R help - Feb 2003 - Using evaluate-deparse-substitute

Being the lazy soul I am, I wish to write a function to replace saying
ls(pattern=...) everytime. Here is what I have:

lsp <- function(x){
  y <- eval(deparse(substitute(x))) 
  print(y)                                   # CHECK

  print( ls(pattern = eval(y))   )                    # TRY 1
  print( ls(pattern = eval(deparse(substitute(x)))) ) # TRY 2
}

Suppose I have 
rubbish.in = rubbish.out = grub <- 1

I get the following when I try
> lsp(rub)
[1] "rub"
character(0)
character(0)

Can someone explain/help with this? Thank you very much.

Adaikalavan Ramasamy wrote:
> Being the lazy soul I am, I wish to write a function to replace saying
> ls(pattern=...) everytime. Here is what I have:
> 
> lsp <- function(x){
>   y <- eval(deparse(substitute(x))) 
>   print(y)                                   # CHECK
> 
>   print( ls(pattern = eval(y))   )                    # TRY 1
>   print( ls(pattern = eval(deparse(substitute(x)))) ) # TRY 2
> }
> 
> Suppose I have 
> rubbish.in = rubbish.out = grub <- 1
> 
> I get the following when I try
> 
> 
>>lsp(rub)
> 
> [1] "rub"
> character(0)
> character(0)
> 
> Can someone explain/help with this? Thank you very much.
> 
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> http://www.stat.math.ethz.ch/mailman/listinfo/r-help

You have to think about environments and scoping rules.
You define another function and ls() is invoked in that functions, thus 
it is called in another environment than you obviously expected.
See ?ls, particularly its argument "name", why the following prints
the
things you expect from ls().

lsp <- function(x){
    y <- deparse(substitute(x))
    print(y)
    print(ls(sys.frame(sys.parent()), pattern = y))
  }

Consider to read
  Venables and Ripley (2000): S Programming, Springer.

Uwe Ligges