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
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