More efficient way to get a data matrix based on some value and index matrices in R?

Question

Assume I have a vector of values like:

M=3;val<-rnorm(M)

and the corresponding index matrix like:

N=20;J=10;ind<-matrix(sample(1:M,N*J,replace=T),nrow=J)

I can easily assign the values with their indices to get the data matrix as:

x<-matrix(val[ind],J,N)

Now I have a matrix of values like:

val<-matrix(rnorm(M*J),nrow=J)

and need to assign the values and indices row by row (i.e., one row in val with one row in ind) to get the data matrix.

I can do this with a for loop as:

x<-ind;
for(j in 1:J){x[j,]<-val[j,ind[j,]]}

But I wonder if there's more efficient way to do this, especially to avoid using the for loop?

I need to do resampling and repeat the assigning process hundreds of thousands of times. So I worry that the for loop will take up a lot of time.

I am slightly uncertain as to 'why' this is. Are you perhaps simply looking for matrix(..., byrow = FALSE), which would fill in the matrix by column. This should let you fill in the matrix by column in the matrix call directly. — Oliver
Sorry there are typos. But the issue is not on x, but to apply val and ind row by row, as shown in the for loop. — Luwei

Martin Morgan Martin Morgan · Accepted Answer · 2019-10-08T10:42:21

In general, a matrix can be subset or subset-assigned using a two-column matrix as row and column indices. So

i_idx = rep(1:J, each = ncol(ind))
x_idx = cbind(i_idx, 1:ncol(ind))
val_idx = cbind(i_idx, as.vector(t(ind[1:J,])))

x[x_idx] = val[val_idx]