[R] Question regarding if statement in while loop

[Stephan Kolassa]

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
> 
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