How to optimize subsetting from a large dataset?

Question

I have this dataset of financial transactions, its pretty big but small enough to keep in memory..

R> str(trans)
'data.frame':   130000000 obs. of  5 variables:
 $ id    : int  5 5 5 5 6 11 11 11 11 11 ...
 $ kod   : int  2 3 2 3 38 2 3 6 7 6 ...
 $ ar    : int  329 329 330 330 7 329 329 329 329 329 ...
 $ belopp: num  1531 -229.3 324 -48.9 0 ...
 $ datum : int  36976 36976 37287 37287 37961 36976 36976 37236 37236 37281 ...

I need to loop through it extracting the transactions for each unique id, and do a bunch of calculations. The trouble is that the subsetting of the dataset is way too slow..

R> system.time(
+ sub <- trans[trans$id==15,]
+ )
   user  system elapsed 
   7.80    0.55    8.36


R> system.time(
+ sub <- subset(trans, id == 15)
+ )
   user  system elapsed 
   8.49    1.05    9.53 

As there are about 10m unique id's in this dataset, such a loop would take forever, any ideas how I might speed it up?

EDIT I've dabbled with ´data.tables´, indexing and sorting with not much luck at all..

library(data.table)
trans2 <- as.data.table(trans)
trans2 <- trans2[order(id)]
trans2 <- setkey(trans2, id)

R> system.time(
+ sub <- trans2[trans2$id==15,]
+ )
   user  system elapsed 
   7.33    1.08    8.41 

R> system.time(
+ sub <- subset(trans2, id == 15)
+ )
   user  system elapsed 
   8.66    1.12    9.78

EDIT2 Awesome.

R> system.time(
+ sub <- trans2[J(15)]
+ )
   user  system elapsed 
      0       0       0 

Answer 1

Note: The post has been edited by changing the function being calculated from rowSums to colSums (using lapply in case of data.table)

I don't think you could get the result faster than data.table. Here's a benchmark between plyr and data.table. Of course if the time-consuming part is your function, then you could use doMC to run in parallel using plyr (assuming you have a lot of cores or you work on a cluster). Else, I'd stick to data.table. Here's an analysis with a huge test data and a dummy function:

# create a huge data.frame with repeating id values
len <- 1e5
reps <- sample(1:20, len, replace = TRUE)
x <- data.frame(id = rep(1:len, reps))
x <- transform(x, v1 = rnorm(nrow(x)), v2 = rnorm(nrow(x)))

> nrow(x) 
[1] 1048534 # 1 million rows

# construct functions for data.table and plyr
# method 1
# using data.table
DATA.TABLE <- function() {
    require(data.table)
    x.dt <- data.table(x, key="id")
    x.dt.out <- x.dt[, lapply(.SD, sum), by=id]
}

# method 2
# using plyr
PLYR <- function() {
    require(plyr)
    x.plyr.out <- ddply(x, .(id), colSums)
}

# let's benchmark
> require(rbenchmark)
> benchmark(DATA.TABLE(), PLYR(), order = "elapsed", replications = 1)[1:5]
          test replications elapsed relative user.self
1 DATA.TABLE()           1  1.006     1.00    .992
2       PLYR()           1  67.755   67.351  67.688

On a data.frame with 1 million rows, data.table takes 0.992 seconds. The speed-up using data.table compared to plyr (admittedly, on computing column sums) is 68x. Depending on the computation time in your function, this speed-up will vary. But data.table will still be way faster. plyr is a split-apply-combine strategy. I don't think you'd get a comparable speed-up compared to using base to split, apply and combine yourself. Of course you can try it.

I ran the code with 10 million rows. data.table ran in 5.893 seconds. plyr took 6300 seconds.

Answer 2

Why not use the split, apply and combine strategy?

Something like this (without a sample data I don't know if this will work):

fastsplit <- function (df) {
  lista <- split(seq(nrow(df)), df$id)
  return(lista)
}

# function to split the data frame into a list by id

lista_split <- fastsplit(trans)

# now, assuming that one of the calculations is, for instance, to sum belopp
# apply the function to each subset

result1 <- lapply(lista_split, function(.indx){

  sum_bellop = sum(trans$belopp[.indx])})

# combine stage
r1 <- do.call(rbind, result1)

Having provided the code above, I'd say that it would be faster and easier if you could use SQL. Maybe the package sqldf could help you here? I never tryed it though. I don't know if it is fast. The code in SQL is quite simple. To do the same as the R code above, just use Something like:

select id
       , sum(belopp) as sum_bellop from trans
group by id

This will return a table with two columns, id and sum of belopp by id