Re: [R] expand.grid() function

From: ONKELINX, Thierry <Thierry.ONKELINX_at_inbo.be>
Date: Mon, 23 Jun 2008 16:07:26 +0200

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
Cel biometrie, methodologie en kwaliteitszorg / Section biometrics, methodology and quality assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium
tel. + 32 54/436 185
Thierry.Onkelinx_at_inbo.be
www.inbo.be

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~ Sir Ronald Aylmer Fisher

The plural of anecdote is not data.
~ Roger Brinner

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-----Oorspronkelijk bericht-----
Van: r-help-bounces_at_r-project.org [mailto:r-help-bounces_at_r-project.org] Namens Gavin Simpson
Verzonden: maandag 23 juni 2008 15:51
Aan: Megh Dal
CC: r-help_at_stat.math.ethz.ch
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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Received on Mon 23 Jun 2008 - 14:31:50 GMT

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