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

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

>>>>> 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 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

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

Hello,

Inline.

?s 11:04 de 03/02/20, Martin Maechler escreveu:
>>>>>> 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!
Yes, and I had noticed it but then, after writing down the answer, 
forgot to mention it.
Thanks for the warning.

Rui Barradas
> 
> Best,
> Martin Maechler
> 
> 
>      >> 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
> 
>      > ______________________________________________
>      > R-devel at r-project.org mailing list
>      > https://stat.ethz.ch/mailman/listinfo/r-devel
>