I am nowaware that I should not post this type of questions on this group.
However, Iwould like to have some clarifications related to the response you've
alreadyprovided. The code you provided yields accurate results, however I still
haveissues grasping the loop process in case 1 & 2.
In case1, the use of "p+1" and "q+1" is still blurry tome? Likewise "0L" and "
i + 1L" in case 2.
Can youplease provide explanations on the loop mechanisms you've used.
Le lundi 3 février 2020 à 03:47:20 UTC−6, Rui Barradas
<ruipbarra...@sapo.pt> a écrit :
Hello,
You can solve the problem in two different ways.
1. Redefine storage1 as a matrix and extract the aic *in* the loop.
storage1 <- matrix(0, 4, 4)
for(p in 0:3){
for(q in 0:3){
storage1[p + 1, q + 1] <- arima(etc)$aic
}
}
2. define storage1 as a list.
storage1 <- vector("list", 16)
i <- 0L
for(p in 0:3){
for(q in 0:3){
i <- i + 1L
storage1[[i]] <- arima(etc)
}
}
lapply(storage1, '[[', "aic") # get the aic's.
Maybe sapply is better it will return a vector.
Hope this helps,
Rui Barradas
Às 06:23 de 03/02/20, ismael ismael via R-devel escreveu:
> Hello
> I am trying to extract AICs from an ARIMA estimation with different
> combinations of p & q ( p =0,1,2,3
> and q=0,1.2,3). I have tried using the following code unsucessfully. Can
> anyone help?
>
> code:
> storage1 <- numeric(16)
> for (p in 0:3){
>
> for (q in 0:3){
>
> storage1[p] <- arima(x,order=c(p,0,q), method="ML")}
> }
> storage1$aic
>
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>
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>
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