R help - Apr 2011 - matrix evaluation using if function

Hi All,

I am trying to create a function which evaluates whether the values (which
are equal to one) of a matrix are the same as their mirror values. Consider
the following matrix:
> n<-matrix(cbind(c(0,1,1),c(1,0,0),c(0,1,0)),3,3)
>
colnames(n)<-cbind("A","B","C");rownames(n)<-cbind("A","B","C")
> n
  A B C
A 0 1 0
B 1 0 1
C 1 0 0

Hence, since n[2,1] and n[1,2] are 1 and the same, the function should
return the name of the row of n[2,1]. I used the following function:

for (i in length(rownames(n))) {

for (j in length(colnames(n))){

if(n[i,j]==n[j,i]){

rownames(n)[[i]]->output} else {}

}

}
> output
NULL

The right answer would have been "B", though. I simply do not see my
mistake. I am very greatful for suggestions.

Thank you.

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On Apr 29, 2011, at 4:27 AM, ivan wrote:
> Hi All,
>
> I am trying to create a function which evaluates whether the values  
> (which
> are equal to one) of a matrix are the same as their mirror values.  
> Consider
> the following matrix:
>
>> n<-matrix(cbind(c(0,1,1),c(1,0,0),c(0,1,0)),3,3)
>>
colnames(n)<-cbind("A","B","C");rownames(n)<-cbind("A","B","C")
>> n
>  A B C
> A 0 1 0
> B 1 0 1
> C 1 0 0
>
> Hence, since n[2,1] and n[1,2] are 1 and the same, the function should
> return the name of the row of n[2,1]. I used the following function:
>
> for (i in length(rownames(n))) {
>
> for (j in length(colnames(n))){
>
> if(n[i,j]==n[j,i]){
>
> rownames(n)[[i]]->output} else {}
>
> }
>
> }
>
>> output
> NULL
>
> The right answer would have been "B", though.
Can you explain why "A" would not be an equally good answer to satisfy
your problem set up?

 > which(n == t(n) & col(n) != row(n) , arr.ind=TRUE)
   row col
B   2   1
A   1   2
 > rownames(which(n == t(n) & col(n) != row(n) , arr.ind=TRUE) )
[1] "B" "A"

# Which would seem to be the correct answer, but
# This adds an additional constraint and also insures no diagonal  
elements

 > rownames(which(n == t(n) & col(n) != row(n) & lower.tri(n),  
arr.ind=TRUE) )
[1] "B"



> I simply do not see my
> mistake.
I would rather program a problem correctly that hash through errors in  
loop logic.
-- 

David Winsemius, MD
West Hartford, CT