# [R] short puzzles

Sundar Dorai-Raj sundar.dorai-raj at pdf.com
Fri Jul 11 14:42:49 CEST 2003

```
Marc Vandemeulebroecke wrote:
> Dear R users,
>
> can someone help with these short puzzles?
>
> 1) Is there a function like outer() that evaluates a three-argument function
> on a threedimensional grid - or else how to define such a function, say,
> outer.3()? E.g., calculate (x/y)^z on (x,y,z) element of {1,2,3}x{3,4}x{4,5} and
> return the results in a 3-dimensional array. I would naively use outer() on
> two of the arguments within a for() loop for the third argument and somehow
> glue the array together. Is there a better way? What about outer.4(), or even
> outer.n(), generalizing outer() to functions with an arbitrary number of
> arguments?
>
> 2)
> Define a function dimnames.outer() such that dimnames.outer(x, y, "*")
> returns, for x <- 1:2, y <- 2:3, the following matrix:
>
>    y
> x   2 3
>   1 2 3
>   2 4 6
>
> (Or does such such a function already exist?)
>
> 3)
>
> How to combine puzzle 1 and puzzle 2? A function dimnames.outer.n() would be
> a nice little tool.
>

Here's what I came up with. If you need the other functions you
mentioned, you can extract them from this example.

outer.3 <- function(x, y, z, FUN, ...) {
print(deparse(substitute(x))) # for question 2
n.x <- NROW(x)
n.y <- NROW(y)
n.z <- NROW(z)
nm.x <- if(is.array(x)) dimnames(x)[] else names(x)
nm.y <- if(is.array(y)) dimnames(y)[] else names(y)
nm.z <- if(is.array(z)) dimnames(z)[] else names(z)
X <- expand.grid(x = x, y = y, z = z)
f <- FUN(X\$x, X\$y, X\$z, ...)
array(f, dim = c(n.x, n.y, n.z),
dimnames = list(nm.x, nm.y, nm.z))
}

a <- 1:3
b <- 3:4
c <- 4:5
names(a) <- a
names(b) <- b
names(c) <- c
outer.3(a, b, c, function(x, y, z) (x/y)^z)
outer.3(as.matrix(a), as.matrix(b), as.matrix(c),
function(x, y, z) (x/y)^z)

> 4)
>
> How can I access, within a function, the name of a variable that I have
> passed to the function? E.g., letting a <- 2, and subsequently calling function
> f(a) as defined below,
>
> f <- function (x) {
>   # How can I get "a" out of x?
> }
>

Use deparse(substitute(x)). See example above.

> 5)
>
> Finally: Letting x <- 2, how can I transform "x+y" into "2+y" (as some
> suitable object), or generally "func(x,y)" into "func(2,y)"?
>

Use substitute(func(x, y), list(x = 2)).

Hope this is useful,

Sundar

```