I have a binary matrix with 0 and 1s and I want each non-zero element substituted by a uniform distribution value on [−0.6, −0.3] ∪ [0.3, 0.6] in R.
Can anyone please help how can I achieve this?
CodePudding user response:
Yes sample(c(1,-1),100,replace=T)*runif(100,0.3,0.6)
.
CodePudding user response:
If your binary matrix looks like this:
set.seed(1)
m <- matrix(rbinom(16, 1, 0.5), 4)
m
#> [,1] [,2] [,3] [,4]
#> [1,] 0 0 1 1
#> [2,] 0 1 0 0
#> [3,] 1 1 0 1
#> [4,] 1 1 0 0
Then you can replace its elements like so:
m[m != 0] <- runif(sum(m), 0.3, 0.6) * sign(rnorm(sum(m)))
m
#> [,1] [,2] [,3] [,4]
#> [1,] 0.0000000 0.0000000 0.3636428 0.4955021
#> [2,] 0.0000000 0.4140106 0.0000000 0.0000000
#> [3,] -0.5152856 -0.5332336 0.0000000 0.3376665
#> [4,] -0.5975718 -0.5804116 0.0000000 0.0000000