Rearrange a matrix in R using two factors

Question

Here is the problem. There is a matrix with N rows and C columns, and two factors: ids and group, both of length N. For example:

m <- matrix( 1:25, nrow= 5, byrow= T )
id <- factor( c( "A", "A", "A", "B", "B" ) )
group <- factor( c( "a", "b", "c", "a", "c" ) )

Not all combinations of factors are present, but each combination of the factors is present only once. The task is to transform the matrix m in such a way that it has length( levels( id ) ) rows and length( levels( group ) ) * C columns. In other words, create a matrix where each variable corresponds to a combination between the original column and all possible levels of factor group. Missing values (for non-existent combinations of id and group) are replaced by NA's. Here is the desired output of the above example:

  a.1 a.2 a.3 a.4 a.5 b.1 b.2 b.3 b.4 b.5 c.1 c.2 c.3 c.4 c.5
A   1   2   3   4   5   6   7   8   9  10  11  12  13  14  15
B  16  17  18  19  20  NA  NA  NA  NA  NA  21  22  23  24  25

I wrote my own function, but it is terribly ineffective and I'm sure it duplicates the functionality of something extremely simple.

matrixReshuffle <- function( m, ids.row, factor.group ) {

  nr <- nrow( m )
  nc <- ncol( m )
  if( is.null( colnames( m ) ) ) colnames( m ) <- 1:nc

  ret <- NULL
  for( id in levels( ids.row ) ) {

    r <- c()

    for( fg in levels( factor.group ) ) {

      d <- m[ ids.row == id & factor.group == fg,, drop= F ]
      if( nrow( d ) > 1 )
        stop( sprintf( "Too many matches for ids.row= %s and factor.group= %s", id, fg ) )
      else if( nrow( d ) < 1 ) {
        r <- c( r, rep( NA, nc ) )
      } else {
        r <- c( r, d[1,] )
      }

    }

    ret <- rbind( ret, r )

  }

  colnames( ret ) <- paste( rep( levels( factor.group ), each= nc ), rep( colnames( m ), length( levels( factor.group ) ) ), sep= "." )
  rownames( ret ) <- levels( ids.row )

  return( ret )
}

Answer 1

Following @Aaron's suggestions:

Using melt and acast from reshape2:

require(reshape2)
df <- as.data.frame(m)
names(df) <- seq_len(ncol(df))
df.m <- melt(df)
df.m$id <- rep(id, nrow(df.m)/length(id))
df.m$group <- rep(group, nrow(df.m)/length(group))

o <- acast(df.m, id ~ group+variable, value.var="value")
colnames(o) <- sub("_", ".", colnames(o))

#   a.1 a.2 a.3 a.4 a.5 b.1 b.2 b.3 b.4 b.5 c.1 c.2 c.3 c.4 c.5
# A   1   2   3   4   5   6   7   8   9  10  11  12  13  14  15
# B  16  17  18  19  20  NA  NA  NA  NA  NA  21  22  23  24  25

You can convert this back to matrix.

Answer 2

For all the matrix indexing fans out there...

C <- ncol(m)
to.row <- matrix(rep(as.numeric(id), C), ncol=C)
to.col <- sweep(col(m),1,(as.numeric(group)-1)*C,`+`)
out <- array(dim=c(nlevels(id), nlevels(group)*C),
             dimnames=list(levels(id), as.vector(t(outer(levels(group), 1:C, paste, sep=".")))))
out[cbind(as.vector(to.row), as.vector(to.col))] <- m
out
#   a.1 a.2 a.3 a.4 a.5 b.1 b.2 b.3 b.4 b.5 c.1 c.2 c.3 c.4 c.5
# A   1   2   3   4   5   6   7   8   9  10  11  12  13  14  15
# B  16  17  18  19  20  NA  NA  NA  NA  NA  21  22  23  24  25