[R] short puzzles

Wiener, Matthew matthew_wiener at merck.com
Fri Jul 11 14:49:30 CEST 2003

I'll try #1.  You can use outer twice:

outer(outer(x,y, "/"),z,function(a,b),"^")

In general I suppose you could write an "outer.n" function that would take
the vectors and the functions to be applied and loop through them.  And I
think that outer will then take care of the dimnames for you.  (Taking care
of #2, and, I think, #3.)

Hope this helps.

Matt Wiener

-----Original Message-----
From: Marc Vandemeulebroecke [mailto:vandemem at gmx.de] 
Sent: Friday, July 11, 2003 7:48 AM
To: r-help at stat.math.ethz.ch
Subject: [R] short puzzles

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}
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
outer.n(), generalizing outer() to functions with an arbitrary number of

Define a function dimnames.outer() such that dimnames.outer(x, y, "*")
returns, for x <- 1:2, y <- 2:3, the following matrix:

x   2 3
  1 2 3
  2 4 6

(Or does such such a function already exist?)


How to combine puzzle 1 and puzzle 2? A function dimnames.outer.n() would be
a nice little tool.


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
f(a) as defined below,

f <- function (x) {
  # How can I get "a" out of x?


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)"?

Many thanks,


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