I'm working with a large data frame called exp (file here) in R. In the interests of performance, it was suggested that I check out the idata.frame() function from plyr. But I think I'm using it wrong.

My original call, slow but it works:

df.median<-ddply(exp, 
                 .(groupname,starttime,fPhase,fCycle), 
                 numcolwise(median), 
                 na.rm=TRUE)

With idata.frame, Error: is.data.frame(df) is not TRUE

library(plyr)
df.median<-ddply(idata.frame(exp), 
                 .(groupname,starttime,fPhase,fCycle), 
                 numcolwise(median), 
                 na.rm=TRUE)

So, I thought, perhaps it is my data. So I tried the baseball dataset. The idata.frame example works fine: dlply(idata.frame(baseball), "id", nrow) But if I try something similar to my desired call using baseball, it doesn't work:

bb.median<-ddply(idata.frame(baseball), 
               开发者_运维百科  .(id,year,team), 
                 numcolwise(median), 
                 na.rm=TRUE)
>Error: is.data.frame(df) is not TRUE

Perhaps my error is in how I'm specifying the groupings? Anyone know how to make my example work?

ETA:

I also tried:

groupVars <- c("groupname","starttime","fPhase","fCycle")
voi<-c('inadist','smldist','lardist')

i<-idata.frame(exp)
ag.median <- aggregate(i[,voi], i[,groupVars], median)
Error in i[, voi] : object of type 'environment' is not subsettable

which uses a faster way of getting the medians, but gives a different error. I don't think I understand how to use idata.frame at all.

---

Given you are working with 'big' data and looking for perfomance, this seems a perfect fit for data.table.

Specifically the lapply(.SD,FUN) and .SDcols arguments with by

Setup the data.table

library(data.table)
DT <- as.data.table(exp)
iexp <- idata.frame(exp)

Which columns are numeric

numeric_columns <- names(which(unlist(lapply(DT, is.numeric))))



dt.median <- DT[, lapply(.SD, median), by = list(groupname, starttime, fPhase, 
    fCycle), .SDcols = numeric_columns]

some benchmarking

library(rbenchmark)
benchmark(data.table = DT[, lapply(.SD, median), by = list(groupname, starttime, 
    fPhase, fCycle), .SDcols = numeric_columns], 
 plyr = ddply(exp, .(groupname, starttime, fPhase, fCycle), numcolwise(median), na.rm = TRUE), 
 idataframe = ddply(exp, .(groupname, starttime, fPhase, fCycle), function(x) data.frame(inadist = median(x$inadist), 
        smldist = median(x$smldist), lardist = median(x$lardist), inadur = median(x$inadur), 
        smldur = median(x$smldur), lardur = median(x$lardur), emptyct = median(x$emptyct), 
        entct = median(x$entct), inact = median(x$inact), smlct = median(x$smlct), 
        larct = median(x$larct), na.rm = TRUE)), 
 aggregate = aggregate(exp[, numeric_columns],
                       exp[, c("groupname", "starttime", "fPhase", "fCycle")], 
              median), 
 replications = 5)

##         test replications elapsed relative user.self 
## 4  aggregate            5    5.42    1.789      5.30   
## 1 data.table            5    3.03    1.000      3.03    
## 3 idataframe            5   11.81    3.898     11.77       
## 2       plyr            5    9.47    3.125      9.45       

---

Strange behaviour, but even in the docs it says that idata.frame is experimental. You probably found a bug. Perhaps you could rewrite the check at the top of ddply that tests is.data.frame().

In any case, this cuts about 20% off the time (on my system):

system.time(df.median<-ddply(exp, .(groupname,starttime,fPhase,fCycle), function(x) data.frame(
inadist=median(x$inadist),
smldist=median(x$smldist),
lardist=median(x$lardist),
inadur=median(x$inadur),
smldur=median(x$smldur),
lardur=median(x$lardur),
emptyct=median(x$emptyct),
entct=median(x$entct),
inact=median(x$inact),
smlct=median(x$smlct),
larct=median(x$larct),
na.rm=TRUE))
) 

Shane asked you in another post if you could cache the results of your script. I don't really have an idea of your workflow, but it may be best to setup a chron to run this and store the results, daily/hourly whatever.