Question

My problem, removing the specific purpose, seems like this:
how to transform a combination like this:
first use combn(letters[1:4], 2) to calculate the combination

     [,1] [,2] [,3] [,4] [,5] [,6]
[1,] "a"  "a"  "a"  "b"  "b"  "c" 
[2,] "b"  "c"  "d"  "c"  "d"  "d" 

use each column to obtain another data frame:

   [,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1    2    3    4   5     6

elements are obtained, for example: the first element, from the first column of the above dataframe

then How can i transform the above dataframe into a matrix, for example result, things like:

   a   b   c  d
a  0   1   2  3
b  1   0   4  5
c  2   4   0  6
d  3   5   6  0

the elements with same col and row names will have zero value where others corresponding to above value

Answer 1

Here is one way that works:

inputs <- letters[1:4]
combs <- combn(inputs, 2)
N <- seq_len(ncol(combs))
nams <- unique(as.vector(combs))
out <- matrix(ncol = length(nams), nrow = length(nams))
out[lower.tri(out)] <- N
out <- t(out)
out[lower.tri(out)] <- N
out <- t(out)
diag(out) <- 0
rownames(out) <- colnames(out) <- inputs

Which gives:

> out
  a b c d
a 0 1 2 3
b 1 0 4 5
c 2 4 0 6
d 3 5 6 0

If I had to do this a lot, I’d wrap those function calls into a function.

Another option is to use as.matrix.dist() to do the conversion for us by setting up a "dist" object by hand. Using some of the objects from earlier:

## Far easier
out2 <- N
class(out2) <- "dist"
attr(out2, "Labels") <- as.character(inputs)
attr(out2, "Size") <- length(inputs)
attr(out2, "Diag") <- attr(out2, "Upper") <- FALSE
out2 <- as.matrix(out2)

Which gives:

> out2
  a b c d
a 0 1 2 3
b 1 0 4 5
c 2 4 0 6
d 3 5 6 0

Again, I’d wrap this in a function if I had to do it more than once.