Question regarding if statement in while loop

Hi Stephanie,

it sounds like R's exception handling may help, something like this:

foo <- try(eblest(i, dir5, sterr5, weight5, aux5))
if ( class(foo) == "try-error" ) next

Take a look at ?try.

HTH,
Stephan


Stephanie Coffey schrieb:

Hi all,
I'm running R version 2.9.2 on a PC.

I'm having a problem with a loop, and have tried using an if statement
within to fix it, but to no avail.
Any advice would be appreciated.

Here is my code:
*****************************************************
eblest <- function(i,dir, sterr, weight, aux) {
n <- nrow(dir)
Y <- as.matrix(dir[,i], ncol=1)
sigma2ei <- as.matrix(sterr[,i]^2, ncol=1)
w <- as.matrix((weight[,3])*(sigma2ei), ncol=1)
X <- as.matrix(subset(aux, select=c(3,5:7,9:10,13)))
a <<- EBLUP.area(Y,cbind(w,1),sigma2ei,n) ? ? ? ? ? ?#The EBLUP.area
function is a function already in R.
}
? ? ? ? ? ? ?# It gives a bunch of output, some of what I need.

#THIS IS THE LOOP I'M HAVING A PROBLEM WITH:
results <- data.frame(length=nrow(dir5))
i <- 3
while (i <=some number) {
eblest(i, dir5, sterr5, weight5, aux5)
out <<- cbind(i, a$EBLUP, a$mse)
results <- cbind(results, out)
i <- i+1
}
***********************************************************************
I have tried running the eblest function for a specific set of input
(i, dir5, etc as in the function) This function runs ok.
However, sometimes eblest does not create the expected output,
sometimes due to the solution being singular or other reasons.
I'm not so concerned about that at this point - it's a function of the
data.

My problem is that, as you can see, after eblest runs, the "out"
pieces of information are bound together in the results matrix.
When the eblest does not run correctly, the loop stops. ?For example,
when i=3 and i=4, eblest runs ok, and I get a maxtrix with the
following columns:

i ? ?EBLUP se.EBLUP i EBLUP se.EBLUP
3 ? ? ? ? ? x ? ? ? ? ? ? ? ?y ? ?4 ? ? ? ?a ? ? ? ? ? ? ? ?bh

...but when it loops back around to start i=5, I get an error because
the solution is singular. ?But I don't want the loop to stop (I have
~100 of these i's for which I need to execute this function.

So I would like to set an if condition that will cause the loop to
step ahead (in this case, to 6), and continue looping...Something like
"if exists("out")=FALSE next, else..." ?However, when I tried that,
the results matrix was empty created at all.

In case it helps, "out" has a "numeric" mode, and is a "matrix"...

I've written these functions and loops using online help and examples
I've found on websites, but clearly I'm missing something.

Thanks for any help you can give!!

~Stephanie Coffey