Adding Diagonals of a matrix using R

Question

I would like to add up the upper part diagonals of a matrix starting from the middle, with increment in column until (1,n), n being the last column and save each sum of every diagonal. My code only add the middle diagonal, how can I loop through the matrix to get the sum of the diagonals

A <- matrix(c(2, 4, 3, 1, 5, 7, 1, 2, 3, 2, 3, 4, 1, 5, 6, 0), # the data elements nrow = 4, # number of rows ncol = 4, # number of columns byrow = TRUE) # fill matrix by rows sum <- 0 print(A) for (a in 1){ for (b in 1:ncol){ if (a<-b){ sum = sum + A[a,b] print (sum) } } }

Here is my result

> print(A) [,1] [,2] [,3] [,4] [1,] 2 4 3 1 [2,] 5 7 1 2 [3,] 3 2 3 4 [4,] 1 5 6 0 for (a in 1){ for (b in 1:ncol){ if (a<-b){ sum = sum + A[a,b] tail(sum, n=1) } } } 12

do you mean if (a==b) ??? otherwise I can't see how this would work ... — Ben Bolker
– Ben Bolker, Commented Mar 22, 2018 at 15:07
I can't tell what your desired result is. Is it c(1,8,13,12,9,5,1), i.e. sums of the off-diagonal elements? — Ben Bolker
– Ben Bolker, Commented Mar 22, 2018 at 15:09

Jilber Urbina · Accepted Answer · 2018-03-22 15:40:06Z

You need diag to extract all main diagonal elements and sum to get the sum of them

sum(diag(A))

I am not sure about what you're asking for, but if you also want to extract the upper triangular matrix, you can use A[upper.tri(A)] which excludes the main diagonal elements, you can also set diag=TRUE to include them A[upper.tri(A, diag = TRUE)]

@shegzter based on your comment, you can use col and row combined with logical comparison == to get the numbers you want.

> A[row(A)==col(A)] # this gives the same out put as `diag(A)` [1] 2 7 3 0 > A[row(A)+1==col(A)] [1] 4 1 4 > A[row(A)+2==col(A)] [1] 3 2 > A[row(A)+3==col(A)] [1] 1

If you want the sum of each of them, so use sum over those elements:

> sum(A[row(A)==col(A)]) [1] 12 > sum(A[row(A)+1==col(A)]) [1] 9 > sum(A[row(A)+2==col(A)]) [1] 5 > sum(A[row(A)+3==col(A)]) [1] 1

If your objective is getting the following sum 12+9+5+1, then you can do it all at once using upper.tri and sum

> sum(A[upper.tri(A, diag = TRUE)]) [1] 27

Or without the diagonal elements:

> sum(A[upper.tri(A)]) [1] 15

I want to be able to get 2 + 7+ 3 + 0 = 12 (which i already have) 4+ 1+ 4 = 9 3+2 = 5 1

Julius Vainora · Accepted Answer · 2018-03-22 23:40:43Z

The following returns the sum for each diagonal:

sapply(split(A, col(A) - row(A)), sum) # -3 -2 -1 0 1 2 3 # 1 8 13 12 9 5 1

Hence, to get only the upper, ones you may use

tail(sapply(split(A, col(A) - row(A)), sum), ncol(A)) # 0 1 2 3 # 12 9 5 1

The drawback of using tail is that we also compute lower diagonal sums. Hence, to save some time when A is large you may want to use

sapply(split(A[upper.tri(A, diag = TRUE)], (col(A) - row(A))[upper.tri(A, diag = TRUE)]), sum) # 0 1 2 3 # 12 9 5 1

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