Partitioning a correlation matrix in R

Question

Please I need help partitioning a symmetric or any matrix in R, say my variables are grouped say the default data in R called longley,

data()
L.df<-longley
cor(L.df)

I need a code which will help me partition this correlation matrix as

So that I can be able to take the average of it partition allowing me to assume equal correlation as the interpolation of each group!

So I can have something like

As my structured correlation matrix.

PS. I obtained the structured manually

Would love to be able to partition it at any Column or row of choice.

Note: The partition assumes variables 1&2 are in group 1, Variables 3,4 & 5 are in group 2 And variables 6&7 are in group 3

Answer 1

You can get the correlation matrix, set the diagonal to NA, iterate through the groups and rewrite their entries with their NA-omitted means, then write 1 into the diagonal:

partition <- function(m, groups = list(1:2, 3:5, 6:7)) {
  crm <- cor(m)
  diag(crm) <- NA
  for(i in groups) {
    for(j in groups) {
      crm[i, j] <- mean(crm[i, j], na.rm = TRUE)
    }
  }
  diag(crm) <- 1
  crm <- round(crm, 3)
  dimnames(crm) <- NULL
  crm
  
}

partition(longley)
#>       [,1]  [,2]  [,3]  [,4]  [,5]  [,6]  [,7]
#> [1,] 1.000 0.992 0.684 0.684 0.684 0.985 0.985
#> [2,] 0.992 1.000 0.684 0.684 0.684 0.985 0.985
#> [3,] 0.684 0.684 1.000 0.291 0.291 0.667 0.667
#> [4,] 0.684 0.684 0.291 1.000 0.291 0.667 0.667
#> [5,] 0.684 0.684 0.291 0.291 1.000 0.667 0.667
#> [6,] 0.985 0.985 0.667 0.667 0.667 1.000 0.971
#> [7,] 0.985 0.985 0.667 0.667 0.667 0.971 1.000

To change the groups, you need to supply them as a list of column indices. For example, if you wanted two groups with columns 1:3 and 4:7, you could do:

partition(longley, list(1:3, 4:7))
#>       [,1]  [,2]  [,3]  [,4]  [,5]  [,6]  [,7]
#> [1,] 1.000 0.739 0.739 0.709 0.709 0.709 0.709
#> [2,] 0.739 1.000 0.739 0.709 0.709 0.709 0.709
#> [3,] 0.739 0.739 1.000 0.709 0.709 0.709 0.709
#> [4,] 0.709 0.709 0.709 1.000 0.694 0.694 0.694
#> [5,] 0.709 0.709 0.709 0.694 1.000 0.694 0.694
#> [6,] 0.709 0.709 0.709 0.694 0.694 1.000 0.694
#> [7,] 0.709 0.709 0.709 0.694 0.694 0.694 1.000

^{Created on 2022-04-11 by the reprex package (v2.0.1)}

Answer 2

Define a grouping vector g and from that create a grouping matrix gm and use that with ave.

g <- c(1, 1, 2, 2, 2, 3, 3)

gm <- outer(g, g, paste)
diag(gm) <- ""
ave(cor(longley), gm)