Using weighted.mean() in aggregate()
On Sat, 21 Jun 2003 18:02:01 -0700
Spencer Graves <spencer.graves at pdf.com> wrote:
> tstdf <- data.frame(Sub =rep(1:2, 2),
+ Length=1:4, Slope=11:14)
> by(tstdf, tstdf$Sub,
+ function(x)weighted.mean(x$Slope, x$Length)) tstdf$Sub: 1 [1] 12.5 ------------------------------------------------------------ tstdf$Sub: 2 [1] 13.33333
>
Does this answer your question? hth. spencer graves
Here are two other ways, using the Hmisc package summarize or mApply functions, which can take a matrix as their first argument. summarize returns a dataframe, mApply a vector g <- function(y) wtd.mean(y[,1],y[,2]) summarize(cbind(y, wts), llist(sex,race), g, stat.name='y') mApply(cbind(y,wts), llist(sex,race), g) Frank Harrell
Aleksey Naumov wrote:
Dear R users, I have a question on using weighted.mean() while aggregating a data frame. I have a data frame with columns Sub, Length and Slope:
x[1:5,]
Sub Length Slope 1 2 351.547 0.0025284969 2 2 343.738 0.0025859390 3 1 696.659 0.0015948968 4 2 5442.338 0.0026132544 5 1 209.483 0.0005304225 and I would like to calculate the weighted.mean of Slope, using Length as weights, for each value of Sub. The obvious way:
aggregate(list(Mean.Slope=x$Slope), by=list(Sub=x$Sub), FUN=weighted.mean,
w=x$Length) does not work. weighted.mean() generates warnings that "longer object length is not a multiple of shorter object length in: x * w", from which I conclude that weights are not supplied as I intend, instead each subset of Sub, when passed to weighted.mean(), receives the whole x$Length as weights, which is not correct. Is there an elegant way to do this, or do I have to have a loop here? Thank you, Aleksey
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--- Frank E Harrell Jr Prof. of Biostatistics & Statistics Div. of Biostatistics & Epidem. Dept. of Health Evaluation Sciences U. Virginia School of Medicine http://hesweb1.med.virginia.edu/biostat