[R] SLOW split() function
Joshua Wiley
jwiley.psych at gmail.com
Tue Oct 11 20:35:05 CEST 2011
I do not know if stripping down functions is generally recommended,
but it is not too difficult to do if you know that you can make
assumptions. Here is an example (I also found a fast way to convert
the data table to a matrix, again if some assumptions can be made).
Using the stripped down function, you can get coefficients and
standard errors in less time than you can get just coefficients using
default lm. It is hugely less flexible.
Cheers,
Josh
##########################################
library(data.table)
## stripped down lm and summary.lm (for standard errors)
minimal.lm <- function(y, x) {
dims <- dim(x)
x <- unlist(x, FALSE, FALSE)
dim(x) <- dims
obj <- lm.fit(x = x, y = y)
resvar <- sum(obj$residuals^2)/obj$df.residual
p <- obj$rank
R <- .Call("La_chol2inv", x = obj$qr$qr[1L:p, 1L:p, drop = FALSE],
size = p, PACKAGE = "base")
m <- min(dim(R))
d <- c(R)[1L + 0L:(m - 1L) * (dim(R)[1L] + 1L)]
se <- sqrt(d * resvar)
cbind(coef = obj$coefficients, se)
}
N <- 1000*100
d <- data.table(data.frame( key= as.integer(runif(N, min=1,
max=N/10)), x=rnorm(N), y=rnorm(N) )) # irregular
## add intercept column
d$int <- 1L
setkey(d, "key"); gc() ## sort and force a garbage collection
cat("N=", N, ". Size of d=", object.size(d)/1024/1024, "MB\n")
print(system.time(si <- split(seq(nrow(d)), d$key)))
cat("\n\t(b) Regressions:\n")
## using lm
print(system.time(all.2b <- lapply(si, function(.indx) { coef(lm(y ~
x, data=d[.indx,])) })))
## using minimal.lm---faster and gives standard errors
print(system.time(all.2c <- lapply(si, function(.indx) { minimal.lm(y
= d[.indx, y], x = d[.indx, list(int, x)]) })))
#### Timings on my system ####
> print(system.time(all.2b <- lapply(si, function(.indx) { coef(lm(y ~
+ x, data=d[.indx,])) })))
user system elapsed
67.87 0.01 68.46
> print(system.time(all.2c <- lapply(si, function(.indx) { minimal.lm(y = d[.indx, y], x = d[.indx, list(int, x)]) })))
user system elapsed
47.72 0.00 48.00
######################################################################
On Tue, Oct 11, 2011 at 8:56 AM, ivo welch <ivo.welch at gmail.com> wrote:
> thanks, josh. in my posting example, I did not need anything except
> coefficients. (when this is the case, I usually do not even use
> lm.fit, but I eliminate all missing obs first and then use solve
> crossprod(y,cbind(1,x)) crossprod(cbind(1,x)).) this is pretty fast.)
>
> alas, I will need to figure how to get coef standard errors faster in
> this case. summary.lm() is really slow.
>
> regards,
>
> /iaw
> ----
> Ivo Welch (ivo.welch at gmail.com)
> http://www.ivo-welch.info/
> J. Fred Weston Professor of Finance
> Anderson School at UCLA, C519
>
>
>
>
>
> On Mon, Oct 10, 2011 at 11:30 PM, Joshua Wiley <jwiley.psych at gmail.com> wrote:
>> 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/
>>
>
--
Joshua Wiley
Ph.D. Student, Health Psychology
Programmer Analyst II, ATS Statistical Consulting Group
University of California, Los Angeles
https://joshuawiley.com/
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