I have like a matrix in R and I want to get:
Max off - diagonal elements
Min off – diagonal elements
Mean off –diagonal elements
With diagonal I used max(diag(A)) , min(diag(A)) , mean(diag(A)) and worked just fine
But for off-diagonal I tried
dataD <- subset(A, V1!=V2)
Error in subset.matrix(A, V1 != V2) : object 'V1' not found
to use:
colMeans(dataD) # get the mean for columns
but I cannot get dataD b/c it says object 'V1' not found
Thanks!
Here the row() and col() helper functions are useful. Using @James A, we can get the upper off-diagonal using this little trick:
> A[row(A) == (col(A) - 1)]
[1] 5 10 15
and the lower off diagonal via this:
> A[row(A) == (col(A) + 1)]
[1] 2 7 12
These can be generalised to give whatever diagonals you want:
> A[row(A) == (col(A) - 2)]
[1] 9 14
and don't require any subsetting.
Then it is a simple matter of calling whatever function you want on these values. E.g.:
> mean(A[row(A) == (col(A) - 1)])
[1] 10
If as per my comment you mean everything but the diagonal, then use
> diag(A) <- NA
> mean(A, na.rm = TRUE)
[1] 8.5
> max(A, na.rm = TRUE)
[1] 15
> # etc. using sum(A, na.rm = TRUE), min(A, na.rm = TRUE), etc..
So this doesn't get lost, Ben Bolker suggests (in the comments) that the above code block can be done more neatly using the row() and col() functions I mentioned above:
mean(A[row(A)!=col(A)])
min(A[row(A)!=col(A)])
max(A[row(A)!=col(A)])
sum(A[row(A)!=col(A)])
which is a nicer solution all round.
In one simple line of code:
For a matrix A if you wish to find the Minimum, 1st Quartile, Median, Mean, 3rd Quartile and Maximum of the upper and lower off diagonals:
summary(c(A[upper.tri(A)],A[lower.tri(A)])).
The diag of a suitably subsetted matrix will give you the off-diagonals. For example:
A <- matrix(1:16,4)
#upper off-diagonal
diag(A[-4,-1])
[1] 5 10 15
#lower off-diagonal
diag(A[-1,-4])
[1] 2 7 12
To get a vector holding the max of the off-diagonal elements of each col or row of a matrix requires a few more steps. I was directed here when searching for help on that. Perhaps others will do the same, so I offer this solution, which I found using what I learned here.
The trick is to create a matrix of only the off-diagonal elements. Consider:
> A <- matrix(c(10,2,3, 4,10,6, 7,8,10), ncol=3)
> A
[,1] [,2] [,3]
[1,] 10 4 7
[2,] 2 10 8
[3,] 3 6 10
> apply(A, 2, max)
[1] 10 10 10
Subsetting using the suggested indexing, A[row(A)!=col(A)] produces a vector of off-diagonal elements, in column-order:
> v <- A[row(A)!=col(A)]
> v
[1] 2 3 4 6 7 8
Returning this to a matrix allows the use of apply() to apply a function of choice to a margin of only off-diagonal elements. Using the max function as an example:
> A.off <- matrix(v, ncol=3)
> A.off
[,1] [,2] [,3]
[1,] 2 4 7
[2,] 3 6 8
> v <- apply(A.off, 2, max)
> v
[1] 3 6 8
The whole operation can be compactly—and rather cryptically—coded in one line:
> v <- apply(matrix(A[row(A)!=col(A)], ncol=ncol(A)), 2, max)
> v
[1] 3 6 8
Just multiply matrix A by 1-diag (nofelements)
for example if A is a 4x4 matrix, then
mean(A*(1-diag(4)) or A*(1-diag(nrow(A)))
This is faster when you need to run the same line of code multiple times
In addition to James' answer, I want to add that you can use the diag function to directly exclude all diagonal elements of a matrix by use of A[-diag(A)]. For example, consider:
summary(A[-diag(A)])
