From lachmann at eva.mpg.de  Wed Aug 17 23:27:39 2011
From: lachmann at eva.mpg.de (Michael Lachmann)
Date: Wed, 17 Aug 2011 23:27:39 +0200
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: <1BA7596D-1AC9-44E9-A165-BB7EF17FB045@eva.mpg.de>


On 17 Aug 2011, at 7:08PM, <cberry at tajo.ucsd.edu> <cberry at tajo.ucsd.edu> wrote:

> 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. 
> 
> 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. 
> 

You are right! 

I was testing running nlogL below 100 times. expm() is then called 2500 times.
The total running time on my machine was 13 seconds.

If you replace in nlogL the part:
--
 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)
--
with:
--
 A<-rbind(
 c(-k[1], k[2]),
 c( k[1], -(k[2]+k[3]))
 )
 A<-as(A,"dgeMatrix")  # <--- this is the difference
  sol<-function(t)100-sum(expm(A*t)%*%x0)
--

this time drops to 1.5 seconds (!).

At that point, expm() takes up much less time than, for example, calculating A*t in sol(), and the sum() - I think because conversions have to be done.

Thus, if m is a 2x2 dgeMatrix, then 
> system.time({for(i in 1:2500) m*3.2})
   user  system elapsed 
  0.425   0.004   0.579 
(i.e. 1/3 of the total time for nlogL() above)

whereas if mm=as.matrix(m), then
> system.time({for(i in 1:2500) mm*3.2})
   user  system elapsed 
  0.004   0.000   0.005 

(!!)

and, similarly:
--
> system.time({for(i in 1:2500) sum(m)})
   user  system elapsed 
  0.399   0.002   0.494 
> system.time({for(i in 1:2500) sum(mm)})
   user  system elapsed 
  0.002   0.000   0.028 
--
whereas
> system.time({for(i in 1:2500) expm(m)})
   user  system elapsed 
  0.106   0.001   0.118 

to sum it up, of 13 seconds, 11.5 were spent on conversions to dgeMatrix
0.5 are spent on multiplying a dgeMatrix by a double
0.5 are spent on summing a dgeMatrix
and 0.1 are actually spent in expm.

Michael

P.S. You could have used Rprof() to see these times, only that interpreting summaryRprof() is a bit hard. (Is there something that does summaryRprof(), but also shows the call graph?)





> 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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> https://stat.ethz.ch/mailman/listinfo/r-devel
> 
> 
> 



[Rd] An example of very slow computation
Michael Lachmann