which is the fastest way to make data.frame out of a three-dimensional array?
Cheat! Arrays are stored in column major order, so you can translate the indexing directly by: Assume dim(yourarray) = c(n1,n2,n3) *** warning: UNTESTED ** yourframe <- data.frame( dat = as.vector(yourarray) , dim1 = rep(seq_len(n1), n2*n3 ,dim2 = rep( rep(seq_len(n2), e=n1), n3) , dim3 = rep(seq_len(n3), e = n1*n2) ) Probably see also the reshape package for more elegant solutions. Cheers, Bert
On Sat, Feb 25, 2012 at 7:54 AM, Hans Ekbrand <hans at sociologi.cjb.net> wrote:
foo <- rnorm(30*34*12)
dim(foo) <- c(30, 34, 12)
I want to make a data.frame out of this three-dimensional array. Each dimension will be a variabel (column) in the data.frame.
I know how this can be done in a very slow way using for loops, like this:
x <- rep(seq(from = 1, to = 30), 34)
y <- as.vector(sapply(1:34, function(x) {rep(x, 30)}))
month <- as.vector(sapply(1:12, function(x) {rep(x, 30*34)}))
my.df <- data.frame(month, x=rep(x, 12), y=rep(y, 12), temp=rep(NA, 30*34*12))
my.counter <- 1
for(month in 1:12){
?for(i in 1:34){
? ?for(j in 1:30){
? ? ?my.df$temp[my.counter] <- foo[j,i,month]
? ? ?my.counter <- my.counter + 1
? ?}
?}
}
str(my.df)
'data.frame': ? 12240 obs. of ?4 variables:
?$ month: int ?1 1 1 1 1 1 1 1 1 1 ...
?$ x ? ?: int ?1 2 3 4 5 6 7 8 9 10 ...
?$ y ? ?: int ?1 1 1 1 1 1 1 1 1 1 ...
?$ temp : num ?0.673 -1.178 0.54 0.285 -1.153 ...
(In the real world problem I had, data was monthly measurements of temperature and x, y was coordinates).
Does anyone care to share a faster and less ugly solution?
TIA
--
Hans Ekbrand
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