Finding the mean of all Duplicates - r

Question

There is a nice explanation here describing how to eliminate duplicates in a data frame by picking the maximum variable.
I can also see how this can be applied to pick the duplicate with the minimum variable.
my question now is how do I display the mean of all duplicates?
for example:
z <- data.frame(id=c(1,1,2,2,3,4),var=c(2,4,1,3,5,2))
# id var
# 1 2
# 1 4
# 2 1
# 2 3
# 3 5
# 4 2
I would like the output:
# id var
# 1 3 mean(2,4)
# 2 2 mean(1,3)
# 3 5
# 4 2
My current code is:
averages<-do.call(rbind,lapply(split(z,z$id),function(chunk) mean(chunk$var)))
z<-z[order(z$id),]
z<-z[!duplicated(z$id),]
z$var<-averages
My code runs very slowly and is takes about 10 times longer than the method for picking the maximum. How do I optimize this code?

Answer 1

Here is a faster solution using data.table
library(data.table)
z <- data.frame(id=sample(letters, 6e5, replace = TRUE),var = rnorm(6e5))
fn1 <- function(z){
z$var <- ave(z$var, z$id, FUN=mean)
return(unique(z))
}
fn2 <- function(z) {
t(sapply(split(z,z$id), function(x) sapply(x,mean)))
}
fn3 <- function(z){
data.table(z)[,list(var = mean(var)), 'id']
}
library(rbenchmark)
benchmark(f1 <- fn1(z), f2 <- fn2(z), f3 <- fn3(z), replications = 2)
est replications elapsed relative user.self sys.self
1 f1 <- fn1(z) 2 3.619 8.455607 3.331 0.242
2 f2 <- fn2(z) 2 0.586 1.369159 0.365 0.220
3 f3 <- fn3(z) 2 0.428 1.000000 0.341 0.086

Answer 2

I think split() and unsplit() is one way.
dupMean <- function(x)
{
result <- split(x[, 2], x[, 1])
result <- lapply(result, mean)
result <- unsplit(result, unique(x[, 1]))
return(result)
}
Or, to save a line with plyr:
require(plyr)
dupMean <- function(x)
{
result <- split(x[, 2], x[, 1])
result <- laply(result, mean)
return(result)
}
Update:
Just for curiosity, here is a comparison of the various functions suggested. Ramnath (fn3) looks to be the winner on my computer.
require(plyr)
require(data.table)
require(rbenchmark)
fn1 <- function(z){
z$var <- ave(z$var, z$id, FUN=mean)
return(unique(z))
}
fn2 <- function(z) {
t(sapply(split(z,z$id), function(x) sapply(x,mean)))
}
fn3 <- function(z){
data.table(z)[,list(var = mean(var)), 'id']
}
fn4 <- function(x)
{
result <- t(sapply(split(x,x$id), function(y) sapply(y,mean)))
return(result)
}
fn5 <- function(x)
{
x$var <- ave(x$var, x$id, FUN=mean)
x <- unique(x)
return(x)
}
fn6 <- function(x)
{
result <- do.call(rbind,lapply(split(x,x$id),function(chunk) mean(chunk$var)))
return(data.frame(id = unique(x[, 1]), var = result))
}
fn7 <- function(x)
{
result <- split(x[, 2], x[, 1])
result <- lapply(result, mean)
result <- unsplit(result, unique(x[, 1]))
return(data.frame(id = unique(x[, 1]), var = result))
}
fn8 <- function(x)
{
result <- split(x[, 2], x[, 1])
result <- laply(result, mean)
return(data.frame(id = unique(x[, 1]), var = result))
}
z <- data.frame(id = rep(c(1,1,2,2,3,4,5,6,6,7), 1e5), var = rnorm(1e6))
benchmark(f1 <- fn1(z), f2 <- fn2(z), f3 <- fn3(z), f4 <- fn4(z), f5 <- fn5(z), f6 <- fn6(z), f7 <- fn7(z), f8 <- fn8(z), replications = 2)
Result:
test replications elapsed relative user.self sys.self
1 f1 <- fn1(z) 2 13.45 20.692308 13.27 0.15
2 f2 <- fn2(z) 2 3.54 5.446154 3.43 0.09
3 f3 <- fn3(z) 2 0.65 1.000000 0.54 0.10
4 f4 <- fn4(z) 2 3.62 5.569231 3.50 0.09
5 f5 <- fn5(z) 2 13.57 20.876923 13.25 0.25
6 f6 <- fn6(z) 2 3.53 5.430769 3.36 0.14
7 f7 <- fn7(z) 2 3.34 5.138462 3.28 0.03
8 f8 <- fn8(z) 2 3.34 5.138462 3.26 0.03

Answer 3

I would use a combination of ave and unique:
z <- data.frame(id=rep(c(1,1,2,2,3,4),1e5),var=rnorm(6e5))
z$var <- ave(z$var, z$id, FUN=mean)
z <- unique(z)
UPDATE: after actually timing the solution, here's something that's a little faster.
z <- data.frame(id=rep(c(1,1,2,2,3,4),1e5),var=rnorm(6e5))
system.time({
averages <- t(sapply(split(z,z$id), function(x) sapply(x,mean)))
})
# user system elapsed
# 1.32 0.00 1.33
system.time({
z$var <- ave(z$var, z$id, FUN=mean)
z <- unique(z)
})
# user system elapsed
# 4.33 0.02 4.37