If I understood your problem correctly, you are just missing a couple of things:
q = which(apply(p.unique,2,function(x)all(x==r)) == TRUE)
Also, you should probably change this line:
if(q>0){c=p.unique[,q]};{c=c(0,0,0)}
To something like this:
if(length(q)>0){c=p.unique[,q]};{c=c(0,0,0)}
Regards,
Gustavo
On Thu, Nov 27, 2008 at 3:02 PM, Salas, Andria Kay <[EMAIL PROTECTED]> wrote:
> I really need help with an if then statement I am having trouble with. A
> brief explanation of what I am trying to do: I have a p matrix of all
> permutations of a vector of length 3 filled with either 1s or -1s (these make
> up the columns). The p.unique matrix is the unique columns of the p matrix;
> here, they are identical but in my real script this will not always be the
> case. I want to be able to take a randomly generated vector and see which
> column in the p.unique matrix, if any, match the random vector. If there is
> a match, I get an integer returned that tells me which column is the match.
> If there is no match, I get the response integer (0). I want to write an if
> then statement that allows me to do different things based upon if there is a
> match to one of the columns in the p.unique matrix or not. Below is a sample
> script plucked from bits of my "master script" that show this problem. The
> vector "r" was made to not match any columns in p.unique. If it did, "c" w!
ou!
> ld equal the integer representing the matching column, and since it does
> not, I want c to equal the vector (0,0,0). My if then statement is not
> working (I keep getting the error that my argument is of length zero), so I
> would really appreciate any help anyone can provide me. Thanks for all of
> your help previously and in advance for the help with this problem!! Happy
> Thanksgiving!
>
> p=as.matrix(do.call(expand.grid,rep(list(c(-1,1)),3)))
> p=t(p)
> p.unique=unique(p,MARGIN=2)
>
> r=c(2,2,2)
> q=which(apply(p.unique,2,function(x)all(x==r)))
> if(q>0){c=p.unique[,q]};{c=c(0,0,0)}
>
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>
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