[R] if then statement problem

[Gustavo Carvalho]

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 <aks2515 at uncw.edu> 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" wou!
>  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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