withTimeout.RdEvaluate an R expression and interrupts it if it takes too long.
withTimeout(expr, substitute=TRUE, envir=parent.frame(), timeout, cpu=timeout,
  elapsed=timeout, onTimeout=c("error", "warning", "silent"), ...)The R expression to be evaluated.
If TRUE, argument expr is
   substitute():ed, otherwise not.
The environment in which the expression should
    be evaluated.
A numeric specifying the maximum number
    of seconds the expression is allowed to run before being
    interrupted by the timeout.  The cpu and elapsed
    arguments can be used to specify whether time should be measured
    in CPU time or in wall time.
A character specifying what action to take if
    a timeout event occurs.
Not used.
Returns the results of the expression evaluated.
 If timed out, NULL is returned if onTimeout was
"warning" or "silent".
 If "error" a TimeoutException is thrown.
This method utilizes setTimeLimit() by first setting the
 timeout limits, then evaluating the expression that may or may not
 timeout.  The method is guaranteed to reset the timeout limits to be
 infinitely long upon exiting, regardless whether it returns normally
 or preemptively due to a timeout or an error.
In order to understand when this function works and when it does not,
 it is useful to know that it utilizes R's built-in time-out mechanism,
 which sets the limits on what is possible and not.
 From setTimeLimit(), we learn that:
"Time limits are checked whenever a user interrupt could occur. This will happen frequently in R code and during Sys.sleep(*), but only at points in compiled C and Fortran code identified by the code author."
More precisely, if a function is implemented in native code (e.g. C) and the developer of that function does not check for user interrupts, then you cannot interrupt that function neither via a user interrupt (e.g. Ctrl-C) nor via the built-in time out mechanism. To change this, you need to contact the developer of that piece of code and ask them to check for R user interrupts in their native code.
Furthermore, it is not possible to interrupt/break out of a "readline"
 prompt (e.g. readline() and readLines()) using
 timeouts; the timeout exception will not be thrown until after the user
 completes the prompt (i.e. after pressing ENTER).
System calls via system() and system2() cannot be
 timed out via the above mechanisms.  However, in R (>= 3.5.0) these
 functions have argument timeout providing their own independent
 timeout mechanism.
Other examples of calls that do not support timeout are "atomic"
 calls that may take very long such as large object allocation and
 rnorm(n) where n is very large.
(*) Note that on Unix and macOS, Sys.sleep(time) will signal a
     timeout error only after time seconds passed,
     regardless of timeout limit (< time).
withTimeout() does not handle the case when the expression
 evaluated temporarily switches the language used by R, e.g.
 assume we run in a non-French locale and call:
 withTimeout({
   olang <- Sys.getenv("LANGUAGE")
   on.exit(Sys.setenv(LANGUAGE=olang))
   Sys.setenv(LANGUAGE="fr")
   repeat Sys.sleep(0.1)
 }, timeout = 1.0, onTimeout = "warning")
 In this case, the error message produced by setTimeLimit() is
 in French, i.e. `la limite de temps est atteinte`.  However, when
 withTimeout() inspects this message, it can not know that
 French was used, and will therefore not check against the French template
 message for timeout errors.  Because of this, withTimeout() fails
 to detect the timeout error (and therefore also deescalate it to a
 warning in this example).
Comment: This appears to only fail on MS Windows and macOS,
 whereas on Linux, withTimeout() appears to work, but it is
 unknown why there is a difference between operating systems in this
 case.
Internally, eval() is used to evaluate the expression and
  setTimeLimit() is used to control for timeout events.
[1] R help thread 'Time out for a R Function' on 2010-12-07.
   https://stat.ethz.ch/pipermail/r-help/2010-December/262316.html 
# - - - - - - - - - - - - - - - - - - - - - - - - -
# Function that takes "a long" time to run
# - - - - - - - - - - - - - - - - - - - - - - - - -
foo <- function() {
  print("Tic")
  for (kk in 1:100) {
    print(kk)
    Sys.sleep(0.1)
  }
  print("Tac")
}
# - - - - - - - - - - - - - - - - - - - - - - - - -
# Evaluate code, if it takes too long, generate
# a timeout by throwing a TimeoutException.
# - - - - - - - - - - - - - - - - - - - - - - - - -
res <- NULL
tryCatch({
  res <- withTimeout({
    foo()
  }, timeout = 0.75)
}, TimeoutException = function(ex) {
  message("Timeout. Skipping.")
})
#> [1] "Tic"
#> [1] 1
#> [1] 2
#> [1] 3
#> [1] 4
#> [1] 5
#> [1] 6
#> [1] 7
#> [1] 8
#> [1] 9
#> [1] 10
#> [1] 11
#> Timeout. Skipping.
# - - - - - - - - - - - - - - - - - - - - - - - - -
# Evaluate code, if it takes too long, generate
# a timeout returning NULL and generate a warning.
# - - - - - - - - - - - - - - - - - - - - - - - - -
res <- withTimeout({
  foo()
}, timeout = 0.75, onTimeout = "warning")
#> [1] "Tic"
#> [1] 1
#> [1] 2
#> [1] 3
#> [1] 4
#> [1] 5
#> [1] 6
#> [1] 7
#> [1] 8
#> [1] 9
#> Warning: reached elapsed time limit [cpu=0.75s, elapsed=0.75s]
# The same using an expression object
expr <- quote(foo())
res <- withTimeout(expr, substitute = FALSE,
                   timeout = 0.75, onTimeout = "warning")
#> [1] "Tic"
#> [1] 1
#> [1] 2
#> [1] 3
#> [1] 4
#> [1] 5
#> [1] 6
#> [1] 7
#> [1] 8
#> [1] 9
#> [1] 10
#> [1] 11
#> Warning: reached elapsed time limit [cpu=0.75s, elapsed=0.75s]
# - - - - - - - - - - - - - - - - - - - - - - - - -
# Evaluate code, if it takes too long, generate
# a timeout, and return silently NULL.
# - - - - - - - - - - - - - - - - - - - - - - - - -
res <- withTimeout({
  foo()
}, timeout = 0.75, onTimeout = "silent")
#> [1] "Tic"
#> [1] 1
#> [1] 2
#> [1] 3
#> [1] 4
#> [1] 5
#> [1] 6
#> [1] 7
#> [1] 8
#> [1] 9
#> [1] 10
#> [1] 11