Combining matrices (above and below diagonal)

Question

Say I have this matrices:

emp <- matrix(nrow = 13, ncol = 13)
emp2 <- matrix(nrow = 13, ncol = 13)

emp[lower.tri(emp)] <- 2
emp2[lower.tri(emp)] <- 3

And I want to combine them so below the diagonal, I have all values = 2 and above all values = 3. I have managed to produce this as such (then the result is just the emp matrix).

emp[upper.tri(emp)] <- t(emp2[lower.tri(emp2)])

However, now I would like to store the result and make a function for this (after all the matrices I want to use are inside elements of a list). Do you know how can I store this in a new matrix? Or perhaps a better alternative.

I hope I have provided a reproducible enough example, otherwise, let me know.

Thank you!

If dimensions are equal consider storing in arrays then!

Friede
– Friede

2025-06-02 17:35:10 +00:00
Commented Jun 2 at 17:35 — Friede
– Friede, Commented Jun 2 at 17:35

ThomasIsCoding · Accepted Answer · 2025-06-05 19:34:25Z

2

I think you can use pmax with option na.rm = TRUE

> pmax(emp, t(emp2), na.rm = TRUE)
      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
 [1,]   NA    3    3    3    3    3    3    3    3     3     3     3     3
 [2,]    2   NA    3    3    3    3    3    3    3     3     3     3     3
 [3,]    2    2   NA    3    3    3    3    3    3     3     3     3     3
 [4,]    2    2    2   NA    3    3    3    3    3     3     3     3     3
 [5,]    2    2    2    2   NA    3    3    3    3     3     3     3     3
 [6,]    2    2    2    2    2   NA    3    3    3     3     3     3     3
 [7,]    2    2    2    2    2    2   NA    3    3     3     3     3     3
 [8,]    2    2    2    2    2    2    2   NA    3     3     3     3     3
 [9,]    2    2    2    2    2    2    2    2   NA     3     3     3     3
[10,]    2    2    2    2    2    2    2    2    2    NA     3     3     3
[11,]    2    2    2    2    2    2    2    2    2     2    NA     3     3
[12,]    2    2    2    2    2    2    2    2    2     2     2    NA     3
[13,]    2    2    2    2    2    2    2    2    2     2     2     2    NA

answered Jun 5 at 19:34

ThomasIsCoding

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Daniel Estévez Jun 5 at 20:27

This is the best answer, thank you

jay.sf · Accepted Answer · 2025-06-02 15:50:58Z

Just wrap the logic in a function and Map over the list.

> fx <- \(e1, e2) {
+   e1[upper.tri(e1)] <- t(e2[lower.tri(e2)])
+   e1
+ }
> lst <- list(list(emp, emp, emp), list(emp2, emp2, emp2))
> Map(fx, lst[[1]], lst[[2]])
[[1]]
      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
 [1,]   NA    3    3    3    3    3    3    3    3     3     3     3     3
 [2,]    2   NA    3    3    3    3    3    3    3     3     3     3     3
 [3,]    2    2   NA    3    3    3    3    3    3     3     3     3     3
 [4,]    2    2    2   NA    3    3    3    3    3     3     3     3     3
 [5,]    2    2    2    2   NA    3    3    3    3     3     3     3     3
 [6,]    2    2    2    2    2   NA    3    3    3     3     3     3     3
 [7,]    2    2    2    2    2    2   NA    3    3     3     3     3     3
 [8,]    2    2    2    2    2    2    2   NA    3     3     3     3     3
 [9,]    2    2    2    2    2    2    2    2   NA     3     3     3     3
[10,]    2    2    2    2    2    2    2    2    2    NA     3     3     3
[11,]    2    2    2    2    2    2    2    2    2     2    NA     3     3
[12,]    2    2    2    2    2    2    2    2    2     2     2    NA     3
[13,]    2    2    2    2    2    2    2    2    2     2     2     2    NA

[[2]]
      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
 [1,]   NA    3    3    3    3    3    3    3    3     3     3     3     3
 [2,]    2   NA    3    3    3    3    3    3    3     3     3     3     3
 [3,]    2    2   NA    3    3    3    3    3    3     3     3     3     3
 [4,]    2    2    2   NA    3    3    3    3    3     3     3     3     3
 [5,]    2    2    2    2   NA    3    3    3    3     3     3     3     3
 [6,]    2    2    2    2    2   NA    3    3    3     3     3     3     3
 [7,]    2    2    2    2    2    2   NA    3    3     3     3     3     3
 [8,]    2    2    2    2    2    2    2   NA    3     3     3     3     3
 [9,]    2    2    2    2    2    2    2    2   NA     3     3     3     3
[10,]    2    2    2    2    2    2    2    2    2    NA     3     3     3
[11,]    2    2    2    2    2    2    2    2    2     2    NA     3     3
[12,]    2    2    2    2    2    2    2    2    2     2     2    NA     3
[13,]    2    2    2    2    2    2    2    2    2     2     2     2    NA

[[3]]
      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
 [1,]   NA    3    3    3    3    3    3    3    3     3     3     3     3
 [2,]    2   NA    3    3    3    3    3    3    3     3     3     3     3
 [3,]    2    2   NA    3    3    3    3    3    3     3     3     3     3
 [4,]    2    2    2   NA    3    3    3    3    3     3     3     3     3
 [5,]    2    2    2    2   NA    3    3    3    3     3     3     3     3
 [6,]    2    2    2    2    2   NA    3    3    3     3     3     3     3
 [7,]    2    2    2    2    2    2   NA    3    3     3     3     3     3
 [8,]    2    2    2    2    2    2    2   NA    3     3     3     3     3
 [9,]    2    2    2    2    2    2    2    2   NA     3     3     3     3
[10,]    2    2    2    2    2    2    2    2    2    NA     3     3     3
[11,]    2    2    2    2    2    2    2    2    2     2    NA     3     3
[12,]    2    2    2    2    2    2    2    2    2     2     2    NA     3
[13,]    2    2    2    2    2    2    2    2    2     2     2     2    NA

Thank you! I hope I do not bother you by asking, but I do not understand the "\" in the function call. Could you explain it? Again thanks
@DanielEstévez \(x) is shorthand for function(x), actually since R4.1.

Friede · Accepted Answer · 2025-06-05 17:32:49Z

-1

I actually found a better solution and I wanted to post it here. In the previous example, it did work because both matrices were composed of single values. But the function is actually not working as it should. It should mirror the lower triangle of the second matrix in the upper triangle of the first one.

You can check this simulating matrices with difference values:

mat1 <- matrix(nrow = 10, ncol = 10)
mat2 <- matrix(nrow = 10, ncol = 10)

mat1[lower.tri(mat1)] <- runif(n = 10, min = 0, max = 1)
mat2[lower.tri(mat2)] <- runif(n = 10, min = 0, max = 1)

fx <- function(mat1, mat2) {
  n <- nrow(mat1)
  for (i in 1:n) {
    for (j in 1:n) {
      if (i > j) {
        mat1[j, i] <- mat2[i, j]
      }
    }
  }
  mat1
}

mat3 <- fx(mat1, mat2)

Note. We can see that in mat3 now the upper triangle corresponds to t(mat2).

edited Jun 5 at 17:32

Friede

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answered Jun 5 at 16:50

Daniel Estévez

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1 Comment

Friede Jun 5 at 17:34

"better"? In R, we do not need a for-for-if logic to do what your fx does. It's inefficient and needless. As a side note, upper.tri() and lower.tri() have diag-arg.

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