I believe that computing covariance matrices takes
much more time than computing their average and so it
does not matter how you do this, but one possibility
is:
z<- lcov[[1]]*0
y <- sapply(lcov,function(x) {z<<-z+x;0;})
y <- y/length(lcov)
--- Rick DeShon <[EMAIL PROTECTED]> wrote:
> Hi All.
>
> I would like to compute a separate covariance matrix
> for a set of
> variables for each of the levels of a factor and
> then compute the
> average covariance matrix over the factor levels. I
> can loop through
> this computation but I need to perform the
> calculation for a large
> number of levels and am looking for something more
> elegant. To be
> concrete....
>
> u <- 3
> n <- 10
>
> x <- rnorm((id*u))
> y <- rnorm((id*u))
> z <- rnorm((id*u))
> id <- gl(u,n)
>
> df <- data.frame(id,x,y,z)
> df.s <- split(xxx,id)
>
> lcov <- lapply(df.s,cov)
> lcov
>
> What's an efficient way to compute the average
> covariance matrix over
> the list members in "lcov"?
>
> Thanks in advance,
>
> Rick DeShon
>
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