R help - Jun 2008 - expand.grid() function

Hi,
   
  I have one question on expand.grid() function.
   
  When I write following syntax :expand.grid(c("u", "l"),
c("u", "l"), c("u", "l")) I get
following as desired :
    Var1 Var2 Var3
1    u    u    u
2    l    u    u
3    u    l    u
4    l    l    u
5    u    u    l
6    l    u    l
7    u    l    l
8    l    l    l

  However I wanted to write that in more concise manner. Therefore I tried :
expand.grid(rep(c("u", "l"), 3)). But I did not get answer
that I previously got. Can people here clarify me why it is not like that? Then
what would be the mose concise way to do that?
   
   
  
 

       
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On Mon, 2008-06-23 at 06:16 -0700, Megh Dal wrote:
> Hi,
>    
>   I have one question on expand.grid() function.
>    
>   When I write following syntax :expand.grid(c("u",
"l"), c("u", "l"),
> c("u", "l")) I get following as desired :
>     Var1 Var2 Var3
> 1    u    u    u
> 2    l    u    u
> 3    u    l    u
> 4    l    l    u
> 5    u    u    l
> 6    l    u    l
> 7    u    l    l
> 8    l    l    l
> 
>   However I wanted to write that in more concise manner. Therefore I
> tried : expand.grid(rep(c("u", "l"), 3)). But I did not
get answer
> that I previously got. Can people here clarify me why it is not like
> that? Then what would be the mose concise way to do that?
In the first case, you have three vectors of length 2 as arguments to
expand.grid, but in the second, you have a single vector of length 6. In
the latter case, expand.grid can't expand a single vector, hence the
single column result.

This is the closest I got to what you want:

as.matrix(expand.grid(split(rep(c("u","l"), times = 3), 
          factor(rep(1:3, each = 2)))))

Which gives:
> as.matrix(expand.grid(split(rep(c("u","l"), times = 3),
factor(rep(1:3, each = 2)))))
     1   2   3  
[1,] "u" "u" "u"
[2,] "l" "u" "u"
[3,] "u" "l" "u"
[4,] "l" "l" "u"
[5,] "u" "u" "l"
[6,] "l" "u" "l"
[7,] "u" "l" "l"
[8,] "l" "l" "l"

But that isn't particularly concise...

HTH

G
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Megh Dal <megh700004 <at> yahoo.com> writes:
>   I have one question on expand.grid() function.
>   When I write following syntax :expand.grid(c("u",
"l"),
>c("u", "l"), c("u", "l")) I get
following as
> desired :
>     Var1 Var2 Var3
> 1    u    u    u
> 2    l    u    u
> 3    u    l    u
> 4    l    l    u
> 5    u    u    l
> 6    l    u    l
> 7    u    l    l
> 8    l    l    l
>   However I wanted to write that in more concise manner. 
>Therefore I tried : expand.grid(rep(c("u", "l"),
> 3)). But I did not get answer that I previously got. 
>Can people here clarify me why it is not like that? >Then
> what would be the mose concise way to do that?
Just to put this under the correct subject heading, 
I repeat it here.

How about

do.call("expand.grid", rep(list(c("u", "l")), 3))
  Var1 Var2 Var3
1    u    u    u
2    l    u    u
3    u    l    u
4    l    l    u
5    u    u    l
6    l    u    l
7    u    l    l
8    l    l    l

Sorry about the incorrect posting...

Ken Knoblauch wrote:
>
> How about
>
> do.call("expand.grid", rep(list(c("u", "l")),
3))
>   Var1 Var2 Var3
> 1    u    u    u
> 2    l    u    u
> 3    u    l    u
> 4    l    l    u
> 5    u    u    l
> 6    l    u    l
> 7    u    l    l
> 8    l    l    l
>
>   
... which can now be nicely generalized and abstracted as

expand.grid.rep = function(x, n=1) do.call(expand.grid, rep(list(x),n))

expand.grid.rep(c("u","l"), 3)

vQ