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43

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2

Hi, I'm trying to transform a 3D array into a matrix. I want the third dimension of the array to form the first row in the matrix, and this third dimension should be read by row (i.e., row 1, then row 2 etc... of dimension 3 should make up the first row of the matrix). I've given an example below, where the array has dimensions of 4, 3, and 5, and the resulting matrix has 5 rows and 12 columns. I have a solution below that achieves what I want, but it seems very cumbersome for large arrays (it first creates vectors from the elements of the array (by row), and then rbinds these to form the matrix). Is there a more elegant way to do this? Thanks in advance for any suggestions.

dat <- array( rnorm(60), dim=c(4, 3, 5) )   

results <- list(1:5)            
for (i in 1:5) {  
    vec <- c( t(dat[, , i]) )  
    results[[i]] <- vec  
    }

datNew <- rbind( results[[1]], results[[2]], results[[3]], results[[4]], results[[5]] )  
A: 

This will give you what you want:

 t(apply(dat, 3, c))
VitoshKa
`all.equal(t(apply(dat, 3, c)), datNew)` says no.
Marek
As Marek noted, the solution above is not correct - it reads the third element of the array by column, rather than row.
Steve
Yap, hurried a little bit too much.
VitoshKa
+1  A: 

Use aperm

X <- aperm(dat,c(3,2,1))
dim(X)<- c(5, 12)
Marek
This works perfectly, thanks.
Steve