R: how to count the number of times two elements have the same ID (perhaps using the outer function) - r

Question

I have the following three dimensional array:
dput(a)
structure(c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 3, 2, 1, 1, 1, 2, 2,
2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 6, 2, 7, 6, 2, 7, 6, 2, 7, 4, 2, 4, 4, 2, 6, 4, 2, 4, 6, 2,
7, 4, 2, 6, 4, 2, 6, 4, 2, 6, 4, 2, 4, 4, 2, 6, 4, 2, 4, 4, 2,
6, 4, 2, 6, 4, 2, 6, 6, 2, 7, 4, 2, 6, 4, 2, 6, 4, 2, 4, 2, 3,
1, 2, 3, 1, 2, 3, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 3, 7, 2, 3,
7, 2, 3, 7, 2, 3, 7, 2, 3, 7, 2, 3, 7, 2, 3, 7, 2, 3, 7, 2, 3,
7, 2, 3, 7, 1, 2, 5, 2, 3, 7, 1, 2, 4, 2, 3, 7, 2, 3, 7, 2, 3,
7, 2, 3, 7, 2, 3, 7, 2, 3, 7, 2, 3, 7, 2, 6, 3, 2, 6, 3, 2, 6,
3, 2, 6, 3, 2, 6, 3, 2, 6, 3, 2, 6, 3, 2, 6, 3, 2, 6, 3, 2, 6,
3, 1, 1, 1, 2, 6, 3, 1, 5, 5, 2, 6, 3, 2, 6, 3, 2, 6, 3, 2, 6,
3, 2, 6, 3, 2, 6, 3, 2, 6, 3, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 13,
2, 3, 13, 2, 3, 5, 2, 3, 5, 2, 15, 17, 2, 15, 17, 2, 15, 17,
2, 3, 5, 2, 15, 17, 2, 3, 13, 2, 15, 17, 2, 15, 17, 2, 3, 13,
2, 3, 5, 2, 15, 17, 2, 15, 17, 2, 3, 5, 2), .Dim = c(3L, 20L,
6L), .Dimnames = list(c("cl.tmp", "cl.tmp", "cl.tmp"), NULL,
NULL))
The dimension of this array (a) is 3x20x6 (after edits).
I wanted to count the proportion of times that a[,i,] matches a[,j,] element-by-element in the matrix. Basically, I wanted to get mean(a[,i,] == a[,j,]) for all i, j, and I would like to do this fast but in R.
It occurred to me that the outer function might be a possibility but I am not sure how to specify the function. Any suggestions, or any other alternative ways?
The output would be a 20x20 symmetric matrix of nonnegative elements with 1 on the diagonals.
The solution given below works (thanks!) but I have one further question (sorry).
I would like to display the coordinates above in a heatmap. I try the following:
n<-dim(a)[2]
xx <- matrix(apply(a[,rep(1:n,n),]==a[,rep(1:n,each=n),],2,sum),nrow=n)/prod(dim(a)[-2])
image(1:20, 1:20, xx, xlab = "", ylab = "")
This gives me the following heatmap.
However, I would like to display (reorder the coordinate) such that I get all the coordinates that have high-values amongst each other together. However, I would not like to bias the results by deciding on the number of groups myself. I tried
hc <- hclust(as.dist(1-xx), method = "single")
but I can not decide how to cut the resulting tree to decide on bunching the coordinates together. Any suggestions? Bascically, in the figure, I would like the coordinate pairs in the top left (and bottom right off-diagonal blocks) to be as low-valued (in this case as red) as possible.
Looking around on SO, I found that there exists a function heatmap which might do this,
heatmap(xx,Colv=T,Rowv=T, scale='none',symm = T)
and I get the following:
which is all right, but I can not figure out how to get rid of the dendrograms on the sides or the axes labels. It does work if I extract out and do the following:
yy <- heatmap(xx,Colv=T,Rowv=T, scale='none',symm = T,keep.dendro=F)
image(1:20, 1:20, xx[yy$rowInd,yy$colInd], xlab = "", ylab = "")
so I guess that is what I will stick with. Here is the result:

Answer 1

Try this:
n<-dim(a)[2]
matrix(apply(a[,rep(1:n,n),]==a[,rep(1:n,each=n),],2,sum),nrow=n)/prod(dim(a)[-2])
It has to be stressed that the memory usage of this method goes with n^2 so you might have trouble to use it with larger arrays.