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

[ismael ismael]

I does help! Thank you for clarifications!


Sent from my iPhone

> On Feb 4, 2020, at 9:36 AM, Rui Barradas <ruipbarradas using 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 <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)$aic
>>   }
>> }
>> 2. define storage1 as a list.
>> storage1 <- vector("list", 16)
>> i <- 0L
>> for(p in 0:3){
>>   for(q in 0:3){
>>     i <- i + 1L
>>     storage1[[i]] <- arima(etc)
>>   }
>> }
>> lapply(storage1, '[[', "aic")  # get the aic's.
>> Maybe sapply is better it will return a vector.
>> Hope this helps,
>> Rui Barradas
>> Às 06:23 de 03/02/20, ismael ismael via R-devel escreveu:
>> > Hello
>> > I am trying to extract AICs from an ARIMA estimation with different
>> > combinations of p & q ( p =0,1,2,3
>> > and q=0,1.2,3). I have tried using the following code unsucessfully. Can
>> > anyone help?
>> >
>> > code:
>> > storage1 <- numeric(16)
>> > for (p in 0:3){
>> >
>> >      for (q in 0:3){
>> >
>> >      storage1[p]  <- arima(x,order=c(p,0,q), method="ML")}
>> > }
>> > storage1$aic
>> >
>> >     [[alternative HTML version deleted]]
>> >
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