delete.response returns a terms object for the same
model but with no response variable.
drop.terms removes variables from the right-hand side of the
model. There is also a "[.terms" method to perform the same
function (with keep.response = TRUE).
reformulate creates a formula from a character vector. If
length(termlabels) > 1, its elements are concatenated with +.
Non-syntactic names (e.g. containing spaces or special characters; see
make.names) must be protected with backticks (see examples).
A non-parseable response still works for now,
back compatibly, with a deprecation warning.
delete.response(termobj)reformulate(termlabels, response = NULL, intercept = TRUE, env = parent.frame())
drop.terms(termobj, dropx = NULL, keep.response = FALSE)
A terms object
character vector giving the right-hand side of a model formula. Cannot be zero-length.
character string, symbol or call giving the left-hand
side of a model formula, or NULL.
logical: should the formula have an intercept?
the environment of the formula
returned.
vector of positions of variables to drop from the right-hand side of the model.
Keep the response in the resulting object?
delete.response and drop.terms return a terms
object.
reformulate returns a formula.
# NOT RUN {
ff <- y ~ z + x + w
tt <- terms(ff)
tt
delete.response(tt)
drop.terms(tt, 2:3, keep.response = TRUE)
tt[-1]
tt[2:3]
reformulate(attr(tt, "term.labels"))
## keep LHS :
reformulate("x*w", ff[[2]])
fS <- surv(ft, case) ~ a + b
reformulate(c("a", "b*f"), fS[[2]])
## using non-syntactic names:
reformulate(c("`P/E`", "`% Growth`"), response = as.name("+-"))
x <- c("a name", "another name")
try( reformulate(x) ) # -> Error ..... unexpected symbol
## rather backquote the strings in x :
reformulate(sprintf("`%s`", x))
stopifnot(identical( ~ var, reformulate("var")),
identical(~ a + b + c, reformulate(letters[1:3])),
identical( y ~ a + b, reformulate(letters[1:2], "y"))
)
# }
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