Adding labels to data with ddply while subsetting.

Question

Let's say I have a data.frame like:

x <- c(1:10,1:10,1:10,1:10,1:10,1:10,1:10,1:10,1:10,1:10)
df <- data.frame(x=x,y=rnorm(100))

and I want to label values that are sorted (descending) in the 80th percentile for each value of x (1:10). I can get the quantiles and order the data, without issue like this:

df <- ddply(df, .(x), subset, y > quantile(y,0.8))
df <- df[with(df, order(x,-y)),]

Now, how could I get ddply to add a column of labels (1,2,3,...n) in a new column of the data.frame for each sorted subset? I can do this now with a for loop by counting nrow(df["x"]), but that seems to lack any sense of eloquence.

Note: This question is a build up from and related to: http://stackoverflow.com/questions/3370621/creating-multiple-subsets-all-in-one-data-frame-possibly-with-ddply

Answer 1

Maybe this function produces what you want:

subno <- function(df, vars, offset=1) {
    id <- do.call("paste", df[,vars, drop=FALSE])
    nr <- seq(along.with=id)
    grpnr <- nr
    grpnr[c(FALSE, id[-1] == id[-length(id)])] <- 0
    subnr <- nr - cummax(grpnr) + offset
    return(subnr)
}

df$label <- subno(df, c('x'))

This function expects a sorted dataframe and vars contains the variable names on which to group.

Answer 2

df <- ddply(df, "x", transform, id = rank(y))

Or, if already sorted:

df <- ddply(df, "x", transform, id = seq_along(y))