Displaying 20 results from an estimated 1000 matches similar to: "question regarding time series packages"
2003 Apr 21
2
Anyone Familiar with Using arima function with exogenous variables?
I've posted this before but have not been able to locate what I'm doing
wrong. I cannot determine how the forecast is made using the estimated
coefficients from a simple AR(2) model when there is an exogenous
variable. Does anyone know what the problem is? The help file for arima
doesn't show the model with any exogenous variables. I haven't been able
to locate any documents
2009 Jan 21
1
forecasting issue
Hello everybody!
I have a problem when I try to perform a forecast of an ARIMA model
produced by an auto.arima function. Here is what I'm doing:
c<-auto.arima(fil[[1]],start.p=0,start.q=0,start.P=0,start.Q=0,stepwise=TRUE,stationary=FALSE,trace=TRUE)
# fil[[1]] is time series of monthly data
ARIMA(0,0,0)(0,1,0)[12] with drift : 1725.272
ARIMA(0,0,0)(0,1,0)[12] with drift
2013 Mar 22
0
predict.Arima error "'xreg' and 'newxreg' have different numbers of columns"
Hello all,
I use arima to fit the model with
fit <- arima(y, order = c(1,0,1), 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
2009 Feb 17
0
What's the predict procedure of ARIMA in R?
Hello,guys:
Recently, I am working on a seasonal ARIMA model. And I met some problem in the forecasting.
Now I just want to know that How does R perform the predict procedure(the predict formula, the initial 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
2010 Mar 31
1
predict.Arima: warnings from xreg magic
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,
2010 May 04
1
How to make predictions with the predict() method on an arimax object using arimax() from TSA library
Hi R Users,
I'm fairly new to R (about 3 months use thus far.)
I wanting to use the arimax function from the TSA library to incorporate some exogenous inputs into the basic underllying arima model.Then with that newly model of type arimax, I would like to make a prediction.
To avoid being bogged down with issues specific to my own work, I would like to refer to readers to the example
2008 Jul 08
0
forecast & xreg
Dear all,
I am fitting an arimax (arima with some extra explanatory variables)
model to a time series. Say, I have a Y (dependent variable) and an X
(explanatory).
Y is 100 observations (time series) and X 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
2012 Apr 26
1
Using the R predict function to forecast a model fit with auto.arima function
Hello R users,
Hope everyone is doing great.
I have a dataset that is in .csv format and consists of two columns: one
named Period (which contains dates in the format yyyy_mm) and goes from
1995_10 to 2007_09 and the second column named pcumsdry which is a
volumetric measure and has been formatted as numeric without any commas or
decimals.
I imported the dataset as pauldataset and made use of
2007 Mar 02
0
R: ARIMA forecasting
Dear all,
I just have a short question regarding the forecasting of ARIMA models with
external regressors.
I tried to program a ARX(1) model
arx.mod <- arima(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
2009 Jul 15
2
storing lm() results and other objects in a list
to clean up some code I would like to make a list of arbitrary length
to store?various objects for use in a loop
sample code:
############ BEGIN SAMPLE ##############
# You can see the need for a loop already
linearModel1=lm(modelSource ~ .,mcReg)
linearModel2=step(linearModel1)
linearModel3=lm(modelSource ~ .-1,mcReg)
linearModel4=step(linearModel3)
#custom
linearModel5=lm(modelSource ~ .
2005 Jul 08
1
help with ARIMA and predict
I'm trying to do the following out of sample
regression with autoregressive terms and additional x
variables:
y(t+1)=const+B(L)*y(t)+C(1)*x_1(t)...+C(K)*x_K(t)
where:
B(L) = lag polynom. for AR terms
C(1..K) = are the coeffs. on K exogenous variables
that have only 1 lag
Question 1:
-----------
Suppose I use arima to fit the model:
2003 Dec 18
0
Help with predict.Arima with external regressor values
Hi all there
I am enjoying R since 2 weeks and I come to my first deadlock, il am trying
to use predict.Arima in the ts package.
I get a "Error in cbind(...) : cannot create a matrix from these types"
-- Start R session -----------------------------------------------------
> fitdiv <- arima(data, c(2, 0, 3), xreg = y ) ; print(fitdiv)
Call:
arima(x = data, order = c(2, 0, 3),
2003 Dec 18
1
Help with predict.Arima with external regressor values [Repalced]
Hi all there
I am enjoying R since 2 weeks and I come to my first deadlock, il am trying
to use predict.Arima in the ts package.
I get a "Error in cbind(...) : cannot create a matrix from these types"
-- Start R session -----------------------------------------------------
> fitdiv <- arima(data, c(2, 0, 3), xreg = y ) ; print(fitdiv)
Call:
arima(x = data, order = c(2, 0, 3),
2012 Nov 14
0
Time Series with External Regressors in R Problems with XReg
Hello everyone,
Hope you all are doing great! I have been fitting arima models and
performing forecasts pretty straightforwardly in R.
However, I wanted to add a couple of regressors to the arima model to see if
it could improve the accuracy of the forecasts but have had a hard time
trying to do so.
I used the following R function:
arima(x, order = c(0, 0, 0),
seasonal = list(order = c(0, 0,
2003 Apr 16
0
arima function - estimated coefficients and forecasts
I'm using the arima function to estimate coefficients and also using
predict.Arima to forecast. This works nicely and I can see that the
results are the same as using SAS's proc arima.
I can also take the coefficent estimates for a simple model like
ARIMA(2,1,0) and manually compute the forecast. The results agree to 5
or 6 decimal places. I can do this for models with and without
2008 Sep 22
0
auto.arima help.
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?
2009 Mar 08
0
ARIMA second order differencing problem
Hi,
I have been using this site (
http://www.stat.pitt.edu/stoffer/tsa2/Rissues.htm) to help me with some
ARIMA modelling in R.
Unfortunately the methods mentioned do not appear to work with second order
differencing; arima(*, 2, *).
I have used some dummy data to illustrate my point.
When I use the xreg=... method, the estimate of intercept is *way* off. This
can be seen by the high s.e but I
2009 Apr 23
0
How to construct confidence bands from a gls fit?
Dear R-list,
I would like to show the implications of estimating a linear trend to
time series,
which contain significant serial correlation.
I want to demonstrate this, comparing lm() and an gls() fits, using
the LakeHuron
data set, available in R.
Now in my particular case I would like to draw confidence bands on the plot and
show that there are differences. Unfortunately, I do not know how to
2008 Apr 30
2
Bug? in summary( ) function base package
There seems to be an error in the summary() function when applied to "ts"
class objects. The results of a call to summary( ), on the R "ts" data set
USAccDeaths , reports the wrong value for Max. The value reported by the
summary function is 11320. The max( ) function returns the correct value
11317, the July 1993 value. Coercing the data to a data.frame and calling
summary
2009 Dec 03
0
Problem with predict() and factors
I am working on a script that takes numeric performance indicators and runs
them against a series of regressors (dummy regressors, yes\no stuff via 0
and 1, e.g. Was is Christmas this week 0=no, 1=yes).
The script is as follows (Written as a function):
-- Begin Script --
doEnv <- function(HOUR,ENVNAME,REPORTNAME) {
library(RODBC)
library(forecast)
library("geneplotter")