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

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
> 
  
	[[alternative HTML version deleted]]

Hello,

Don't worry, we've seen worst questions :).
Inline.

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

I does help! Thank you for clarifications!


Sent from my iPhone
> On Feb 4, 2020, at 9:36 AM, Rui Barradas <ruipbarradas at sapo.pt>
wrote:
> 
> ?Hello,
> 
> Don't worry, we've seen worst questions :).
> Inline.
> 
> ?s 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 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 <mailto:R-devel at r-project.org>
mailing list
>> > https://stat.ethz.ch/mailman/listinfo/r-devel
>> >