CODEWITHSUNDEEP
rsimulationmatrix
Sarah
Sarah

Reputation: 23

Putting generated data in a matrix format

I have one question about putting a simulated data in a matrix format, but I cannot suitably write its program in R, and constantly receive an error, I guess my "rep" definition and final "Matrix" expression are somehow wrong, but I do not know how to fix them. Here my specific question is:

I would like to produce a matrix contains generated values. I have 20000 generated values for x and y. As the output, I like to have a (2000 by 10) matrix that each column of the matrix contains the output of following for loop.

My R.code:

x=rnorm(2e4,5,6)
vofdiv=quantile(x,probs=seq(0,1,0.1))
y=rnorm(2e4,4,6)
Matrix=rep(NULL,2000)
for(i in 1:10)
{
Matrix[i]=y[(x>=vofdiv[i] & x<vofdiv[i+1])]  #The i(th) col of matrix
}
Matrix   # A 2000*10 Matrix, as the final output

I highly appreciate that someone helps me!

Upvotes: 1

Views: 70

Answers (1)

David Arenburg
David Arenburg

Reputation: 92292

You have several problems here.

First of all, the correct way to define an empty matrix of size 2e4*10, would be

Matrix <- matrix(NA, 2e4, 10)

Although you could potentially create a matrix using your way(rep) and then use dim, something like

Matrix <- rep(NA, 2e5)
dim(Matrix) <- c(2e4, 10)

Second problem is, when trying to insert into a column in a matrix, you need to index it correctly, i.e., 

Matrix[, i] <-

instead of

Matrix[i] <-

The latter will index Matrix as if it was a vector (which is it basically is). In other words, it will convert a 2000*10 matrix to a 20000 length single vector and index it.

The third problem is, that when your loop reaches i = 11 and you are running x<vofdiv[i+1] you are always excluding the last values which are x == vofdiv[11], thus you are always getting less than 2000 values:

for(i in 1:10)
{
  print(length(y[ (x >= vofdiv[i] & x < vofdiv[i+1])]))
}
# [1] 2000
# [1] 2000
# [1] 2000
# [1] 2000
# [1] 2000
# [1] 2000
# [1] 2000
# [1] 2000
# [1] 2000
# [1] 1999 <----

Thus, it will give you an error if you will try to replace 2000 length vector with 1999 length one, because a matrix in R can't contain different dimensions for each column.

The workaround would be to add = to your last statement, such as

Matrix <- matrix(NA, 2e4, 10)
for(i in 1:10)
  {
    Matrix[, i] <- y[x >= vofdiv[i] & x <= vofdiv[i + 1]] 
  }

Upvotes: 3

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