On Mon, 10 May 2004, Christophe Pallier wrote:
The use of tapply(x,f,mean)[match(f,unique(f))] assumes a particular
order in the result of tapply, no? It seems a bit dangerous to me.
My original code for the group means problem used rowsum(,reorder=FALSE)
rather than tapply(), and we do know
Liaw, Andy wrote:
Suppose I
define the function:
fun - function(x, f) {
m - tapply(x, f, mean)
ans - x - m[match(f, unique(f))]
names(ans) - names(x)
ans
}
May I ask what is the purpose of match(f,unique(f)) ?
To remove the group means, I have be using:
x-tapply(x,f,mean)[f]
Both of you might have missed my question from Friday: For very long `x'
(e.g., length=5), indexing by names can take a long time. See that
thread for detail. (For small data, you can hardly tell the difference.)
Also, I'm trying to write the function in a way that one can pass in more
On Mon, 10 May 2004, Liaw, Andy wrote:
Both of you might have missed my question from Friday: For very long `x'
(e.g., length=5), indexing by names can take a long time. See that
thread for detail. (For small data, you can hardly tell the difference.)
That's solved in R-devel as of
On Sat, 8 May 2004, Gabor Grothendieck wrote:
predict(lm(AV~as.factor(GROUP)))
If Felix actually has a huge data frame this will be slow. Instead
try
groupmeans-rowsum(AV,GROUP,reorder=FALSE)
individual.means- groupmeans[match(GROUP, unique(GROUP)]
It uses hashing and takes roughly
predict(lm(AV~as.factor(GROUP)))
Felix Eschenburg Atropin75 at t-online.de writes:
:
: Hello list !
:
: I have a huge data.frame with several variables observed on about 3000
: persons. For every person (row) there is variable called GROUP which indices
: the group the person belongs to.
