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.

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"