[no subject]

[Erin Hodgess <hodgess@uhdux2.dt.uh.edu>]

Hi there!

Here's another variation on the theme: 

> grid1
function(x, n = length(x))
{
        y <- rep(cbind.data.frame(x), n)
        z <- expand.grid(y)
        z
}

> xa
[1] "a" "b"
> grid1(xa,3)
  X1 X1 X1 
1  a  a  a
2  b  a  a
3  a  b  a
4  b  b  a
5  a  a  b
6  b  a  b
7  a  b  b
8  b  b  b


Sincerely,
Erin M. Hodgess, Ph.D.
Assistant Professor
Department of Computer and Mathematical Sciences
University of Houston - Downtown
One Main Street 
Houston, TX 77002
e-mail: hodgess@uhdux2.dt.uh.edu


Subject: Re2: [splus-users,11557] [S] Combinatory function
In-Reply-To: <199906221414.XAA25328@cabbage.math.keio.ac.jp>


Hi!

>dd1<-seq(from=1, by=1, length=5)
>comb <- expand.grid(x1 =dd1 , x2 =dd1, x3 =dd1, x4 = dd1, x5=dd1,
>xx6=dd1)
>This program gives you all combinations of c(1,2,3,4,5).
>If you replace "1" by "a", "2" by "b",-----, "5" by "e",
>it results in all combinations of a,b,c,d,e.


   Program like below may be simpler; you do not have
to replace numbers by letters :

function()
{
        dd1 <- c("a", "b")
        comb <- expand.grid(x1 = dd1, x2 = dd1, x3 = dd1)
        print(comb)
}


Result is :

  x1 x2 x3 
1  a  a  a
2  b  a  a
3  a  b  a
4  b  b  a
5  a  a  b
6  b  a  b
7  a  b  b
8  b  b  b

K. Takezawa

   *****    Kunio Takezawa, Ph.D. (takezawa@affrc.go.jp)    *****
  *****            Research Information Section              *****
 **** Hokuriku National Agricultural Experiment Station, JAPAN ****
*****     <http://www.inada.affrc.go.jp/~takezawa/patent-e.html>    *****

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