On 05-08-2013, at 10:32, Salaam Batur <swordligh...@gmail.com> wrote:
> Dear Berend,
>
> I am using R version 3.0.1 and I loaded code packages below
>
> library(TSA)
> library(lattice)
> library(stats)
> library(utils)
> library(forecast)
>
> I reisntalled R and loaded packages 'steries' and 'forecast' as you
> suggested. Everything works out fine. Maybe it was the 'TSA' package which
> gave me the warinig masseges
>
>
Wonderful.
Reinstalling R was most likely unnecessary.
But please in future reply to the list only so that others can follow any
discussion and also offer help.
And show a working reproducible example.
Berend
>
>
> On Mon, Aug 5, 2013 at 3:03 PM, Berend Hasselman <b...@xs4all.nl> wrote:
>
> On 05-08-2013, at 07:47, Salaam Batur <swordligh...@gmail.com> wrote:
>
> > Dear R users,
> >
> > I am having a problem using forecast.Arima fuction. Here is the whole code:
> >
> > d=scan("D:/Data.txt")
> > d
> > D=ts(data=d, start=1981,end=2012, frequency=1)
> > D
> > Time Series:
> > Start = 1981
> > End = 2012
> > Frequency = 1
> > [1] 384 403 427 450 499 550 575 615 640 680 702 730 760 790
> > [15] 790 830 870 871 906 920 968 1010 1060 1111 1165 1191 1217 1221
> > [29] 1089 1089 1090 1103
> >
>
> Please use dput to show objects.
>
> > lnD=log(D)
> > lnD3=diff(lnD, differences=3)
> > adf.test(lnD3)
> > Augmented Dickey-Fuller Test
> >
> > data: lnD3
> > Dickey-Fuller = -3.5315, Lag order = 3, p-value = 0.05795
> > alternative hypothesis: stationary
> >
> > #########d parameter is determined by ADF test, which is 3#############
> >
> > Now choosing p and q parameters
> >
> > par(mfrow=c(2,1))
> > acf(lnD3, lag.max=20)
> > pacf(lnD3, lag.max=20)
> >
> > ######### from acf and pacf, p=2, q=1##############
> >
> > Now fitting Arima
> >
> > arima1=arima(lnD3, order=c(2,0,1))
> > arima1
> > Series: x
> > ARIMA(2,0,1) with non-zero mean
> >
> > Coefficients:
> > ar1 ar2 ma1 intercept
> > -0.5189 -0.2033 -1.0000 -1e-04
> > s.e. 0.1806 0.1770 0.0993 5e-04
> >
> > sigma^2 estimated as 0.00118: log likelihood=54.24
> > AIC=-100.48 AICc=-97.87 BIC=-93.64
> >
> > ####### Which looks good#########
> >
> > Using auto.arima() to see what R have in mind
> >
> > autoarima=auto.arima(lnD, d=3)
> > autoarima
> > Series: lnD
> > ARIMA(2,3,0)
> >
> > Coefficients:
> > ar1 ar2
> > -1.0282 -0.5851
> > s.e. 0.1524 0.1560
> >
> > sigma^2 estimated as 0.001731: log likelihood=50.37
> > AIC=-94.73 AICc=-93.77 BIC=-90.63
> >
> > ###### From AIC and BIC, I prefer arima1 instead of autoarima######
> > Now using forecast.Arima
> >
> > forecastArima1=foreca.Arima(arima1, h=5)
> > *Warining message
> > Error in ts(x) : 'ts' object must have one or more observations*
> >
>
> I assume that this should be forecast.Arima
>
> > But forecasting autoarima is no problem
> > forecastAutoArima=forecast.Arima(autoarima, h=5, c=(0.95))
> > forecastAutoArima
> >
> > Point Forecast Lo 80 Hi 80 Lo 95 Hi 95
> > 2013 7.084688 7.031373 7.138004 7.003150 7.166227
> > 2014 7.167079 7.049206 7.284951 6.986808 7.347349
> > 2015 7.285478 7.069813 7.501142 6.955648 7.615308
> > 2016 7.443636 7.085484 7.801789 6.895890 7.991383
> > 2017 7.618670 7.081729 8.155612 6.797489 8.439852
> >
> > Why???? Is there a bug probelm with arima() function itself???
> >
>
> This not reproducible code.
> Which packages have you loaded?
> adf.test, auto.arima and forecast.Arima are not standard.
>
> So using package sos told me that I needed package tseries and forecast.
>
> Using your data and your code gave me this for the code starting at
> auto.arima:
>
> autoarima=auto.arima(lnD, d=3)
> > autoarima
> Series: lnD
> ARIMA(2,3,0)
>
> Coefficients:
> ar1 ar2
> -1.0282 -0.5851
> s.e. 0.1524 0.1560
>
> sigma^2 estimated as 0.001731: log likelihood=50.37
> AIC=-94.73 AICc=-93.77 BIC=-90.63
> >
> > forecastArima1=forecast.Arima(arima1, h=5)
> > forecastArima1
> Point Forecast Lo 80 Hi 80 Lo 95 Hi 95
> 2013 -0.0212777938 -0.06606177 0.02350618 -0.08976898 0.04721339
> 2014 0.0088824273 -0.07130043 0.08906529 -0.11374668 0.13151153
> 2015 -0.0004052589 -0.08451693 0.08370641 -0.12904295 0.12823243
> 2016 -0.0017177972 -0.08583008 0.08239448 -0.13035643 0.12692083
> 2017 0.0008517381 -0.08342997 0.08513345 -0.12804602 0.12974949
> >
> > forecastAutoArima=forecast.Arima(autoarima, h=5, c=(0.95))
> > forecastAutoArima
> Point Forecast Lo 80 Hi 80 Lo 95 Hi 95
> 2013 7.084688 7.031373 7.138004 7.003150 7.166227
> 2014 7.167079 7.049206 7.284951 6.986808 7.347349
> 2015 7.285478 7.069813 7.501142 6.955648 7.615308
> 2016 7.443636 7.085484 7.801789 6.895890 7.991383
> 2017 7.618670 7.081729 8.155612 6.797489 8.439852
>
> So you must be doing something other than what you have shown.
>
> Berend
>
>
> > If anyone knows the problem, or shows me a right direction, I would really
> > appreciate it!!!
> > Many many thanks!!!
> >
> > Chintemur Batur
> >
> > [[alternative HTML version deleted]]
> >
> > ______________________________________________
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> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
>
>
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