Expand.grid works with lists too.
> 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
This is probably as concise as is can get.
HTH,
Thierry
------------------------------------------------------------------------
----
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and Forest
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-----Oorspronkelijk bericht-----
Van: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Namens Gavin Simpson
Verzonden: maandag 23 juni 2008 15:51
Aan: Megh Dal
CC: [EMAIL PROTECTED]
Onderwerp: Re: [R] expand.grid() function
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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