[R] short puzzles

[Sundar Dorai-Raj]

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)[[1]] else names(x)
   nm.y <- if(is.array(y)) dimnames(y)[[1]] else names(y)
   nm.z <- if(is.array(z)) dimnames(z)[[1]] 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