Question

Consider the following matrix:

sequence <- structure(list(C1 = c(2L, 9L, 3L, 9L, 1L, 8L, 9L, 6L, 4L, 5L, 
3L, 2L), C2 = c(3L, 6L, 5L, 8L, 8L, 7L, 3L, 7L, 2L, 1L, 4L, 1L
), C3 = c(8L, 2L, 6L, 4L, 6L, 5L, 7L, 4L, 5L, 9L, 1L, 7L)), .Names = c("C1", 
"C2", "C3"), class = "data.frame", row.names = c(NA, -12L))

Each row has a combination of 3 numbers. I’m trying to recombine all the triads into pairs, with each triad row being divided into three rows (each containing a possible pair). For example, row 1 (2, 3, 8) should be transformed into row 1 (2, 3), row 2 (3, 8) and row 3 (2, 8). The result should look like this:

result <- structure(list(Col1 = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 
3L, 3L), .Label = c("Row 1", "Row 2", "Row 3"), class = "factor"), 
    Col2 = c(2L, 3L, 2L, 9L, 6L, 9L, 3L, 5L, 3L), Col3 = c(3L, 
    8L, 8L, 6L, 2L, 2L, 5L, 6L, 6L)), .Names = c("Col1", "Col2", 
"Col3"), class = "data.frame", row.names = c(NA, -9L))

(the table repeats until all rows are recombined)

I’ve tried to do this with combn function: t(combn(unlist(t(sequence)),2)) but this is recombining all elements of the matrix amongst themselves, rather than recombining only the elements of each row. Any light?

Answer 1

I’m sure there’s a much cleaner way but you could just use cbind to get the pairs of interest three times, then use rbind to put them together.

sequence <- structure(list(C1 = c(2L, 9L, 3L, 9L, 1L, 8L, 9L, 6L, 4L, 5L, 
3L, 2L), C2 = c(3L, 6L, 5L, 8L, 8L, 7L, 3L, 7L, 2L, 1L, 4L, 1L
), C3 = c(8L, 2L, 6L, 4L, 6L, 5L, 7L, 4L, 5L, 9L, 1L, 7L)), .Names = c("C1", 
"C2", "C3"), class = "data.frame", row.names = c(NA, -12L))

# Essentially what you wanted
temp.result <- with(sequence, rbind(cbind(C1, C2), cbind(C2, C3), cbind(C1, C3)))
# Identify which rows we're talking about
id <- rep(seq(nrow(sequence)), 3)
# Put it all together
result <- cbind(id, temp.result)
# Order it the way you imply in your question
result <- result[order(result[,1]),]
# Give it the colnames you want
colnames(result) <- c("Col1", "Col2", "Col3")
head(result)
#     Col1 Col2 Col3
#[1,]    1    2    3
#[2,]    1    3    8
#[3,]    1    2    8
#[4,]    2    9    6
#[5,]    2    6    2
#[6,]    2    9    2