ls() with different defaults: Solution;
On Tue, Mar 12, 2013 at 12:59 PM, Szumiloski, John
<john_szumiloski at merck.com> wrote:
Dear useRs, Some time ago I queried the list as to an efficient way of building a function which acts as ls() but with a different default for all.names: http://tolstoy.newcastle.edu.au/R/e6/help/09/03/7588.html I have struck upon a solution which so far has performed admirably. In particular, it uses ls() and not its explicit source code, so only has a dependency on its name and the name of its all.names argument. Here is my solution: lsall <- function(...) { thecall <- as.call(c(as.name('ls'), list(...))) newcall <- match.call(definition=ls, call=thecall) if( !('all.names' %in% names(newcall)) ) newcall[['all.names']] <- TRUE eval(newcall, envir=parent.frame()) }#### end lsall
Why not just do:
lsall <- function(..., all.names = TRUE) {
ls(..., all.names = all.names)
}
? Then the function practically documents itself.
Hadley
Chief Scientist, RStudio http://had.co.nz/