, se.fit = TRUE, ...) { myNCOL <- function(x) if (is.null(x)) 0 else NCOL(x) rsd <- object$residuals xr <- object$call$xreg xreg <- if (!is.null(xr)) eval.parent(xr) else NULL ncxreg <- myNCOL(xreg) if (myNCOL(newxreg) != ncxreg)...

...0.0686 0.0582 0.1098 0.1558 0.1568 0.1007 0.0716 sar2 sma1 drift -0.0711 -0.8963 -780.9456 s.e. 0.0747 0.0608 403.2112 sigma^2 estimated as 10202381: log likelihood = -1100.61 AIC = 2206.69 AICc = 2209.23 BIC = 2236.98 >for<-forecast(mods,h=12,newxreg=(1+length(x[[1]])):(length(x[[1]]+12))) #forecast and as I said before, the results dont seem to be right. In fact, when I restrict the search of the model on the auto.arima function to stationary models only an I perform the forecast (without the newxreg-option) the results are very much acceptab...

...is 100 + 20 (20 to use for the forecast horizon). I can not make xreg work with the forecast function for an arima fit. The "predict" function seems to be working but the "forecast" returns > Error in predict.Arima(object, n.ahead = h) : 'xreg' and 'newxreg' have different numbers of columns... Xreg is a time series of explanatory variables of the same length of the dependent and newxreg should be as long as the forecast horizon (same number of explanatory variables). Is this right? Or should xreg be: X[1:100] and newxreg X[101:20], i.e....

...a(reihe.lern, order = c(1, 0, 0), seasonal = list(order = c(0, 0, 0), period = 52), xreg = lern.design, include.mean = TRUE) for which I need to estimate the next (105th) value. Xreg=lern.design is - at this time - 104 rows long. I tried to use the following function: predict(arx.mod, n.ahead = 1, newxreg=lern.design[105,]) Here, lern.design is 105 rows long, same number of columns. Now the problem is the following error: Error in predict.Arima(arx.mod, n.ahead = 1, newxreg = lern.design[105, : 'xreg' and 'newxreg' have different numbers of columns If I use the total &qu...

...r=arma5.fit$order,seasonal=list(order=arma5.fit$seasonal),xreg=newmcReg,include.drift=F) #All the drift terms ones are broke as the drift term doesn't match up to the # of columns in fReg #Still don't know how to fix at this time. forecast1.b <-predict(newFit1.b,n.ahead=modLength-modMax,newxreg=mfReg) forecast2.b <-predict(newFit2.b,n.ahead=modLength-modMax,newxreg=newmfReg) forecast3.b <-predict(newFit3.b,n.ahead=modLength-modMax,newxreg=mfReg) forecast4.b <-predict(newFit4.b,n.ahead=modLength-modMax,newxreg=newmfReg) forecast5.b <-predict(newFit5.b,n.ahead=modLength-modMax,n...

...ARIMA(3,0,0)(2,1,1)[12] with drift : 1e+20 * ARIMA(4,0,0)(2,1,0)[12] with drift : 1e+20 Best model: ARIMA(3,0,0)(2,1,0)[12] with drift # this is the output of the function fore<-forecast(c,h=12) # and this is the error Error en predict.Arima(object, n.ahead = h, newxreg = xreg) : 'xreg' and 'newxreg' have different numbers of columns This error only occurs when the output of the auto.arima function contanis drift. Any help is aprecciated!!!! [[alternative HTML version deleted]]

....1836 0.1134 -0.1742 0.0236 -0.0482 s.e. 0.0588 0.0251 0.0363 0.0278 0.0907 0.0528 0.0860 0.0516 0.1518 0.1025 0.0591 0.0470 sigma^2 estimated as 1.258: log likelihood = -762.3, aic = 1584.59 > > fordiv <- predict(fitdiv, n.ahead = 1, newxreg = newregy , se.fit = TRUE) Error in cbind(...) : cannot create a matrix from these types > > > str(data) num [1:497] -0.34 -1.36 -0.5 -0.46 0.01 0.1 0.68 0.06 0.16 0.48 ... > > str(newregy) num [1, 1:23] -0.6 -0.3 0.15 1.08 -1.8 3 2 3 0 5 ... - attr(*, "dimnames")=List...

Good afternoon! I'm trying to learn time series but i have a bit of of a problem using R packages for this. 1. > LakeHuron > sample(500:600, 98) > sample(500:600, 98)->t > fit<-arima(LakeHuron, order=c(2,1,1), xreg=t) > fit > predict(fit, n.ahead=1, newxreg=t) Now, my problem is this: is it ok to use the same t in predict function or should my newxreg contain 99 values in order to properly predict the 99th value for LakeHuron? Another problem is that i don't receive the 99th value on the screen, as i should (running the example in ?predict &...

....1836 0.1134 -0.1742 0.0236 -0.0482 s.e. 0.0588 0.0251 0.0363 0.0278 0.0907 0.0528 0.0860 0.0516 0.1518 0.1025 0.0591 0.0470 sigma^2 estimated as 1.258: log likelihood = -762.3, aic = 1584.59 > > fordiv <- predict(fitdiv, n.ahead = 2, newxreg = newregy , se.fit = TRUE) Error in cbind(...) : cannot create a matrix from these types > > > str(data) num [1:497] -0.34 -1.36 -0.5 -0.46 0.01 0.1 0.68 0.06 0.16 0.48 ... > > str(newregy) num [1, 1:23] -0.6 -0.3 0.15 1.08 -1.8 3 2 3 0 5 ... - attr(*, "dimnames")=List...

...h(names(linearModel[[i]]$coeff[-1]),paste(names(mfReg),"1",sep=""))] newFit[[i]] <- Arima(modelSource,order=arima_Fit[[i]]$order,seasonal=list(order=arima_Fit[[i]]$seasonal),xreg=newmcReg[[i]],include.drift=F) newForecast[[i]] <-predict(newFit[[i]],n.ahead=modLength-modMax,newxreg=newmfReg[[i]]) } #################### # END FOR EACH MODEL #################### dev.off() return("AUTOBOT") } -- End Script -- The problem exists in the line: newForecast[[i]] <-predict(newFit[[i]],n.ahead=modLength-modMax,newxreg=newmfReg[[i]]) resulting in the error: Error in as...

...BIC=4555.25 After this, I wanted to make some predictions based on the model suggested by the auto.arima function. I tried to use the following code to get the predictions: variableA.pred<-predict(paulfit, n.ahead=36) *Error in predict.Arima(paulfit, n.ahead = 36) : 'xreg' and 'newxreg' have different numbers of columns: 1 != 0*> Now, I have understood, and, according to my research the xreg comes into play if you add external regressors to your arima fit, but since I am just working with a univariate time series, I really do not understand the reason for that error mess...

...e the forecast. The results agree to 5 or 6 decimal places. I can do this for models with and without means (i.e., using the include.mean argument). When I include an exogenous variable using the xreg argument, I can't manually reproduce the forecast I get when using predict.Arima with the newxreg argument, even with just one exogenous variable with an AR(1) model for the response. There does not appear to be any information in the arima help file explaining the model when xreg is used. I would have thought it was as simple as including an extra term containing the product of the coeffi...

...s.dat) arima.dat<-arima(ts.dat,order=c(1,2,0),xreg=1:length(dat)) arima.dat tvalue<-(arima.dat$coef)/(diag(sqrt(arima.dat$var.coef))) tvalue deg.free<-length(dat)-length(arima.dat$coef) pvalue<-round(2*pt(-abs(tvalue),deg.free),digits=4) pvalue fore.arima.dat<-predict(arima.dat, 20, newxreg=(length(ts.dat)+1):(length(ts.dat)+20)) ts.plot(ts.dat, fore.arima.dat$pred) Using the diff(diff(...)) method produces the correct t and p values but the forecast looks terrible: dat<-c(4,5,9,7,10,15,12,13,17,25,20,29,26,31,32) ts.dat=ts(dat) ts.plot(ts.dat) diff.ts.dat<-diff(ts.dat) diff.t...

Hello, I am calling the auto.arima method in the forecast package at it returns what seems to be valid Arima output. But when I feed this output to 'predict' I get: Error in predict.Arima(catall.fit[[.index]], n.ahead = 12) : 'xreg' and 'newxreg' have different numbers of columns Is there a way to tell what is being supplied to xreg from the Arima output? Any ideas? Thank you. Kevin

...able to solve. I use this function on time series that contains NA values, but every time that the resulting model contains drift I can't perform a forecasting (using forecast.Arima function). The printed error (when I try to forecast the resulting model) claims a dimension mismatch "nreg=newxreg" Here is what I've been triyng to do >ser<-auto.arima(x,start.p=0,start.q=0,start.P=0,start.Q=0,trace=TRUE,stepwise=TRUE); >forecast(ser,h=12) x is ts object (-monthly data-) Thanks [[alternative HTML version deleted]]

...gt; var1.fit Now the weird thing here is that R was actually able to fit the model, however, when I wanted to make predictions (see below): > var1.pred<-predict(var1.fit, n.ahead=12) R popped the following error: [b]Error in predict.Arima(var1.fit, n.ahead = 12) : 'xreg' and 'newxreg' have different numbers of columns > Which is weird beaucse 1. I made sure that my regressor matrix had the same amount of rows as my univariate time series and 2. R was able to fit the model I was suggesting. Has anyone done arima models including external regressors in R? Does anyone h...