Finding the first value without warning in a loop

Thanks for that. I am having some difficulty figuring out how it works. The
help page tells me (I think) that "y<-log(x)" gets matched to expr. Am I
correct in thinking that if either an error or warning is produced, that
tryCatch silently returns <something>  (or <nothing>?) to the for loop, and
 only otherwise goes on to evaluate the rest of the second assignment and
break call? So any warning would abort processing of the remaining
expressions?

Are the assignment "xx<-x" and the break call considered "handlers" as
described in the help page?

I see the purpose of the last argument. Taking out the warning =
function(w{}) changes the behavior when none of the elements in the vector
passed to x were valid.

for (x in c(-2, -1)) {

+       tryCatch({
+               y <- log(x)
+               xx <- x
+               break
+       })
+ }
Warning message:
In log(x) : NaNs produced

print(xx)

[1] -2


--

David Winsemius

On Dec 7, 2008, at 12:21 PM, Gabor Grothendieck wrote:

Try this:

for (x in c(-2, 2, 4)) {
       tryCatch({
               y <- log(x)
               xx <- x
               break
       }, warning = function(w) {})
}
print(xx) # 2



On Sun, Dec 7, 2008 at 2:38 AM, Andreas Wittmann
<andreas_wittmann at gmx.de> wrote:

Dear R useRs,

with the following piece of code i try to find the first value which can
be
calculated without warnings

`test` <- function(a)
{
repeat
{
 ## hide warnings
 suppressWarnings(log(a))

 if (exists("last.warning", envir = .GlobalEnv))
 {
  a <- a + 0.1
  ## clear existing warnings
  rm("last.warning", envir = .GlobalEnv)
 }

if(a > 5 || !exists("last.warning", envir = .GlobalEnv))
 break
}

return(a)
}

if i run this with test(-3), i would expect a=0 as return value.
Is it also possible to hide warnings during my function, i guess i use
suppressWarnings in a wrong way here?

Thanks and best regards

Andreas