I does help! Thank you for clarifications!
Sent from my iPhone
> On Feb 4, 2020, at 9:36 AM, Rui Barradas <ruipbarra...@sapo.pt> wrote:
>
> Hello,
>
> Don't worry, we've seen worst questions :).
> Inline.
>
> Às 13:20 de 04/02/20, ismael ismael escreveu:
>> I am now aware that I should not post this type of questions on this group.
>> However, I would like to have some clarifications related to the response
>> you've already provided. The code you provided yields accurate results,
>> however I still have issues grasping the loop process in case 1 & 2.
>> In case 1, the use of "p+1" and "q+1" is still blurry to me?
>
> 1. R indexes starting from 1, both your orders p and q are 0:3. So to assign
> the results to the results matrix, add 1 and get indices 1:4.
> You could also set the row and column names after, to make it more clear:
>
> dimnames(storage1) <- list(paste0("p", 0:3), paste0("q", 0:3))
>
> 2. 0L is an integer, just 0 is a floating-point corresponding to the C
> language double.
>
> class(0L) # "integer"
> class(0) # "numeric"
>
> typeof(0L) # "integer"
> typeof(0) # "double"
>
> Indices are integers, so I used integers and added 1L every iteration through
> the inner loop.
>
> This also means that in point 1. I should have indexed the matrix with p + 1L
> and q + 1L, see the output of
>
> class(0:3)
>
>
> Hope this helps,
>
> Rui Barradas
>
> Likewise
>> "0L" and " i + 1L" in case 2.
>> Can you please 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
>> >
>> > [[alternative HTML version deleted]]
>> >
>> > ______________________________________________
>> > R-devel@r-project.org <mailto:R-devel@r-project.org> mailing list
>> > https://stat.ethz.ch/mailman/listinfo/r-devel
>> >
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