Cast function argument as a character string?

Question

Super quick question...

How do you take a some particular function's (user-defined) argument and cast it as a character-string?

If for a simple example,

foo <- function(x) { ... }

I want to simply return x's object name. So,

foo(testing123)

returns "testing123" (and testing123 could just be some random numeric vector)

Apologies if this question has been asked before--searched, but couldn't find it! Thanks!!

score 42 · Accepted Answer · answered Nov 05 '10 at 17:56

42

foo <- function(x) deparse(substitute(x))

answered Nov 05 '10 at 17:56

JD Long

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score 20 · Answer 2 · edited Jul 21 '13 at 20:10

Meta-answer: if you know R does something and you want to do it, check the source. For example, you may have spotted that plot(foo) sticks 'foo' in the ylab, so plot can do it. How? Start by looking at the code:

> plot
function (x, y, ...) 
{
    if (is.function(x) && is.null(attr(x, "class"))) {
        if (missing(y)) 
            y <- NULL
        hasylab <- function(...) !all(is.na(pmatch(names(list(...)), 
            "ylab")))
        if (hasylab(...)) 
            plot.function(x, y, ...)
        else plot.function(x, y, ylab = paste(deparse(substitute(x)), 
            "(x)"), ...)
    }
    else UseMethod("plot")
}

And there's some deparse(substitute(x)) magic.

Not sure if it really matters, as the answer itself is great, but `plot` doesn't look like this anymore — Rich Scriven, Jan 17 '15 at 00:41

score 2 · Answer 3 · answered Nov 05 '10 at 17:56

2

Whoops, apparently I didn't search hard enough...

foo <- function(x) {return(as.character(substitute(x)))}

Well that's easy...

answered Nov 05 '10 at 17:56

Ray

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`deparse(substitute(x))` would be the usual way of doing this, as per JD's answer. Compare your version with JD's on this `foo(testing * bar)` to see why. – Gavin Simpson Nov 05 '10 at 18:24

Cast function argument as a character string?

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