Dear Petr,
Your codes are so beautiful!
When I have a matrix s, with each column represents a sequence. I want to
recover all equivalent sequences from the sequences/columns in s. I used this
command
do.call(cbind,apply(s,2,function(x) getEquivalent(x,tt)))
This did a good job when ncol(s) >1, but when ncol(s)=1, there is an error.
How to getter a better coding which could deal with either the case of ncol(s)
>1 or ncol(s)=1 by itself?
Thanks a lot,
Wei
________________________________
From: Petr Savicky <savi...@cs.cas.cz>
To: r-help@r-project.org
Sent: Monday, February 13, 2012 5:38 PM
Subject: Re: [R] non-isomorphic sequences
On Mon, Feb 13, 2012 at 02:04:51PM -0800, zheng wei wrote:
> Dear Petr,
>
> This is fantastic!
>
> I have one more question, when p=4, tt=4. We have 15 non-isomorphic sequences
> as you have generated. Among these 15, I selected 2 sequences. How do I
> recover all the members of the equivalent classes corresponding to these 2
> sequences? For example, corresponding to the sequence of 1111, I would like
> to recover 1111,2222,3333,4444 from this sequence.
Dear Wei:
Try the following.
getEquivalent <- function(a, tt)
{
b <- as.matrix(rev(expand.grid(rep(list(1:tt), times=max(a)))))
ok <- apply(b, 1, function(x) length(unique(x))) == ncol(b)
b <- b[ok, , drop=FALSE]
dimnames(b) <- NULL
t(apply(b, 1, function(x) x[a]))
}
getEquivalent(c(1, 1, 1, 1), 4)
[,1] [,2] [,3] [,4]
[1,] 1 1 1 1
[2,] 2 2 2 2
[3,] 3 3 3 3
[4,] 4 4 4 4
getEquivalent(c(1, 1, 1, 2), 4)
[,1] [,2] [,3] [,4]
[1,] 1 1 1 2
[2,] 1 1 1 3
[3,] 1 1 1 4
[4,] 2 2 2 1
[5,] 2 2 2 3
[6,] 2 2 2 4
[7,] 3 3 3 1
[8,] 3 3 3 2
[9,] 3 3 3 4
[10,] 4 4 4 1
[11,] 4 4 4 2
[12,] 4 4 4 3
Hope this helps.
Petr.
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and provide commented, minimal, self-contained, reproducible code.
