Question

Suppose I have a following formula:

fr <- formula(y~myfun(x)+z)

Given object fr is there a function in R which returns myfun(x)? I’ve written my own function (code below) which basically does what I need, but maybe there is some standard way of doing that?

The code for my function:

selectmds <- function(expr,funcn) {
    if(length(expr)>2) {
        a <- expr[[2]]
        b <- expr[[3]]
        if(length(a)>1) {
            if(as.name(a[[1]])==funcn) {
                if(length(grep(funcn,all.names(b)))>0) {
                    return(list(a,selectmds(b,funcn)))
                }
                else return(list(a))
            }
        }
        if(length(b)>1) {
            if(as.name(b[[1]])==funcn) {
                if(length(grep(funcn,all.names(a)))>0) {
                    return(list(b,selectmds(a,funcn)))
                }
                else return(list(b))
            }
        }
        for(i in 2:length(expr)) {
            if(length(grep(funcn,all.names(expr[[i]])))>0)return(selectmds(expr[[i]],funcn))
        }
    }
    return(NULL)
}

Here are several examples:

> selectmds(formula(y~myfun(x)+z),"myfun")
[[1]]
myfun(x)


> unlist(selectmds(formula(y~myfun(x)+z+myfun(zz)),"myfun"))
[[1]]
myfun(zz)

[[2]]
myfun(x)

Answer 1

Not sure this is best, but you can do it by:

f <- function(fm, fun) {
  l <- as.list(attr(terms(fm), "variables"))[-1]
  l[grep(fun, l)]
}

then,

> f(formula(y~myfun(x)+z),"myfun")
[[1]]
myfun(x)

> f(formula(y~myfun(x)+z+myfun(zz)),"myfun")
[[1]]
myfun(x)

[[2]]
myfun(zz)