Hi,
I've tried to put together a simpler example where I'm having the issue.
I've built a foo package by only including a single .R file with the two
functions listed below: trt and cmt. The second function calls the first.
In the namespace file, if I only export(cmt), I get the following error
message when running this
library(foo)
set.seed(1)
dd <- data.frame(y = rbinom(100, 1, 0.5), treat = rbinom(100, 1, 0.5), x =
rnorm(100),
f = gl(4, 250, labels = c("A", "B", "C", "D")))
dd2 <- cmt(y ~ x + f + trt(treat), data =dd)
> Error could not find function "trt"
The problem is solved by doing export(cmt, trt) in the namespace. However,
I'd like to avoid exporting trt and should not be required. Sorry I can't
seem to figure this out by myself, and so I'd appreciate your help.
Thanks,
Axel.
----
#mycodefiles <- c("cmt.R")
#package.skeleton(name = "foo", code_files = mycodefiles)
#promptPackage("foo")
#where cmt.R includes the code below:
trt <- function(x) x
cmt <- function(formula, data, subset, na.action = na.pass) {
if (!inherits(formula, "formula"))
stop("Method is only for formula objects.")
mf <- match.call(expand.dots = FALSE)
args <- match(c("formula", "data", "subset", "na.action"),
names(mf), 0)
mf <- mf[c(1, args)]
mf$drop.unused.levels <- TRUE
mf[[1]] <- as.name("model.frame")
special <- "trt"
mt <- if(missing(data)) terms(formula, special) else terms(formula,
special, data = data)
browser()
mf$formula <- mt
mf <- eval.parent(mf)
Terms <- attr(mf, "terms")
attr(Terms, "intercept") <- 0
trt.var <- attr(Terms, "specials")$trt
ct <- mf[, trt.var]
y <- model.response(mf, "numeric")
var_names <- attributes(Terms)$term.labels[-(trt.var-1)]
x <- model.matrix(terms(reformulate(var_names)),
mf, contrasts)
intercept <- which(colnames(x) == "(Intercept)")
if (length(intercept > 0)) x <- x[, -intercept]
return(x)
}
On Mon, Jan 27, 2014 at 2:42 AM, Henrik Bengtsson <h...@biostat.ucsf.edu>wrote:
> On Sun, Jan 26, 2014 at 6:34 AM, Axel Urbiz <axel.ur...@gmail.com> wrote:
> > Hi Duncan,
> >
> > My most sincere apologies. It's really not my intention to waste anyones
> > time. More the opposite...for some reason I thought that the problem had
> to
> > do with my call to options() and thought that would be enough. Here's
> > something reproducible:
> >
> > I built a foo package based on the code under the "----" below. In the
> > namespace file, I've only exported: trt and cmt (not contr.none and
> > contr.diff). Notice that cmt calls contr.none and contr.diff by default.
>
> As a start, try to export everything, particularly 'contr.none' and
> 'contr.diff' and see if that works. Just a guess, but worth trying
> out.
>
> My $.02
>
> /Henrik
>
> >
> > Then in R, I run this code and I get this error message:
> >
> > library(foo)
> > set.seed(1)
> > dd <- data.frame(y = rbinom(100, 1, 0.5), treat = rbinom(100, 1, 0.5), x
> =
> > rnorm(100),
> > f = gl(4, 250, labels = c("A", "B", "C", "D")))
> > dd2 <- cmt(y ~ x + f + trt(treat), data =dd)
> >> Error in get(ctr, mode = "function", envir = parent.frame()) :
> > object 'contr.none' of mode 'function' was not found
> >
> > Thanks,
> > Axel.
> >
> > --------------------------------------------
> >
> > trt <- function(x) x
> >
> > cmt <- function(formula, data, subset, na.action = na.pass, cts = TRUE)
> {
> >
> > if (!inherits(formula, "formula"))
> > stop("Method is only for formula objects.")
> > mf <- match.call(expand.dots = FALSE)
> > args <- match(c("formula", "data", "subset", "na.action"),
> > names(mf), 0)
> > mf <- mf[c(1, args)]
> > mf$drop.unused.levels <- TRUE
> > mf[[1]] <- as.name("model.frame")
> > special <- "trt"
> > mt <- if(missing(data)) terms(formula, special) else terms(formula,
> > special, data = data)
> > mf$formula <- mt
> > mf <- eval.parent(mf)
> > Terms <- attr(mf, "terms")
> > attr(Terms, "intercept") <- 0
> > trt.var <- attr(Terms, "specials")$trt
> > ct <- mf[, trt.var]
> > y <- model.response(mf, "numeric")
> > var_names <- attributes(Terms)$term.labels[-(trt.var-1)]
> > treat.names <- levels(as.factor(ct))
> > oldcontrasts <- unlist(options("contrasts"))
> > if (cts)
> > options(contrasts = c(unordered = "contr.none", ordered =
> "contr.diff"))
> > x <- model.matrix(terms(reformulate(var_names)),
> > mf, contrasts)
> > options(contrasts = oldcontrasts)
> > intercept <- which(colnames(x) == "(Intercept)")
> > if (length(intercept > 0)) x <- x[, -intercept]
> > return(x)
> > }
> >
> > #######################################
> > # An alternative contrasts function for unordered factors
> > # Ensures symmetric treatment of all levels of a factor
> > #######################################
> > contr.none <- function(n, contrasts) {
> > if (length(n) == 1)
> > contr.treatment(n, contrasts = n<=2)
> > else
> > contr.treatment(n, contrasts = length(unique(n))<=2)
> > }
> >
> > #######################################
> > # An alternative contrasts function for ordered factors
> > # Ensures use of a difference penalty for such factors
> > #######################################
> > contr.diff <- function (n, contrasts = TRUE)
> > {
> > if (is.numeric(n) && length(n) == 1) {
> > if (n > 1)
> > levs <- 1:n
> > else stop("not enough degrees of freedom to define contrasts")
> > }
> > else {
> > levs <- n
> > n <- length(n)
> > }
> > contr <- array(0, c(n, n), list(levs, paste(">=", levs, sep="")))
> > contr[outer(1:n,1:n, ">=")] <- 1
> > if (n < 2)
> > stop(gettextf("contrasts not defined for %d degrees of freedom",
> > n - 1), domain = NA)
> > if (contrasts)
> > contr <- contr[, -1, drop = FALSE]
> > contr
> > }
> >
> >
> >
> > On Sun, Jan 26, 2014 at 1:21 PM, Duncan Murdoch <
> murdoch.dun...@gmail.com>wrote:
> >
> >> On 14-01-25 6:05 PM, Axel Urbiz wrote:
> >>
> >>> Thanks again all. Essentially, this is the section of the code that is
> >>> causing trouble. This is part of the (exported) function which calls
> >>> contr.none (not exported). As mentioned, when I call the exported
> function
> >>> it complains with the error described before.
> >>>
> >>>
> >>> oldcontrasts <- unlist(options("contrasts"))
> >>> if (cts)
> >>> options(contrasts = c(unordered = "contr.none", ordered =
> >>> "contr.diff"))
> >>> x <- model.matrix(terms(reformulate(var_names)), mf, contrasts)
> >>> options(contrasts = oldcontrasts)
> >>>
> >>
> >> This is hugely incomplete. Please stop wasting everyone's time, and
> post
> >> something reproducible.
> >>
> >> Duncan Murdoch
> >>
> >>
> >
> > [[alternative HTML version deleted]]
> >
> > ______________________________________________
> > R-devel@r-project.org mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-devel
>
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