SLOW split() function
As another followup, given that you are doing numerous regression models and (I presume) working with finance/stock data that is strictly numeric (no need for special contrast coding, etc.), you can substantially reduce the time spent estimating the coefficients. A simple way is to use lm.fit directly instead of lm. For lm.fit, you pass the y and x (design) matrices directly. This skips a good deal of overhead. Here is one naive way, I imagine more speedups could be gained by incorporating the intercept (1 vector) into d instead of cbind()ing it. The catch it that lm.fit requires matrices, not data tables, so what you gain may be lost in having to do an extra conversion. In any case, here are the times on my system for the two options (note I used N = 1000 * 100 because I am presently on a glorified netbook).
print(system.time(all.2b <- lapply(si, function(.indx) { coef(lm(y ~
+ x, data=d[.indx,])) }))) user system elapsed 69.00 0.00 69.56
print(system.time(all.2c <- lapply(si, function(.indx) { coef(lm.fit(y = d[.indx, y], x = cbind(1, d[.indx, x]))) })))
user system elapsed 37.83 0.03 38.36 the column names for the coeficients will not be the same as from lm, but the estimates should be identical. While this is not recommended in typical usage, in an application like regressions on rolling time windows, etc. where you know the data are not changing, I think it makes sense to bypass the clever determine your data and best methods to use, and go straight to passing the design matrix. Since you do not need residuals, variances, etc. it may be possible to speed this up even more, perhaps bypassing dqrls altogether. Cheers, Josh
On Mon, Oct 10, 2011 at 9:56 PM, ivo welch <ivo.welch at gmail.com> wrote:
thank you, everyone. ?this was very helpful to my specific task and
understanding. ?for the benefit of future googlers, I thought I would
post some experiments and results here.
ultimately, I need to do a by() on an irregular matrix, and I now know
how to speed up by() on a single-core, and then again on a multi-core
machine.
library(data.table)
N <- 1000*1000
d <- data.table(data.frame( key= as.integer(runif(N, min=1,
max=N/10)), x=rnorm(N), y=rnorm(N) )) ?# irregular
setkey(d, "key"); gc() ## sort and force a garbage collection
cat("N=", N, ". ?Size of d=", object.size(d)/1024/1024, "MB\n")
cat("\nStandard by() Function:\n")
print(system.time( all.1 <- by( d, d$key, function(d) coef(lm(y ~ x, data=d)))))
cat("\n\nPreSplit Function [aka Jim H]\n\t(a) Splitting Operation:\n")
print(system.time(si <- split(seq(nrow(d)), d$key)))
cat("\n\t(b) Regressions:\n")
print(system.time(all.2 <- lapply(si, function(.indx) {
coef(lm(d$y[.indx] ~ d$x[.indx])) })))
print(system.time(all.2b <- lapply(si, function(.indx) { coef(lm(y ~
x, data=d[.indx,])) })))
cat("\n\nNaive Split Data Frame\n\t(a) Splitting Operation:\n")
print(system.time(ds <- split(d, d$key)))
cat("\n\t(b) Regressions:\n")
print(system.time(all.3a <- lapply(ds, function(ds) { coef(lm(ds$y ~ ds$x)) })))
print(system.time(all.3b <- lapply(ds, function(ds) { coef(lm(y ~ x,
data=ds)) })))
the first and the last ways (all.1 and all.3) are "naive" ways of
doing this, and take about 400-500 seconds on a Mac Air, core i5.
Jim's suggestion (all.2) cuts this roughly into half by speeding up
the split to take almost no time.
and now,
library(multicore)
print(system.time(all.4 <- mclapply(si, function(.indx) { coef(lm(y ~
x, data=d[.indx,])) })))
on my dual-core (quad-thread) i5, all four pseudo cores become busy,
and the time roughly halves again from 230 seconds to 120 seconds.
maybe the by() function should use Jim's approach, and multicore
should provide mcby(). ?of course, knowing how to do this myself fast
now by hand, this is not so important for me. ?but it may help some
other novices.
thanks again everybody.
regards,
/iaw
----
Ivo Welch (ivo.welch at gmail.com)
On Mon, Oct 10, 2011 at 9:31 PM, William Dunlap <wdunlap at tibco.com> wrote:
The following avoids the overhead of data.frame methods
(and assumes the data.frame doesn't include matrices
or other data.frames) and relies on split(vector,factor)
quickly splitting a vector into a list of vectors.
For a 10^6 row by 10 column data.frame split in 10^5
groups this took 14.1 seconds while split took 658.7 s.
Both returned the same thing.
Perhaps something based on this idea would help your
parallelized by().
mysplit.data.frame <-
function (x, f, drop = FALSE, ...)
{
? ?f <- as.factor(f)
? ?tmp <- lapply(x, function(xi) split(xi, f, drop = drop, ...))
? ?rn <- split(rownames(x), f, drop = drop, ...)
? ?tmp <- unlist(unname(tmp), recursive = FALSE)
? ?tmp <- split(tmp, factor(names(tmp), levels = unique(names(tmp))))
? ?tmp <- lapply(setNames(seq_along(tmp), names(tmp)), function(i) {
? ? ? ?t <- tmp[[i]]
? ? ? ?names(t) <- names(x)
? ? ? ?attr(t, "row.names") <- rn[[i]]
? ? ? ?class(t) <- "data.frame"
? ? ? ?t
? ?})
? ?tmp
}
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf Of Jim Holtman
Sent: Monday, October 10, 2011 7:29 PM
To: ivo welch
Cc: r-help
Subject: Re: [R] SLOW split() function
instead of spliting the entire dataframe, split the indices and then use these to access your data:
try
system.time(s <- split(seq(nrow(d)), d$key))
this should be faster and less memory intensive. ?you can then use the indices to access the subset:
result <- lapply(s, function(.indx){
? ? doSomething <- sum(d$someCol[.indx])
})
Sent from my iPad
On Oct 10, 2011, at 21:01, ivo welch <ivo.welch at gmail.com> wrote:
dear R experts: ?apologies for all my speed and memory questions. ?I
have a bet with my coauthors that I can make R reasonably efficient
through R-appropriate programming techniques. ?this is not just for
kicks, but for work. ?for benchmarking, my [3 year old] Mac Pro has
2.8GHz Xeons, 16GB of RAM, and R 2.13.1.
right now, it seems that 'split()' is why I am losing my bet. ?(split
is an integral component of *apply() and by(), so I need split() to be
fast. ?its resulting list can then be fed, e.g., to mclapply().) ?I
made up an example to illustrate my ills:
? ?library(data.table)
? ?N <- 1000
? ?T <- N*10
? ?d <- data.table(data.frame( key= rep(1:T, rep(N,T)), val=rnorm(N*T) ))
? ?setkey(d, "key"); gc() ## force a garbage collection
? ?cat("N=", N, ". ?Size of d=", object.size(d)/1024/1024, "MB\n")
? ?print(system.time( s<-split(d, d$key) ))
My ordered input data table (or data frame; doesn't make a difference)
is 114MB in size. ?it takes about a second to create. ?split() only
needs to reshape it. ?this simple operation takes almost 5 minutes on
my computer.
with a data set that is larger, this explodes further.
am I doing something wrong? ?is there an alternative to split()?
sincerely,
/iaw
----
Ivo Welch (ivo.welch at gmail.com)
______________________________________________ R-help at r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
______________________________________________ R-help at r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
______________________________________________ R-help at r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Joshua Wiley Ph.D. Student, Health Psychology Programmer Analyst II, ATS Statistical Consulting Group University of California, Los Angeles https://joshuawiley.com/