Finding non diagonal elements in a matrix

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Here is what I want to do. Starting form this matrix :

  A B C D E F
A 3 0 2 1 2 0
B 4 4 0 0 4 1
C 2 0 3 0 2 0
D 1 0 0 5 1 0
E 5 0 0 3 6 1
F 0 0 0 0 2 3

I would like to determine all combinations containing 0 elements. For instance : (B,C,D) as :

  1. column B has zeros for C and D
  2. column C has zeros for B and D
  3. column D has zeros for B and C
  4. row B has zeros for C and D
  5. row C has zeros for B and D
  6. row D has zeros for B and C

column B has zeros for C and D, C

I did the following in R :

library(gtools)
mat <- matrix(c(3, 0, 2, 1, 2, 0, 4, 4, 0, 0, 4, 1, 2, 0, 3, 0, 2, 0, 1, 0, 0, 5,
                1, 0, 5, 0, 0, 3, 6, 1, 0, 0, 0, 0, 2, 3), nrow=6, ncol=6)
rownames(mat) <- colnames(mat) <- c("A", "B", "C", "D", "E", "F")
mat <- t(mat) # start matrix (as shown above)


non_diag <- mat - diag(diag(mat))  # matrix with non diagonal only
mat_check <- t(non_diag) + non_diag
colnames(mat_check) <- colnames(mat)
rownames(mat_check) <- rownames(mat)
diag(mat_check) <- diag(mat) # Needed for later


res <- apply(mat_check, 1, function(x) x[x == 0]) # Get elements that are zero (for each col)
j <- 1
list_seq <- list()
for (i in res) {
  tmp <- list(sequences=mixedsort(c(names(res)[j], names(res[[j]]))))
  list_seq [j] <- tmp
  j <- j + 1
}

unique(list_seq)
unique(list_seq)

and this outputs

[[1]]
[1] "A" "F"

[[2]]
[1] "B" "C" "D"

[[3]]
[1] "B" "C" "D" "F"

[[4]]
[1] "E"

[[5]]
[1] "A" "C" "D" "F"

which is not ok as

[[5]]
[1] "A" "C" "D" "F"

is not among possibilities.

Is there some way to do this ?

matrix r
1 Answers

Try this.

diag(mat) <- NA

r <- lapply(colnames(mat), \(x) {
  sort(c(x, intersect(names(which(mat[x, ] == 0)), names(which(mat[, x] == 0)))))
})

for (i in seq_along(r)) {
  r[-(1:i)] <- lapply(r[-(1:i)], setdiff, r[[i]])
}

lapply(r[lengths(r) > 0], toString)
# [[1]]
# [1] "A, F"
# 
# [[2]]
# [1] "B, C, D"
# 
# [[3]]
# [1] "E"
jay.sf0
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