From rvaradhan at jhmi.edu  Wed Aug 17 20:32:50 2011
From: rvaradhan at jhmi.edu (Ravi Varadhan)
Date: Wed, 17 Aug 2011 18:32:50 +0000
Subject: [Rd] An example of very slow computation
In-Reply-To: <87aab82cfy.fsf@tajo.ucsd.edu>
References: <4E4B7B69.6030106@uottawa.ca> <87aab82cfy.fsf@tajo.ucsd.edu>
Message-ID: <2F9EA67EF9AE1C48A147CB41BE2E15C3059E2B@DOM-EB-MAIL2.win.ad.jhu.edu>

Yes, the culprit is the evaluation of expm(A*t).  This is a lazy way of solving the system of ODEs, where you save analytic effort, but you pay for it dearly in terms of computational effort!

Even in this lazy approach, I am sure there ought to be faster ways to evaluating exponent of a matrix than that in "Matrix" package.

Ravi.

-------------------------------------------------------
Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric Medicine and Gerontology School of Medicine Johns Hopkins University

Ph. (410) 502-2619
email: rvaradhan at jhmi.edu

-----Original Message-----
From: r-devel-bounces at r-project.org [mailto:r-devel-bounces at r-project.org] On Behalf Of cberry at tajo.ucsd.edu
Sent: Wednesday, August 17, 2011 1:08 PM
To: r-devel at stat.math.ethz.ch
Subject: Re: [Rd] An example of very slow computation

John C Nash <nashjc at uottawa.ca> writes:

> This message is about a curious difference in timing between two ways of computing the
> same function. One uses expm, so is expected to be a bit slower, but "a bit" turned out to
> be a factor of >1000. The code is below. We would be grateful if anyone can point out any
> egregious bad practice in our code, or enlighten us on why one approach is so much slower
> than the other. 

Looks like A*t in expm(A*t) is a "matrix".

'getMethod("expm","matrix")' coerces it arg to a "Matrix", then calls
expm(), whose method coerces its arg to a "dMatrix" and calls expm(),
whose method coerces its arg to a 'dgeMatrix' and calls expm(), whose
method finally calls '.Call(dgeMatrix_exp, x)' 

Whew!

The time difference between 'expm( diag(10)+1 )' and 'expm( as( diag(10)+1,
"dgeMatrix" ))' is a factor of 10 on my box. 

Dunno 'bout the other factor of 100.

Chuck




> The problem arose in an activity to develop guidelines for nonlinear
> modeling in ecology (at NCEAS, Santa Barbara, with worldwide participants), and we will be
> trying to include suggestions of how to prepare problems like this for efficient and
> effective solution. The code for nlogL was the "original" from the worker who supplied the
> problem.
>
> Best,
>
> John Nash
>
> --------------------------------------------------------------------------------------
>
> cat("mineral-timing.R == benchmark MIN functions in R\n")
> #  J C Nash July 31, 2011
>
> require("microbenchmark")
> require("numDeriv")
> library(Matrix)
> library(optimx)
> # dat<-read.table('min.dat', skip=3, header=FALSE)
> # t<-dat[,1]
> t <- c(0.77,  1.69,  2.69,  3.67,  4.69,  5.71,  7.94,  9.67, 11.77, 17.77,
>  23.77, 32.77, 40.73, 47.75, 54.90, 62.81, 72.88, 98.77, 125.92, 160.19,
>  191.15, 223.78, 287.70, 340.01, 340.95, 342.01)
>
> # y<-dat[,2] # ?? tidy up
>  y<- c(1.396, 3.784, 5.948, 7.717, 9.077, 10.100, 11.263, 11.856, 12.251, 12.699,
> 12.869, 13.048, 13.222, 13.347, 13.507, 13.628, 13.804, 14.087, 14.185, 14.351,
> 14.458, 14.756, 15.262, 15.703, 15.703, 15.703)
>
>
> ones<-rep(1,length(t))
> theta<-c(-2,-2,-2,-2)
>
>
> nlogL<-function(theta){
>   k<-exp(theta[1:3])
>   sigma<-exp(theta[4])
>   A<-rbind(
>   c(-k[1], k[2]),
>   c( k[1], -(k[2]+k[3]))
>   )
>   x0<-c(0,100)
>   sol<-function(t)100-sum(expm(A*t)%*%x0)
>   pred<-sapply(dat[,1],sol)
>   -sum(dnorm(dat[,2],mean=pred,sd=sigma, log=TRUE))
> }
>
> getpred<-function(theta, t){
>   k<-exp(theta[1:3])
>   sigma<-exp(theta[4])
>   A<-rbind(
>   c(-k[1], k[2]),
>   c( k[1], -(k[2]+k[3]))
>   )
>   x0<-c(0,100)
>   sol<-function(tt)100-sum(expm(A*tt)%*%x0)
>   pred<-sapply(t,sol)
> }
>
> Mpred <- function(theta) {
> # WARNING: assumes t global
> kvec<-exp(theta[1:3])
> k1<-kvec[1]
> k2<-kvec[2]
> k3<-kvec[3]
> #   MIN problem terbuthylazene disappearance
>     z<-k1+k2+k3
>     y<-z*z-4*k1*k3
>     l1<-0.5*(-z+sqrt(y))
>     l2<-0.5*(-z-sqrt(y))
>     val<-100*(1-((k1+k2+l2)*exp(l2*t)-(k1+k2+l1)*exp(l1*t))/(l2-l1))
> } # val should be a vector if t is a vector
>
> negll <- function(theta){
> # non expm version JN 110731
>   pred<-Mpred(theta)
>   sigma<-exp(theta[4])
>   -sum(dnorm(dat[,2],mean=pred,sd=sigma, log=TRUE))
> }
>
> theta<-rep(-2,4)
> fand<-nlogL(theta)
> fsim<-negll(theta)
> cat("Check fn vals: expm =",fand,"   simple=",fsim,"  diff=",fand-fsim,"\n")
>
> cat("time the function in expm form\n")
> tnlogL<-microbenchmark(nlogL(theta), times=100L)
> tnlogL
>
> cat("time the function in simpler form\n")
> tnegll<-microbenchmark(negll(theta), times=100L)
> tnegll
>
> ftimes<-data.frame(texpm=tnlogL$time, tsimp=tnegll$time)
> # ftimes
>
>
> boxplot(log(ftimes))
> title("Log times in nanoseconds for matrix exponential and simple MIN fn")
>

-- 
Charles C. Berry                         cberry at tajo.ucsd.edu

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[Rd] An example of very slow computation
Ravi Varadhan