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I would like the text plotted with 'mtext' to be alighned like it is for printing on the console. Here is what I have: > print(emt) ME RMSE MAE MPE MAPE MASE original -1.034568e+07 1.097695e+08 2.433160e+07 -31.30554 37.47713 1.5100050 xreg 1.561235e+01 2.008599e+03 9.089473e+02 267.05490 280.66734

Hello, sorry for posting this independently of the original thread, but it is not that easy to answer to mails, when receiving the r-help as digest... ... The question was: > I compare each row of a matrix with each row of another matrix. > > testmat1 <- matrix(c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16), nrow=4) > testmat2 <- matrix(c(1,2,3,5,5,6,7,8,9,10,11,12,13,14,15,16),

Hi, I am trying to understand exactly what xreg does in arima. The documentation for xreg says:"xreg Optionally, a vector or matrix of external regressors, which must have the same number of rows as x." What does this mean with regard to the action of xreg in arima? Apparently somehow xreg made the following two arima fit equivalent in R: arima(x, order=c(1,1,1), xreg=1:length(x)) is

Dear R-help-archive.. I am trying to figure out how to make arima prediction when I have a process involving multivariate time series input, and one output time series (output is to be predicted) .. (thus strictly speaking its an ARMAX process). I know that the arima function of R was not designed to handle multivariate analysis (there is dse but it doesnt handle arma multivariate analysis, only

As I am reading ?arima, only NA entries in the argument fixed= imports. The following seems to indicate otherwise: x <- arima.sim(model=list(ar=0.8), n=100) + (1:100)/50 > t <- 1:100 > mod1 <- lm(x ~ t) > > init1 <- c(0, coef(mod1)[2]) > fixed1 <- c(as.numeric(NA), 0) > > arima(x, order=c(1,0,0), xreg=t, include.mean=FALSE, init=init1, fixed=fixed1)

When I run predict.Arima in my code, I get warnings like: Warning message: In cbind(intercept = rep(1, n), xreg) : number of rows of result is 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,

Is it possible to fit a lagged regression, "y[t]=b0+b1*x[t-1]+e", using the function "lag"? If so, how? If not, of what use is the function "lag"? I get the same answer from y~x as y~lag(x), whether using lm or arima. I found it using y~c(NA, x[-length(x)])). Consider the following: > set.seed(1) > x <- rep(c(rep(0, 4), 9), len=9) > y <-

I am using arima to develop a time series regression model, I am using arima b/c I have autocorrelated errors. Several of my independent variables are categorical and I have coded them as factors . When I run ARIMA I don't get any warning or error message, but I do not seem to get estimates for all the levels of the factor. Can/how does ARIMA handle factors in xreg? here is some example

I noticed that the text() command adds text to a plot. Is there a way to either make the plot blank or add text to a "blank sheet". I would like to "plot" a page that contains just text, no plot lines, labels, etc. Suggestions? Kevin [[alternative HTML version deleted]]

Hello I want to fit an AR model were two of the coefficients are fixed to zero (the second and third ar-coefficients). I used the "arima" function with the "fixed" argument but the ar3 coefficient is not set to zero: ============================================== > arima(Y, order=c(4,0,0), xreg=1:23, fixed=c(NA,0,0,NA,NA,NA)) Call: arima(x = Y, order = c(4, 0, 0), xreg =

Hello, Does anybody know if the function arima with an external regressor (xreg) applies the auto correlation on the dependant variable or on the residuals. In comparison with SAS (proc autoreg), it seems that the auto correlation applies on the residuals but i'd like to have the confirmation. I want to estimate: Y[t] = a[1]*X[t] + a[2] + E[t] with E[t]=b[1]*E[t-1] Should I use : arima(Y,

Given a Date object or simply a year is there an R function to tell me if the it is a leap year or not? I was hoping for something like 'is.leapyear'. I probably can build my own function (year divisible by 4 etc.) but I would rather use an existing function if it is available. Thank you. Kevin

Hello, Does anyone know how the parameter estimates are calculated for xreg variables when called as part of an arima() command, or know of any literature that provides this info? In particular, I was wondering if there is a quick way to compare different combinations of "xreg" variables in the arima() fit in the same way that you would in multiple regression (using AIC & R^2

Hello! If you run the whole code below, it'll produce a stacked diagram. And it looks good - because the tick-marks are aligned with the grid. However, if I stretch the graph window, grid becomes misaligned with the tickmarks. Or, rather, it seems aligned for the first and the last tick mark, but not for tickmarks in between. Can it be addressed? Thank you! Dimitri ### Creating a data set

Why does: (shape/scale) * (1:365/scale)^(shape - 1) return a vector of numbers but calling a function hasard(1:365,shape,scale) defined like: hazard <- function(x,shape,scale) { return (shape/scale) * (x/scale)^(shape - 1) } Only return a single value? It is like x becomes a single value passed as an argument. Thank you. Kevin

On 21 May 2015, at 12:49 , Martin Maechler <maechler at lynne.stat.math.ethz.ch> wrote: >>>>>> peter dalgaard <pdalgd at gmail.com> >>>>>> on Thu, 21 May 2015 11:03:05 +0200 writes: > >> On 21 May 2015, at 10:35 , Martin Maechler <maechler at lynne.stat.math.ethz.ch> wrote: > >>>> >>>> I noticed that

lag is a very confusing function i try to plot lag(x,-h) and y , the results are the same figures for all the h lags. then i saw online people use lag(x,-h) as xreg in arima that won't work in my eyes if lag function is consistent, isnce it returns the same value for all the lags i don't know in which occasion people will use this function besides I suppose Lag(x,h) is the same as

I had a question about the basic power functions in R. For example from the R console I enter: -1 ^ 2 [1] -1 but also -1^3 [1] -1 -0.1^2 [1] -0.01 Normally pow(-1, 2) return either -Infinity or NaN. Has R taken over the math functions? If so I would think that -1^2 is 1 not -1 and -0.1^2 is 0.01 not -0.01. Thank you. Kevin

I noticed that when I fit a linear model using 'lm' there is an attribute called "factors" that is added to the "term". It doesn't seem to appear for 'model.matrix', just 'lm'. I have been unable to find where it gets constructed or what it means? It looks like a two dimensional array that I may be able to use so I would just like to get some

i am using sarima() function as below ___________________________________________________________________________________________ sarima=function(data,p,d,q,P=0,D=0,Q=0,S=-1,tol=.001){ n=length(data) constant=1:n xmean=matrix(1,n,1) if (d>0 & D>0) fitit=arima(data, order=c(p,d,q), seasonal=list(order=c(P,D,Q), period=S),