Reputation: 729
Assuming I have a matrix looks like below, the values up or down the diagonal are the same. In other words, [,1] x [2,] and [,2] x [1,] both are 2 in the matrix.
> m = cbind(c(1,2,3),c(2,4,5),c(3,5,6))
> m
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 2 4 5
[3,] 3 5 6
Then I have real name for 1, 2, and 3 as well.
>Real_name
A B C # A represents 1, B represents 2, and C represents 3.
If I would like to convert the matrix to 3 columns containing corresponding real name for each pair, and the pair must be unique, A x B is the same as B x A, so we keep A x B only. How can I achieve it using R?
A A 1
A B 2
A C 3
B B 4
B C 5
C C 6
Upvotes: 2
Views: 170
Reputation: 214957
colnames(m) <- c("A", "B", "C")
rownames(m) <- c("A", "B", "C")
m[lower.tri(m)] = NA # replace lower triangular elements with NA
data.table::melt(m, na.rm = TRUE) # melt and remove NA
# Var1 Var2 value
#1 A A 1
#4 A B 2
#5 B B 4
#7 A C 3
#8 B C 5
#9 C C 6
Or you can do it in a single line: melt(replace(m, lower.tri(m), NA), na.rm = TRUE)
Upvotes: 3
Reputation: 23101
This will also work:
g <- expand.grid(1:ncol(m), 1:ncol(m))
g <- g[g[,2]>=g[,1],]
cbind.data.frame(sapply(g, function(x) Real_name[x]), Val=m[as.matrix(g)])
Var1 Var2 Val
1 A A 1
2 A B 2
3 B B 4
4 A C 3
5 B C 5
6 C C 6
Upvotes: 0
Reputation: 73295
The following is straightforward:
m <- cbind(c(1,2,3), c(2,4,5), c(3,5,6))
## read `?lower.tri` and try `v <- lower.tri(m, diag = TRUE)` to see what `v` is
## read `?which` and try `which(v, arr.ind = TRUE)` to see what it gives
ij <- which(lower.tri(m, diag = TRUE), arr.ind = TRUE)
Real_name <- LETTERS[1:3]
data.frame(row = Real_name[ij[, 1]], col = Real_name[ij[, 2]], val = c(m[ij]))
# row col val
#1 A A 1
#2 B A 2
#3 C A 3
#4 B B 4
#5 C B 5
#6 C C 6
Upvotes: 3