R help - Sep 2009 - creating Surv object from character vector

Hello, 

I have

srv <- Surv(sample(1:10), sample(0:1, 10, replace = TRUE))
srv

 [1]  1  10   2+  8   6+  7+  3   5+  4+  9+
srv.char <- as.character(srv)
srv.char

 [1] " 1 " "10 " " 2+" " 8 " "
6+" " 7+" " 3 " " 5+" " 4+" "
9+"

Is there an inverse to as.character(srv).  That is, I would like 

identical(srv, ???(srv.char)) to return TRUE, where ??? is some unknown
function.  I don't think it exists, but maybe I'm wrong. I suppose it
would be easy enough to roll my own...

Thanks!
Erik Iverson

On Sep 29, 2009, at 3:08 PM, Erik Iverson wrote:
> Hello,
>
> I have
>
> srv <- Surv(sample(1:10), sample(0:1, 10, replace = TRUE))
> srv
>
> [1]  1  10   2+  8   6+  7+  3   5+  4+  9+
> srv.char <- as.character(srv)
> srv.char
>
> [1] " 1 " "10 " " 2+" " 8 " "
6+" " 7+" " 3 " " 5+" " 4+" "
9+"
>
> Is there an inverse to as.character(srv).  That is, I would like
>
> identical(srv, ???(srv.char)) to return TRUE, where ??? is some  
> unknown function.  I don't think it exists, but maybe I'm wrong. I
> suppose it would be easy enough to roll my own...
Exactly. as.character strips off all the attributes and adds "+"'s
to
the (numeric) times that have status=0 and " "'s (or even
"-"'s) to
others:

 > str(as.character(srv))
  chr [1:10] " 4+" " 2 " " 8 " " 3 "
" 6 " " 1 " " 5+" " 7+" "10 "
...


Whereas:
 > str(srv)
  int [1:10, 1:2]  4+  2   8   3   6   1   5+  7+ 10   9+
  - attr(*, "dimnames")=List of 2
   ..$ : NULL
   ..$ : chr [1:2] "time" "status"
  - attr(*, "type")= chr "right"
  - attr(*, "units")= chr "Day"
  - attr(*, "time.label")= chr "sample(1:10)"
  - attr(*, "event.label")= chr "sample(0:1, 10, replace =
TRUE)"

 > srv[ , 1]
  [Day]
  [1]  4  2  8  3  6  1  5  7 10  9
 > srv[ , 2]
  [Day]
  [1] 0 1 1 1 1 1 0 0 1 0


 > methods(as.character)
Shows that there is an as.character.Surv function

David Winsemius, MD
Heritage Laboratories
West Hartford, CT

> 
> identical(srv, ???(srv.char)) to return TRUE, where ??? is some unknown
> function.  I don't think it exists, but maybe I'm wrong. I suppose
it
> would be easy enough to roll my own...
> 
FYI, This works in my simple case for right-censored data... it takes as input a
vector like c("1", "2+", "3+", "5")

as.Surv <- function(x) {
  Surv(as.integer(sub("\\+", "", x)),
       as.integer(ifelse(regexpr("\\+", x) > -1, 0, 1)))
}
> identical(srv, as.Surv(srv.char))
[1] TRUE

> 
> as.Surv <- function(x) {
>   Surv(as.integer(sub("\\+", "", x)),
>        as.integer(ifelse(regexpr("\\+", x) > -1, 0, 1)))
> }
> 
> > identical(srv, as.Surv(srv.char))
> [1] TRUE
> 
But Do NOT use "as.integer" with real data, which may contain
fractions of course.  I was too eager to get identical to return TRUE because I
had only used integers in my sample dataset! Oops.