delete.response

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Modify Terms Objects

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.

Keywords
programming
Usage
delete.response(termobj)

reformulate(termlabels, response = NULL, intercept = TRUE)

drop.terms(termobj, dropx = NULL, keep.response = FALSE)

Arguments
termobj

A terms object

termlabels

character vector giving the right-hand side of a model formula. Cannot be zero-length.

response

character string, symbol or call giving the left-hand side of a model formula, or NULL.

intercept

logical: should the formula have an intercept?

dropx

vector of positions of variables to drop from the right-hand side of the model.

keep.response

Keep the response in the resulting object?

Value

delete.response and drop.terms return a terms object.

reformulate returns a formula.

See Also

terms

Aliases
  • reformulate
  • drop.terms
  • delete.response
  • [.terms
Examples
library(stats) # 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("+-")) stopifnot(identical( ~ var, reformulate("var")), identical(~ a + b + c, reformulate(letters[1:3])), identical( y ~ a + b, reformulate(letters[1:2], "y")) ) # }
Documentation reproduced from package stats, version 3.5.0, License: Part of R 3.5.0

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