You can do it directly from the X matrix like so:
> by(X, as.list(as.data.frame(X[,3:5])), function(R)weighted.mean(R
[1], R[2]))
A: 0
B: 0
C: 0
[1] 0.4912458
--------------------------------------------------------------------
A: 1
B: 0
C: 0
[1] NA
--------------------------------------------------------------------
A: 0
B: 1
C: 0
[1] -0.2694550
--------------------------------------------------------------------
A: 1
B: 1
C: 0
[1] NA
--------------------------------------------------------------------
A: 0
B: 0
C: 1
[1] NA
--------------------------------------------------------------------
A: 1
B: 0
C: 1
[1] 0.01102718
--------------------------------------------------------------------
A: 0
B: 1
C: 1
[1] NA
--------------------------------------------------------------------
A: 1
B: 1
C: 1
[1] -0.5622041
Or you can poke about a bit with it first:
> Z = as.data.frame(X)
> names(Z) = c("x", "w", "A", "B",
"C")
> by(Z[,c("x", "w")], list(paste("A", Z$A,
"B", Z$B, "C", Z$C)),
function(R)weighted.mean(R$x, R$w))
: A 0 B 0 C 0
[1] 0.4912458
--------------------------------------------------------------------
: A 0 B 1 C 0
[1] -0.2694550
--------------------------------------------------------------------
: A 1 B 0 C 1
[1] 0.01102718
--------------------------------------------------------------------
: A 1 B 1 C 1
[1] -0.5622041
-Alex
On 10 Oct 2006, at 16:13, Young Cho wrote:
> HI,
>
> I am trying to figure out an efficient way to calculate group means
> and
> associate each entry with it. I made up an example:
>
> A = rep(rep(0:1,each=2),3)
> B = rep(rep(0:1,4),3)
> C = rep(rep(c(0,0,1,1),2),3)
> X =cbind(rnorm(24,0,1),runif(24,0,1),A,B,C)
> A B C
> [1,] -1.92926469 0.32213127 0 0 0
> [2,] -0.83935617 0.77794096 0 1 0
> [3,] -1.27799751 0.26276934 1 0 1
>
> Suppose I want to compute a weighted mean of X[,1] by for each
> group, which
> is defined by unique vector (A,B,C) with weights are X[,2]. And
> then add a
> column for the weighted group mean. How can I do this ? My matrix
> is fairly
> large (few thousands) but, luckily, I have only a few factors (<10).
>
> Thanks a lot,
>
> Young.
>
> [[alternative HTML version deleted]]
>
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