# [Rd] Stroring and extracting AICs from an ARIMA model using a nested loop

Rui Barradas ru|pb@rr@d@@ @end|ng |rom @@po@pt
Tue Feb 4 16:36:30 CET 2020

```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,

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
> <ruipbarradas using 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,
>
>
>
>
>
> 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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>
>  >

```