Why is plyr so slow?

Viewed 6395

I think I am using plyr incorrectly. Could someone please tell me if this is 'efficient' plyr code?

require(plyr)
plyr <- function(dd) ddply(dd, .(price), summarise, ss=sum(volume)) 

A little context: I have a few large aggregation problems and I have noted that they were each taking some time. In trying to solve the issues, I became interested in the performance of various aggregation procedures in R.

I tested a few aggregation methods - and found myself waiting around all day.

When I finally got results back, I discovered a huge gap between the plyr method and the others - which makes me think that I've done something dead wrong.

I ran the following code (I thought I'd check out the new dataframe package while I was at it):

require(plyr)
require(data.table)
require(dataframe)
require(rbenchmark)
require(xts)

plyr <- function(dd) ddply(dd, .(price), summarise, ss=sum(volume)) 
t.apply <- function(dd) unlist(tapply(dd$volume, dd$price, sum))
t.apply.x <- function(dd) unlist(tapply(dd[,2], dd[,1], sum))
l.apply <- function(dd) unlist(lapply(split(dd$volume, dd$price), sum))
l.apply.x <- function(dd) unlist(lapply(split(dd[,2], dd[,1]), sum))
b.y <- function(dd) unlist(by(dd$volume, dd$price, sum))
b.y.x <- function(dd) unlist(by(dd[,2], dd[,1], sum))
agg <- function(dd) aggregate(dd$volume, list(dd$price), sum)
agg.x <- function(dd) aggregate(dd[,2], list(dd[,1]), sum)
dtd <- function(dd) dd[, sum(volume), by=(price)]

obs <- c(5e1, 5e2, 5e3, 5e4, 5e5, 5e6, 5e6, 5e7, 5e8)
timS <- timeBasedSeq('20110101 083000/20120101 083000')

bmkRL <- list(NULL)

for (i in 1:5){
  tt <- timS[1:obs[i]]

  for (j in 1:8){
    pxl <- seq(0.9, 1.1, by= (1.1 - 0.9)/floor(obs[i]/(11-j)))
    px <- sample(pxl, length(tt), replace=TRUE)
    vol <- rnorm(length(tt), 1000, 100)

    d.df <- base::data.frame(time=tt, price=px, volume=vol)
    d.dfp <- dataframe::data.frame(time=tt, price=px, volume=vol)
    d.matrix <- as.matrix(d.df[,-1])
    d.dt <- data.table(d.df)

    listLabel <- paste('i=',i, 'j=',j)

    bmkRL[[listLabel]] <- benchmark(plyr(d.df), plyr(d.dfp), t.apply(d.df),     
                         t.apply(d.dfp), t.apply.x(d.matrix), 
                         l.apply(d.df), l.apply(d.dfp), l.apply.x(d.matrix),
                         b.y(d.df), b.y(d.dfp), b.y.x(d.matrix), agg(d.df),
                         agg(d.dfp), agg.x(d.matrix), dtd(d.dt),
          columns =c('test', 'elapsed', 'relative'),
          replications = 10,
          order = 'elapsed')
  }
}

The test was supposed to check up to 5e8, but it took too long - mostly due to plyr. The 5e5 the final table shows the problem:

$`i= 5 j= 8`
                  test  elapsed    relative
15           dtd(d.dt)    4.156    1.000000
6        l.apply(d.df)   15.687    3.774543
7       l.apply(d.dfp)   16.066    3.865736
8  l.apply.x(d.matrix)   16.659    4.008422
4       t.apply(d.dfp)   21.387    5.146054
3        t.apply(d.df)   21.488    5.170356
5  t.apply.x(d.matrix)   22.014    5.296920
13          agg(d.dfp)   32.254    7.760828
14     agg.x(d.matrix)   32.435    7.804379
12           agg(d.df)   32.593    7.842397
10          b.y(d.dfp)   98.006   23.581809
11     b.y.x(d.matrix)   98.134   23.612608
9            b.y(d.df)   98.337   23.661453
1           plyr(d.df) 9384.135 2257.972810
2          plyr(d.dfp) 9384.448 2258.048123

Is this right? Why is plyr 2250x slower than data.table? And why didn't using the new data frame package make a difference?

The session info is:

> sessionInfo()
R version 2.15.1 (2012-06-22)
Platform: x86_64-apple-darwin9.8.0/x86_64 (64-bit)

locale:
[1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

other attached packages:
[1] xts_0.8-6        zoo_1.7-7        rbenchmark_0.3   dataframe_2.5    data.table_1.8.1     plyr_1.7.1      

loaded via a namespace (and not attached):
[1] grid_2.15.1    lattice_0.20-6 tools_2.15.1 
1 Answers
Related