Hi Henrique,

Thank you for your help, but it doesn't produce what I'm looking for...

On Sun, Mar 2, 2008 at 10:50 AM, Henrique Dallazuanna <[EMAIL PROTECTED]>
wrote:

> One options is:
>
> data.1 <- matrix(seq(from=1,to=9,by=1),nrow=3,ncol=3)
> data.2 <- matrix(seq(from=11,to=19,by=1),nrow=3,ncol=3)
> data.3 <- matrix(seq(from=21,to=29,by=1),nrow=3,ncol=3)
> data.4 <- matrix(seq(from=31,to=39,by=1),nrow=3,ncol=3)
> data.5 <- matrix(seq(from=41,to=49,by=1),nrow=3,ncol=3)
> data.6 <- matrix(seq(from=51,to=59,by=1),nrow=3,ncol=3)
>
> patt <- ls(patt='^data\\.[0-9]')
> nsubs <- 2
> ngroups <- 3
> apply(array(as.vector(sapply(patt, get)), dim = c(dim(get(patt[1])),
> ngroups, nsubs)),          3, mean)


generated [1] 20 30 40

but what I need is an average per cell...so in the above, there are 6 (3 x
3) matrices across 3 groups ( 2 subjects per group)..what I need is 1 matrix
per group, not one mean.  Moreover, I'm storing the data in a single
multi-dimensional array, not as a collection of arrays since I'm creating
each data matrix within a loop (I read a file, create the data matrix, store
it in a larger structure) so that I can perform operations on all the data
outside of the loop...

for(i in 1:6)
{
     str <- paste("test",i,sep=".")
     str <- matrix(seq(from=1,to=9,by=1),nrow=3,ncol=3) # works, creates 6
matrices
}
pattern <- ls(patt='^test\\.[0-9]') # doesn't work variables only exist in
the loop

which is why I had:
subject.data[,,1,1] <- data.1               # subject 1 group 1
subject.data[,,2,1] <- data.2               # subject 1 group 2
subject.data[,,3,1] <- data.3               # subject 1 group 3
subject.data[,,3,2] <- data.4               # subject 2 group 3
subject.data[,,2,2] <- data.5               # subject 2 group 2
subject.data[,,1,2] <- data.6               # subject 2 group 1

to demonstrate all the data was in a meta structure so that I could access
them later...am I making this more complicated than it needs to be? How can
I create the matrices inside a loop and have them available outside of it?

thanks again,

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