How to reliably get dependent variable name from formula object?

Question

Let's say I have the following formula:

myformula<-formula("depVar ~ Var1 + Var2")

How to reliably get dependent variable name from formula object?

I failed to find any built-in function that serves this purpose. I know that as.character(myformula)[[2]] works, as do

sub("^(\\w*)\\s~\\s.*$","\\1",deparse(myform))

It just seems to me, that these methods are more a hackery, than a reliable and standard method to do it.

---

Does anyone know perchance what exactly method the e.g. lm use? I've seen it's code, but it is a little to cryptic to me... here is a quote for your convenience:

    > lm
function (formula, data, subset, weights, na.action, method = "qr", 
    model = TRUE, x = FALSE, y = FALSE, qr = TRUE, singular.ok = TRUE, 
    contrasts = NULL, offset, ...) 
{
    ret.x <- x
    ret.y <- y
    cl <- match.call()
    mf <- match.call(expand.dots = FALSE)
    m <- match(c("formula", "data", "subset", "weights", "na.action", 
        "offset"), names(mf), 0L)
    mf <- mf[c(1L, m)]
    mf$drop.unused.levels <- TRUE
    mf[[1L]] <- as.name("model.frame")
    mf <- eval(mf, parent.frame())
    if (method == "model.frame") 
        return(mf)
    else if (method != "qr") 
        warning(gettextf("method = '%s' is not supported. Using 'qr'", 
            method), domain = NA)
    mt <- attr(mf, "terms")
    y <- model.response(mf, "numeric")
    w <- as.vector(model.weights(mf))
    if (!is.null(w) && !is.numeric(w)) 
        stop("'weights' must be a numeric vector")
    offset <- as.vector(model.offset(mf))
    if (!is.null(offset)) {
        if (length(offset) != NROW(y)) 
            stop(gettextf("number of offsets is %d, should equal %d (number of observations)", 
                length(offset), NROW(y)), domain = NA)
    }
    if (is.empty.model(mt)) {
        x <- NULL
        z <- list(coefficients = if (is.matrix(y)) matrix(, 0, 
            3) else numeric(), residuals = y, fitted.values = 0 * 
            y, weights = w, rank = 0L, df.residual = if (!is.null(w)) sum(w != 
            0) else if (is.matrix(y)) nrow(y) else length(y))
        if (!is.null(offset)) {
            z$fitted.values <- offset
            z$residuals <- y - offset
        }
    }
    else {
        x <- model.matrix(mt, mf, contrasts)
        z <- if (is.null(w)) 
            lm.fit(x, y, offset = offset, singular.ok = singular.ok, 
                ...)
        else lm.wfit(x, y, w, offset = offset, singular.ok = singular.ok, 
            ...)
    }
    class(z) <- c(if (is.matrix(y)) "mlm", "lm")
    z$na.action <- attr(mf, "na.action")
    z$offset <- offset
    z$contrasts <- attr(x, "contrasts")
    z$xlevels <- .getXlevels(mt, mf)
    z$call <- cl
    z$terms <- mt
    if (model) 
        z$model <- mf
    if (ret.x) 
        z$x <- x
    if (ret.y) 
        z$y <- y
    if (!qr) 
        z$qr <- NULL
    z
}

Answer 1

Try using all.vars:

all.vars(myformula)[1]

Answer 2

I suppose you could also cook your own function to work with terms():

getResponse <- function(formula) {
    tt <- terms(formula)
    vars <- as.character(attr(tt, "variables"))[-1] ## [1] is the list call
    response <- attr(tt, "response") # index of response var
    vars[response] 
}

R> myformula <- formula("depVar ~ Var1 + Var2")
R> getResponse(myformula)
[1] "depVar"

It is just as hacky as as.character(myformyula)[[2]] but you have the assurance that you get the correct variable as the ordering of the call parse tree isn't going to change any time soon.

This isn't so good with multiple dependent variables:

R> myformula <- formula("depVar1 + depVar2 ~ Var1 + Var2")
R> getResponse(myformula)
[1] "depVar1 + depVar2"

as they'll need further processing.

Answer 3

I found an useful package 'formula.tools' which is suitable for your task.

code Example:

f <- as.formula(a1 + a2~a3 + a4)

lhs.vars(f) #get dependent variables

[1] "a1" "a2"

rhs.vars(f) #get independent variables

[1] "a3" "a4"