I am trying to take a very large set of records with multiple indices, calculate an aggregate statistic on groups determined by a subset of the indices, and then insert that into every row in the table. The issue here is that these are very large tables - over 10M rows each.
Code for reproducing the data is below.
The basic idea is that there are a set of indices, say ix1, ix2, ix3, ..., ixK. Generally, I am choosing only a couple of them, say ix1 and ix2. Then, I calculate an aggregation of all the rows with matching ix1 and ix2 values (over all combinations that appear), for a column called val. To keep it simple, I'll focus on a sum.
I have tried the following methods
Via sparse matrices: convert the values to a coordinate list, i.e. (ix1, ix2, val), then create a sparseMatrix - this nicely sums up everything, and then I need only convert back from the sparse matrix representation to the coordinate list. Speed: good, but it is doing more than is necessary and it doesn't generalize to higher dimensions (e.g. ix1, ix2, ix3) or more general functions than a sum.
Use of lapply and split: by creating a new index that is unique for all (ix1, ix2, ...) n-tuples, I can then use split and apply. The bad thing here is that the unique index is converted by split into a factor, and this conversion is terribly time consuming. Try system({zz <- as.factor(1:10^7)}).
I'm now trying data.table, via a command like sumDT <- DT[,sum(val),by = c("ix1","ix2")]. However, I don't yet see how I can merge sumDT with DT, other than via something like DT2 <- merge(DT, sumDT, by = c("ix1","ix2"))
Is there a faster method for this data.table join than via the merge operation I've described?
[I've also tried bigsplit from the bigtabulate package, and some other methods. Anything that converts to a factor is pretty much out - as far as I can tell, that conversion process is very slow.]
Code to generate data. Naturally, it's better to try a smaller N to see that something works, but not all methods scale very well for N >> 1000.
N <- 10^7
set.seed(2011)
ix1 <- 1 + floor(rexp(N, 0.01))
ix2 <- 1 + floor(rexp(N, 0.01))
ix3 <- 1 + floor(rexp(N, 0.01))
val <- runif(N)
DF <- data.frame(ix1 = ix1, ix2 = ix2, ix3 = ix3, val = val)
DF <- DF[order(DF[,1],DF[,2],DF[,3]),]
DT <- as.data.table(DF)
Well, it's possible you'll find that doing the merge isn't so bad as long as your keys are properly set.
Let's setup the problem again:
N <- 10^6 ## not 10^7 because RAM is tight right now
set.seed(2011)
ix1 <- 1 + floor(rexp(N, 0.01))
ix2 <- 1 + floor(rexp(N, 0.01))
ix3 <- 1 + floor(rexp(N, 0.01))
val <- runif(N)
DT <- data.table(ix1=ix1, ix2=ix2, ix3=ix3, val=val, key=c("ix1", "ix2"))
Now you can calculate your summary stats
info <- DT[, list(summary=sum(val)), by=key(DT)]
And merge the columns "the data.table way", or just with merge
m1 <- DT[info] ## the data.table way
m2 <- merge(DT, info) ## if you're just used to merge
identical(m1, m2)
[1] TRUE
If either of those ways of merging is too slow, you can try a tricky way to build info at the cost of memory:
info2 <- DT[, list(summary=rep(sum(val), length(val))), by=key(DT)]
m3 <- transform(DT, summary=info2$summary)
identical(m1, m3)
[1] TRUE
Now let's see the timing:
#######################################################################
## Using data.table[ ... ] or merge
system.time(info <- DT[, list(summary=sum(val)), by=key(DT)])
user system elapsed
0.203 0.024 0.232
system.time(DT[info])
user system elapsed
0.217 0.078 0.296
system.time(merge(DT, info))
user system elapsed
0.981 0.202 1.185
########################################################################
## Now the two parts of the last version done separately:
system.time(info2 <- DT[, list(summary=rep(sum(val), length(val))), by=key(DT)])
user system elapsed
0.574 0.040 0.616
system.time(transform(DT, summary=info2$summary))
user system elapsed
0.173 0.093 0.267
Or you can skip the intermediate info table building if the following doesn't seem too inscrutable for your tastes:
system.time(m5 <- DT[ DT[, list(summary=sum(val)), by=key(DT)] ])
user system elapsed
0.424 0.101 0.525
identical(m5, m1)
# [1] TRUE