R devel - Feb 2020 - Stroring and extracting AICs from an ARIMA model using a nested loop

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

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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




?s 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 at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-devel
>

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

Sent from my iPhone
> On Feb 3, 2020, at 3:47 AM, Rui Barradas <ruipbarradas at 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)$aic
>  }
> }
> 
> 
> 2. define storage1 as a list.
> 
> storage1 <- vector("list", 16)
> i <- 0L
> for(p in 0:3){
>  for(q in 0:3){
>    i <- i + 1L
>    storage1[[i]] <- arima(etc)
>  }
> }
> 
> lapply(storage1, '[[', "aic")  # get the aic's.
> 
> Maybe sapply is better it will return a vector.
> 
> 
> Hope this helps,
> 
> Rui Barradas
> 
> 
> 
> 
> ?s 06:23 de 03/02/20, ismael ismael via R-devel escreveu:
>> Hello
>> I am trying to extract AICs from an ARIMA estimation with different
>> combinations of p & q ( p =0,1,2,3
>> and q=0,1.2,3). I have tried using the following code unsucessfully.
Can
>> anyone help?
>> code:
>> storage1 <- numeric(16)
>> for (p in 0:3){
>>     for (q in 0:3){
>>       storage1[p]  <- arima(x,order=c(p,0,q),
method="ML")}
>> }
>> storage1$aic
>>    [[alternative HTML version deleted]]
>> ______________________________________________
>> R-devel at r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-devel

I am nowaware that I should not post this type of questions on this group.
However, Iwould like to have some clarifications related to the response
you've?alreadyprovided. The code you provided yields accurate results,
however I still haveissues grasping the loop process in case 1 & 2.

In case1,?the use of?"p+1" and "q+1" is still blurry tome?
Likewise "0L" and " i + 1L" in case 2.

?

Can youplease provide explanations on the loop mechanisms you've used.?




    Le lundi 3 f?vrier 2020 ? 03:47:20 UTC?6, Rui Barradas <ruipbarradas at
sapo.pt> a ?crit :
 
 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




?s 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 at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-devel
> 
  
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