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.
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.