> It's not possible to be perfect in this scenario, but you could use
> some heuristics:
>
> survfit <- function(x, ...) {
> xq <- substitute(x)
> if (!is.call(xq)) {
> stop("First argument to survfit must be a function call")
> }
It would be good to add another logical argument, say substitute=TRUE or FALSE,
to let the caller avoid those heuristics and say exactly what is wanted. C.f.
library's
character.only argument.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
> -----Original Message-----
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf
> Of Hadley Wickham
> Sent: Friday, March 07, 2014 7:46 AM
> To: Therneau, Terry M., Ph.D.
> Cc: R-help; Thomas Lumley
> Subject: Re: [R] Survfit error
>
> On Fri, Mar 7, 2014 at 6:46 AM, Therneau, Terry M., Ph.D.
> <thern...@mayo.edu> wrote:
> >>> The help page for the survfit function says it expects a formula as its
> >>> first argument so try:
> >>>
> >>> > sleepfit <- survfit(Surv(timeb, death)~1, data = sleep)
> >>>
> >>> David
> >>> Sent from my iPhone ... so unable to test.
> >>>
> >>>
> >> This was a recent (well, 2007) change in behaviour. Previously the
> >> function
> >> did some tricks to make either approach work, which could be described as
> >> 'clever' or 'too clever by half'.
> >>
> >> -thomas
> >
> >
> > Certainly the latter, a design mistake that I finally admitted and
> > corrected. The trouble is this usage
> > fit <- survfit(Surv(time, status), data=mine)
> >
> > The data, subset, weights, and na.action arguments are all meant to work
> > with formulas. The first argument above isn't a formula. This left survfit
> > with the quandry of how to recognize that the first argument is a "Surv"
> > object and not a formula, and do something differently. Which leads to a
> > stalemate
> > a. One doesn't know what type of object the first argument is until
> > it is evaluated
> > b. Evaluation will fail, however, unless it is evaluated in the
> > "data=mine" context
> > c. The function that does that process (model.frame) expects a
> > formula as its first arg
> >
> > My program needed to know the answer to a) in order to transform the first
> > arg into a formula and correctly execute c), but needed the answer from c)
> > to know if that was necessary.
> >
> > The old code peeked back at the actual string that was typed by the user,
> > looking for the letters "Surv", but was easily fooled.
>
> It's not possible to be perfect in this scenario, but you could use
> some heuristics:
>
>
> survfit <- function(x, ...) {
> xq <- substitute(x)
> if (!is.call(xq)) {
> stop("First argument to survfit must be a function call")
> }
>
> if (identical(xq[[1]], quote(Surv))) {
> "surv"
> } else if (identical(xq[[1]], quote(`~`))) {
> "formula"
> } else {
> "don't know"
> }
> }
>
> survfit(Surv(time, status), data=mine)
> survfit(Surv(timeb, death)~1, data = sleep)
>
>
> You'd need a different heuristic if is.name(xq) is TRUE, indicating
> that the user passed in the name of an existing variable.
>
> Hadley
>
>
> --
> http://had.co.nz/
>
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