Create a similarity matrix of integers, using R

Question

I have a matrix with diagonals equal to zero and off-diagonals all equal to one (the inverse of an identity matrix):

mat1 <- matrix(c(0,1,1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,0), 5, 5)

I also have a vector that is always the same length as the dims of the matrix and always starts at zero:

vec1 <- c(0,1,2,3,4)

using these two objects I want to create a matrix that looks like this:

mat2 <- matrix(c(0,1,2,3,4,1,0,1,2,3,2,1,0,1,2,3,2,1,0,1,4,3,2,1,0), 5, 5)

     [,1] [,2] [,3] [,4] [,5]
[1,]    0    1    2    3    4
[2,]    1    0    1    2    3
[3,]    2    1    0    1    2
[4,]    3    2    1    0    1
[5,]    4    3    2    1    0

I want an operation that will generalize so that if I have a matrix of dims 9 by 9, for example, and a vector of 0:8 I can achieve the equivalent result. Any ideas for how to approach this?

Answer 1

As vec1 starts with a zero, then you can do :

MakeMatrix <- function(x){
  n <- length(x)
  id <- abs(rep(1:n,n)-rep(1:n,each=n)) + 1
  matrix(x[id],ncol=n)
}

MakeMatrix(vec1)

So there's no need to take the mat1 in the input, as that one is actually redundant. You can just construct the matrix within the function.

The trick is in providing a sequence of id values to select from the vector, and then transform everything to a matrix.

---

Edit : If you're only going to use sequences, you could as well do :

MakeMatrix <- function(n){
  id <- abs(rep(1:n,n)-rep(1:n,each=n))
  matrix(id,ncol=n)
}

MakeMatrix(7)

Answer 2

The following solution makes use of upper.tri and lower.tri to isolate the upper and lower triangular matrix. In addition, it makes use of sequence to create the desired vector sequence.

n <- 9
vec <- (1:n)-1
m <- matrix(0, n, n)
m[lower.tri(m, diag=TRUE)] <- vec[sequence(n:1)]  #### Edit
m <- t(m)
m[lower.tri(m, diag=TRUE)] <- vec[sequence(n:1)]  #### Edit
m

      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
 [1,]    0    1    2    3    4    5    6    7    8
 [2,]    1    0    1    2    3    4    5    6    7
 [3,]    2    1    0    1    2    3    4    5    6
 [4,]    3    2    1    0    1    2    3    4    5
 [5,]    4    3    2    1    0    1    2    3    4
 [6,]    5    4    3    2    1    0    1    2    3
 [7,]    6    5    4    3    2    1    0    1    2
 [8,]    7    6    5    4    3    2    1    0    1
 [9,]    8    7    6    5    4    3    2    1    0

Answer 3

How about:

genMat <- function(n){
  mat <- outer(1:n,1:n,"-")%%n
  tmp <- mat[lower.tri(mat)]
  mat <- t(mat)
  mat[lower.tri(mat)] <- tmp
  mat
}

> genMat(5)
     [,1] [,2] [,3] [,4] [,5]
[1,]    0    1    2    3    4
[2,]    1    0    1    2    3
[3,]    2    1    0    1    2
[4,]    3    2    1    0    1
[5,]    4    3    2    1    0

Edit

For arbitrary vec1:

genMat2 <- function(vec){
  n <- length(vec)
  mat <- outer(1:n,1:n,"-")%%n
  tmp <- mat[lower.tri(mat)]
  mat <- t(mat)
  mat[lower.tri(mat)] <- tmp
  matrix(vec[mat+1],n,n)
}

> genMat2(c(0,2,4,3,9))
     [,1] [,2] [,3] [,4] [,5]
[1,]    0    2    4    3    9
[2,]    2    0    2    4    3
[3,]    4    2    0    2    4
[4,]    3    4    2    0    2
[5,]    9    3    4    2    0

Edit 2 In fact, there's no need to use the modulus and then play with the matrix, abs will work fine to make the original matrix definition a 1-liner:

abs(outer(1:n,1:n,"-"))

So,

genMat <- function(n){
  abs(outer(1:n,1:n,"-"))
}

and

genMat2 <- function(vec){
  n <- length(vec)
  matrix(vec[abs(outer(1:n,1:n,"-"))+1],n,n)
}