Hi,
Thanks Rui, but the problem is at this point, I don't want to predict.
I want to see for the time being, how good my model fits the previous data
points.
So, if I specify n.ahead=-1 (or any -ve value) , R shows an error like:
Error in rep(1, n.ahead) : invalid 'times' argument
How do I get this comparison done?
Another problem I am facing here is (and this is about prediction):
To get the model, I entered : arima=arima((y,order=c(2,0,2),xreg=g);
Suppose, I want tot predict the next value of y, but when I type
predict=predict(arima,n.ahead=1,newxreg=g[16]), the following
warning appears:
Warning message:
In predict = predict(arima, n.ahead = 1, newxreg = g[16]) :
number of items to replace is not a multiple of replacement length
So these are the 2 issues I am facing.Any help is welcome.
Thanks,
Preetam
On Sun, May 19, 2013 at 3:58 AM, Rui Barradas <ruipbarra...@sapo.pt> wrote:
> Hello,
>
> At an R prompt type
>
> ?predict.Arima
>
> Hope this helps,
>
> Rui Barradas
>
> Em 18-05-2013 21:42, Preetam Pal escreveu:
>
> Hi all,
>>
>> I have a time series Y which I have modelled as ARIMA(2,0,2) by using the
>> arima function .
>> I want to know the model predicted values so that I can compare them with
>> the actual values.
>> Which function do I need to use to get the predicted values? I have tried
>> to find out, but with no luck till now.
>> Appreciate your help.
>>
>> Thanks,
>> Preetam
>>
>>
>>
>>
--
Preetam Pal
(+91)-9432212774
M-Stat 2nd Year, Room No. N-114
Statistics Division, C.V.Raman
Hall
Indian Statistical Institute, B.H.O.S.
Kolkata.
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