It works!!!
Thank you so much for your help!
Sent from my iPhone
> On Feb 3, 2020, at 3:47 AM, Rui Barradas <ruipbarra...@sapo.pt> wrote:
>
> 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
>> [[alternative HTML version deleted]]
>> ______________________________________________
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>> https://stat.ethz.ch/mailman/listinfo/r-devel
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