How can I identify and summarize sets of data from matching groups in a dataframe?

Question

Here is an example dataframe:

set.seed(0)
x1 <- c(1, 1, 1, 1, 1, 2, 2, 2, 2)
x2 <- c(1, 1, 0, 0, 0, 1, 1, 1, 1)
x3 <- c(1, 1, 2, 2, 4, 1, 1, 2, 1)
n  <- c(1, 1, 1, 5, 5, 1, 1, 1, 1)
y <- rnorm(9)

mydf <- data.frame(x1, x2, x3, n, y)

What I would like to do is

1. identify rows with n=1 and which share identical values of (x1, x2, x3)
2. return a single row for each subset with y = mean(y) and n = length(y)
3. keep other rows the same.

for example, the new dataframe would be

x1 <- c(1,            1,    1,    1,    2,                 2)
x2 <- c(1,            0,    0,    0,    1,                 1)
x3 <- c(1,            2,    2,    4,    1,                 2)
n  <- c(2,            1,    5,    5,    3,                 1)
y  <- c(mean(y[1:2]), y[3], y[4], y[5], mean(y[c(6:7,9)]), y[8])

newdf <- data.frame(x1, x2, x3, n, y)

I can figure this out with conditionals and loops, but I would prefer to learn more elegant way to do this.

Answer 1

By "identical values in other columns", I take it you mean that each subset is defined by the same value of x1 in each of the rows of the subset, not that x1 is equal to x2 . Thanks for the example to see what you meant.

library("plyr")

To get parts one and two

ddply(mydf[mydf$n==1,], .(x1, x2, x3), summarise, n = length(y), y = mean(y))

This can be rbind -ed with the part of mydf where n!=1 to get what you said

rbind(
  ddply(mydf[mydf$n==1,], .(x1, x2, x3), summarise, n = length(y), y = mean(y)),
  mydf[mydf$n!=1,]
)

This doesn't have the same order as you listed. If that is really important, you can add some auxiliary sorting variables.

mydf$order = seq(length=nrow(mydf))
newdf <- rbind(
  ddply(mydf[mydf$n==1,], .(x1, x2, x3), summarise, 
    n = length(y), y = mean(y), order=min(order)),
  mydf[mydf$n!=1,]
)
newdf <- newdf[order(newdf$order),]
newdf$order <- NULL