[R] if ((x >.2 || x<(-.2)) && (col(x)!=row(x))) {x=x[, -col(x)]}

[David Winsemius]

You are trying to test the equality of a matrix to a scalar, which  
will produce a logical vector. You are also using && in an apparent  
attempt to conjoin a complex object which will probably not give you  
the results you expect in that context either since it would only  
return a single TRUE or FALSE. Use & in situations where you want  
element-wise comparisons of vectors.

You might start by experimenting on a much smaller object and seeing  
how your efforts at indexing could be improved.

 > X <- matrix(c(runif(9)),nrow=3)
 > X
           [,1]      [,2]      [,3]
[1,] 0.4151688 0.2116687 0.6049845
[2,] 0.7924464 0.6624862 0.8444203
[3,] 0.2634175 0.3357537 0.6923846

 > X>.8
       [,1]  [,2]  [,3]
[1,] FALSE FALSE FALSE
[2,] FALSE FALSE  TRUE
[3,] FALSE FALSE FALSE

#Use apply to create a logical vector that flags the unwanted rows:
 > apply(X,1,function(x) max(x) > 0.8)
[1] FALSE  TRUE FALSE

#Now use that construction on both rows and colums
 > X[-apply(X,1,function(x) max(x) > 0.8), -apply(X,2,function(x)  
max(x) > 0.8)]
           [,1]      [,2]
[1,] 0.6624862 0.8444203
[2,] 0.3357537 0.6923846


-- 
David


On May 8, 2009, at 1:39 AM, onyourmark wrote:

>
> Hi. I have a correlation matrix 'x' which is of size 923x923
>
> I need to remove variables that are highly correlated. I don't have a
> sophisticated way of selecting which of the two in a highly  
> correlated pair
> to remove. I thought I would just go through each entry of the  
> correlation
> matrix and if it is greater than 0.6 (or less than -0.6) I will  
> remove that
> column and then redo the check from scratch with the matrix (where  
> now the
> matrix has one less column).
> As a test, I tried if ((x >.2 || x<(-.2)) && col(x)!=row(x)) {x=x[,- 
> col(x)]}
> but it does not remove any columns.
>
> Also, I realized that I actually need to pull out the row that is  
> associated
> with that variable as well, and so perhaps the section inside {}  
> needs to be
> something like:
> {x=x[-col(x),-col(x)]
> Any idea on how to do this?
> Thank you.
> -- 
> View this message in context: http://www.nabble.com/if-%28%28x-%3E.2-%7C%7C-x%3C%28-.2%29%29----%28col%28x%29%21%3Drow%28x%29%29%29-%7Bx%3Dx-%2C-col%28x%29-%7D-tp23440419p23440419.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

David Winsemius, MD
Heritage Laboratories
West Hartford, CT