R: matrix to indexes

Question

I have a matrix like

      [,1] [,2]
 [1,]    1    3
 [2,]    4    6
 [3,]   11   12
 [4,]   13   14

I want to convert this matrix to a vector

Answer 1

Throwing this one here, it uses base R and should be somewhat fast since the inevitable loop is handled by rep:

zero.lengths <- m[,1] - c(0, head(m[,2], -1)) - 1
one.lengths  <- m[,2] - m[,1] + 1

rep(rep(c(0, 1), nrow(m)),
    as.vector(rbind(zero.lengths, one.lengths)))

Or another solution using sequence:

out <- integer(m[length(m)])    # or `integer(20)` following OP's edit.
one.starts  <- m[,1]
one.lengths <- m[,2] - m[,1] + 1
one.idx <- sequence(one.lengths) + rep(one.starts, one.lengths) - 1L
out[one.idx] <- 1L

Answer 2

@ Arun's answer seems better.

Now that I understand the problem (or do I?). Here is a solution in base R that makes use of the idea that only contiguous sequences of zeroes need to be kept.

find.ones <- function (mat) {
  ones <- rep(0, max(mat))
  ones[c(mat)] <- 1
  ones <- paste0(ones, collapse="")
  ones <- gsub("101", "111", ones)
  ones <- as.numeric(strsplit(ones, "")[[1]])
  ones
}

On the OP's original example:

m <- matrix(c(1, 3, 4, 6, 11, 12, 13, 14), ncol=2, byrow=TRUE)
find.ones(m)
[1] 1 1 1 1 1 1 0 0 0 0 1 1 1 1

To benchmark the solution, let's make a matrix big enough:

set.seed(10)
m <- sample.int(n=1e6, size=5e5)                                              
m <- matrix(sort(m), ncol=2, byrow=TRUE)                                           

head(m)                                                           
     [,1] [,2]
[1,]    1    3
[2,]    4    5
[3,]    9   10
[4,]   11   13
[5,]   14   18
[6,]   22   23

system.time(ones <- find.ones(m))

 user  system elapsed 
1.167   0.000   1.167 

Answer 3

Here's an answer using IRanges package:

require(IRanges)
xx.ir <- IRanges(start = xx[,1], end = xx[,2])
as.vector(coverage(xx.ir))
# [1] 1 1 1 1 1 1 0 0 0 0 1 1 1 1

If you specify a min and max value of your entire vector length, then:

max.val <- 20
min.val <- 1
c(rep(0, min.val-1), as.vector(coverage(xx.ir)), rep(0, max.val-max(xx)))