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

Martin Maechler m@ech|er @end|ng |rom @t@t@m@th@ethz@ch
Mon Feb 3 12:04:45 CET 2020

```>>>>> ismael ismael via R-devel
>>>>>     on Mon, 3 Feb 2020 04:09:24 -0600 writes:

> It works!!!
> Thank you so much for your help!

and it was an "R help" question which  Rui  so generously answered.

But this is the R-devel mailing list.
Please do *NOT* misuse it for  R-help questions in the future:

These should go to the R-help mailing list instead!

Best,
Martin Maechler

>> On Feb 3, 2020, at 3:47 AM, Rui Barradas <ruipbarradas using 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,
>>
>>
>>
>>
>>
>> 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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```