R: indexes to matrix

Question

I have a matrix call res initially like the following:

      [,1] [,2]
[1,]     0    0
[2,]     0    0
[3,]     0    0
[4,]     0    0
[5,]     0    0
[6,]     0    0
[7,]     0    0
[8,]     0    0
[9,]     0    0
[10,]    0    0 

I have a matrix of indexes (indexes) like the following:

     [,1] [,2]
 [1,]   2    3
 [2,]   7    9

I want the resulting res matrix to be like the following:

      [,1] [,2]
[1,]     0    0
[2,]     1    1
[3,]     1    1
[4,]     0    0
[5,]     0    0
[6,]     0    0
[7,]     1    1
[8,]     1    1
[9,]     1    1
[10,]    0    0 

I have a big matrix, it takes a long time to loop through the indexes matrix. Please let me know if there is a better approach to do this. I am hoping to do something like mat[indexes,] <- 1. However, this does not work I wanted.

Answer 1

If res if your main matrix and indexes is the matrix of indices:

This could help:

idx  <- do.call("c",apply(indexes,1,function(x){seq(x[1],x[2])}))

res[idx,] <- 1

As to timing, first create a large indices matrix:

> set.seed(42)
> indexes <- t(matrix(sort(sample(1:10000,1000)),2,500))
> head(indexes)
     [,1] [,2]
[1,]    3    4
[2,]   14   16
[3,]   23   33
[4,]   40   63
[5,]   67   74
[6,]   79   83

and time them:

> system.time(idx  <- do.call("c",apply(indexes,1,function(x){seq(x[1],x[2])})))   user  system elapsed 
  0.008   0.000   0.007 

> system.time( idx2 <- unlist( apply( indexes , 1 , FUN = function(x){ seq.int(x[1],x[2])}) ))
   user  system elapsed 
  0.004   0.000   0.002

It would appear that the second method is slightly faster.

Answer 2

Use an answer to your previous question to create a vector of row indexes ridx, and then

res[as.logical(ridx),] = 1L

Answer 3

EDIT: I did misunderstand, this should help:

test <- matrix(rep(0,1E7), ncol=2)
Index <- matrix(sort(sample(1:(1E7*0.5), size=10000)), ncol=2, byrow=TRUE)
test[unlist(apply(Index, 1, function(x){x[1]:x[2]})),] <- 1