[R] Dataframe of factors transform speed?
jim holtman
jholtman at gmail.com
Sat Jul 21 23:33:18 CEST 2007
One of the problems is that you are probably paging on your system
with an object that size (240000 x 1000). This is about 1GB for a
single object:
> set.seed(123)
> n <- 240000
> system.time({
+ genoT <- lapply(1:n, function(i) factor(sample(c("AA",
+ "AB", "BB"), 1000, prob=c(1000, 1, 1), rep=T)))
+ })
user system elapsed
95.00 0.61 104.71
> names(genoT) = paste("snp", 1:n, sep="")
>
> object.size(genoT)
[1] 1045258752
>
I can create it on my 2GB machine as a list, but have problems
converting it to a dataframe because I don't have enough memory.
So unless you have at least 4GB on your system, it might take a long
time. Look at your performance measurements on your system and see if
you have run out of physical memory and are paging.
On 7/21/07, Latchezar Dimitrov <ldimitro at wfubmc.edu> wrote:
> Hi,
>
> Thanks for the help. My 1st question still unanswered though :-) Please
> see bellow
>
> > -----Original Message-----
> > From: Benilton Carvalho [mailto:bcarvalh at jhsph.edu]
> > Sent: Friday, July 20, 2007 3:30 AM
> > To: Latchezar Dimitrov
> > Cc: r-help at stat.math.ethz.ch
> > Subject: Re: [R] Dataframe of factors transform speed?
> >
> > set.seed(123)
> > genoT = lapply(1:240000, function(i) factor(sample(c("AA",
> > "AB", "BB"), 1000, prob=sample(c(1, 1000, 1000), 3), rep=T)))
> > names(genoT) = paste("snp", 1:240000, sep="") genoT =
> > as.data.frame(genoT)
>
> Now this _is the problem. Everything before converting to data.frame
> worked almost instantaneously however as.data.frame runs forever.
> Obviously there is some scalability memory management issue. When I
> tried my own method but creating a new result (instead of modifying the
> old) dataframe it worked like a charm for the 1st 100 cols ~ .3s. I
> figured 300,000 cols should be ~1000s. Nope! It ran for about 50,000(!)s
> to finish about 42,000 cols only.
>
> BTW, what ver. of R is yours?
>
> Now here's what I "discovered" further.
>
> #-- create a 1-col frame:
> geno <-
> data.frame(c(geno.GASP[[1]],geno.JAG[[1]]),row.names=c(rownames(geno.GAS
> P),rownames(geno.JAG)))
>
> #-- main code I repeated it w/ j in 1:1000, 2001:3000, and 3001:4000,
> i.e., adding a 1000 of cols to geno each time
>
> system.time(
> # for(j in 1:(ncol(geno.GASP ))){
> for(j in 3001:(4000 )){
> gt.GASP<-geno.GASP[[j]]
> for(l in 1:length(gt.GASP at levels)){
> levels(gt.GASP)[l] <-
> switch(gt.GASP at levels[l],AA="0",AB="1",BB="2")
> }
> gt.JAG <-geno.JAG [[j]]
> # for(l in 1:length(gt.JAG @levels)){
> # levels(gt.JAG )[l] <- switch(gt.JAG
> @levels[l],AA="0",AB="1",BB="2")
> # }
> geno[[j]]<-factor(c(as.numeric(factor(gt.GASP,levels=0:2))-1
> ### factor(c(as.numeric(factor(gt.GASP,levels=0:2))-1
> ,as.numeric(factor(gt.JAG, levels=0:2))-1
> )
> ,levels=0:2
> )
> }
> )
>
> Times (each one is for a 1000 cols!):
> [1] 26.673 0.032 26.705 0.000 0.000
> [1] 77.186 0.037 77.225 0.000 0.000
> [1] 128.165 0.042 128.209 0.000 0.000
> [1] 180.940 0.047 180.989 0.000 0.000
>
> See the big diff and the scaling I mentioned above?
>
> Further more I removed geno[[j]] assignment leaving the operation
> though, i.e., replaced it with ### line above. Times:
>
> [1] 0.857 0.008 0.865 0.000 0.000
>
> Huh!? What the heck! That's my second question :-) Any ideas?
>
> I still believe my method is near optimal. Of course I have to somehow
> get rid of the assignment bottleneck.
>
> For now the lesson is: "God bless lists"
>
> Here is my final solution:
>
> > system.time({
> + geno.GASP.L<-lapply(geno.GASP
> + ,function(x){
> + for(l in 1:length(x at levels)){levels(x)[l] <-
> switch(x at levels[l],AA="0",AB="1",BB="2")}
> + factor(x,levels=0:2)
> + }
> + )
> + geno.JAG.L <-lapply(geno.JAG
> + ,function(x){
> + # for(l in 1:length(x at levels)){levels(x)[l] <-
> switch(x at levels[l],AA="0",AB="1",BB="2")}
> + factor(x,levels=0:2)
> + }
> + )
> + })
> [1] 192.800 1.566 194.413 0.000 0.000 !!!!!!!!! :-)))))
> > system.time({
> + class (geno.GASP.L)<-"data.frame"
> + row.names(geno.GASP.L)<-row.names(geno.GASP)
> + class (geno.JAG.L )<-"data.frame"
> + row.names(geno.JAG.L )<-row.names(geno.JAG )
> + })
> [1] 12.156 0.001 12.155 0.000 0.000
> > system.time({
> + geno<-rbind(geno.GASP.L,geno.JAG.L)
> + })
> [1] 1542.340 9.072 2066.310 0.000 0.000
>
> I logged my notes here as I was trying various things. Partly the reason
> is my two questions:
>
> "What was wrong with me?" and
> "What the heck?!" remember above? :-)))
>
> which still remain unanswered :-(
>
> I would have had a lot of fun if I had not to have this done by ...
> Yesterday :-))
>
> Thanks a lot for the help
>
> Latchezar
>
> > dim(genoT)
> > class(genoT)
> > system.time(out <- lapply(genoT, function(x) match(x, c("AA", "AB",
> > "BB"))-1))
> > ##
> > ##
> > user system elapsed
> > 119.288 0.004 119.339
> >
> > (for all 240K)
> >
> > best,
> > b
> >
> > ps: note that "out" is a list.
> >
> > On Jul 20, 2007, at 2:01 AM, Latchezar Dimitrov wrote:
> >
> > > Hi,
> > >
> > >> -----Original Message-----
> > >> From: Benilton Carvalho [mailto:bcarvalh at jhsph.edu]
> > >> Sent: Friday, July 20, 2007 12:25 AM
> > >> To: Latchezar Dimitrov
> > >> Cc: r-help at stat.math.ethz.ch
> > >> Subject: Re: [R] Dataframe of factors transform speed?
> > >>
> > >> it looks like that whatever method you used to genotype the
> > >> 1002 samples on the STY array gave you a transposed matrix of
> > >> genotype calls. :-)
> > >
> > > It only looks like :-)
> > >
> > > Otherwise it is correctly created dataframe of 1002 samples X (big
> > > number) of columns (SNP genotypes). It worked perfectly until I
> > > decided to put together to cohorts independently processed in R
> > > already. I got stuck with my lack of foreseeing. Otherwise I would
> > > have put 3 dummy lines w/ AA,AB, and AB on each one to make
> > sure all 3
> > > genotypes are present and that's it! Lesson for the future :-)
> > >
> > > Maybe I am not using columns and rows appropriately here but the
> > > dataframe is correct (I have not used FORTRAN since FORTRAN IV ;-)
> > > - as
> > > str says 1002 observ. of (big number) vars.
> > >
> > >>
> > >> i'd use:
> > >>
> > >> genoT = read.table(yourFile, stringsAsFactors = FALSE)
> > >>
> > >> as a starting point... but I don't think that would be
> > efficient (as
> > >> you'd need to fix one column at a time - lapply).
> > >
> > > No it was not efficient at all. 'matter of fact nothing is more
> > > efficient then loading already read data, alas :-(
> > >
> > >>
> > >> i'd preprocess yourFile before trying to load it:
> > >>
> > >> cat yourFile | sed -e 's/AA/1/g' | sed -e 's/AB/2/g' | sed -e
> > >> 's/BB/3/ g' > outFile
> > >>
> > >> and, now, in R:
> > >>
> > >> genoT = read.table(outFile, header=TRUE)
> > >
> > > ... Too late ;-) As it must be clear now I have two
> > dataframes I want
> > > to put together with rbind(geno1,geno2). The issue again is
> > > "uniformization" of factor variables w/ missing factors -
> > they ended
> > > up like levels AA,BB on one of the and levels AB,BB on the
> > other which
> > > means as.numeric of AA is 1 on the 1st and as.numeric of AB is 1 on
> > > the second - complete mess. That's why I tried to make both
> > uniform,
> > > i.e.
> > > levels "AA","AB", and "BB" for every SNP and then rbind works.
> > >
> > > In any case my 1st questions remains: "What's wrong with me?" :-)
> > >
> > > Thanks,
> > > Latchezar
> > >
> > >>
> > >> b
> > >>
> > >> On Jul 19, 2007, at 11:51 PM, Latchezar Dimitrov wrote:
> > >>
> > >>> Hello,
> > >>>
> > >>> This is a speed question. I have a dataframe genoT:
> > >>>
> > >>>> dim(genoT)
> > >>> [1] 1002 238304
> > >>>
> > >>>> str(genoT)
> > >>> 'data.frame': 1002 obs. of 238304 variables:
> > >>> $ SNP_A.4261647: Factor w/ 3 levels "0","1","2": 3 3 3 3 3
> > >> 3 3 3 3 3
> > >>> ...
> > >>> $ SNP_A.4261610: Factor w/ 3 levels "0","1","2": 1 1 3 3 1
> > >> 1 1 2 2 2
> > >>> ...
> > >>> $ SNP_A.4261601: Factor w/ 3 levels "0","1","2": 1 1 1 1 1
> > >> 1 1 1 1 1
> > >>> ...
> > >>> $ SNP_A.4261704: Factor w/ 3 levels "0","1","2": 3 3 3 3 3
> > >> 3 3 3 3 3
> > >>> ...
> > >>> $ SNP_A.4261563: Factor w/ 3 levels "0","1","2": 3 1 2 1 2
> > >> 3 2 3 3 1
> > >>> ...
> > >>> $ SNP_A.4261554: Factor w/ 3 levels "0","1","2": 1 1 NA
> > 1 NA 2 1 1
> > >>> 2 1
> > >>> ...
> > >>> $ SNP_A.4261666: Factor w/ 3 levels "0","1","2": 1 1 2 1 1
> > >> 1 1 1 1 2
> > >>> ...
> > >>> $ SNP_A.4261634: Factor w/ 3 levels "0","1","2": 3 3 2 3 3
> > >> 3 3 3 3 2
> > >>> ...
> > >>> $ SNP_A.4261656: Factor w/ 3 levels "0","1","2": 1 1 2 1 1
> > >> 1 1 1 1 2
> > >>> ...
> > >>> $ SNP_A.4261637: Factor w/ 3 levels "0","1","2": 1 3 2 3 2
> > >> 1 2 1 1 3
> > >>> ...
> > >>> $ SNP_A.4261597: Factor w/ 3 levels "AA","AB","BB": 2 2 3 3 3 2 1
> > >>> 2 2 3
> > >>> ...
> > >>> $ SNP_A.4261659: Factor w/ 3 levels "AA","AB","BB": 3 3 3 3 3 3 3
> > >>> 3 3 3
> > >>> ...
> > >>> $ SNP_A.4261594: Factor w/ 3 levels "AA","AB","BB": 2 2 2 1 1 1 2
> > >>> 2 2 2
> > >>> ...
> > >>> $ SNP_A.4261698: Factor w/ 2 levels "AA","AB": 1 1 1 1 1 1 1 1 1
> > >>> 1 ...
> > >>> $ SNP_A.4261538: Factor w/ 3 levels "AA","AB","BB": 2 3 2 2 3 2 2
> > >>> 1 1 2
> > >>> ...
> > >>> $ SNP_A.4261621: Factor w/ 3 levels "AA","AB","BB": 1 1 1 1 1 1 1
> > >>> 1 1 1
> > >>> ...
> > >>> $ SNP_A.4261553: Factor w/ 3 levels "AA","AB","BB": 1 1 2 1 1 1 1
> > >>> 1 1 1
> > >>> ...
> > >>> $ SNP_A.4261528: Factor w/ 2 levels "AA","AB": 1 1 1 1 1 1 1 1 1
> > >>> 1 ...
> > >>> $ SNP_A.4261579: Factor w/ 3 levels "AA","AB","BB": 1 1 1 1 1 2 1
> > >>> 1 1 2
> > >>> ...
> > >>> $ SNP_A.4261513: Factor w/ 3 levels "AA","AB","BB": 2 1 2
> > >> 2 2 NA 1 NA
> > >>> 2
> > >>> 1 ...
> > >>> $ SNP_A.4261532: Factor w/ 3 levels "AA","AB","BB": 1 2 2 1 1 1 3
> > >>> 1 1 1
> > >>> ...
> > >>> $ SNP_A.4261600: Factor w/ 2 levels "AB","BB": 2 2 2 2 2 2 2 2 2
> > >>> 2 ...
> > >>> $ SNP_A.4261706: Factor w/ 2 levels "AA","BB": 1 1 1 1 1 1 1 1 1
> > >>> 1 ...
> > >>> $ SNP_A.4261575: Factor w/ 3 levels "AA","AB","BB": 1 1 1 1 1 1 1
> > >>> 2 2 1
> > >>> ...
> > >>>
> > >>> Its columns are factors with different number of levels
> > >> (from 1 to 3 -
> > >>> that's what I got from read.table, i.e., it dropped missing
> > >> levels). I
> > >>> want to convert it to uniform factors with 3 levels. The
> > >> 1st 10 rows
> > >>> above show already converted columns and the rest are not yet
> > >>> converted.
> > >>> Here's my attempt wich is a complete failure as speed:
> > >>>
> > >>>> system.time(
> > >>> + for(j in 1:(10 )){ #-- this is to try 1st
> > 10 cols and
> > >>> measure the time, it otherwise is ncol(genoT) instead of 10
> > >>>
> > >>> + gt<-genoT[[j]] #-- this is to avoid 2D indices
> > >>> + for(l in 1:length(gt at levels)){
> > >>> + levels(gt)[l] <-
> > >> switch(gt at levels[l],AA="0",AB="1",BB="2")
> > >>> #-- convert levels to "0","1", or "2"
> > >>> + genoT[[j]]<-factor(gt,levels=0:2) #-- make a 3-level
> > >>> factor
> > >>> and put it back
> > >>> + }
> > >>> + }
> > >>> + )
> > >>> [1] 785.085 4.358 789.454 0.000 0.000
> > >>>
> > >>> 789s for 10 columns only!
> > >>>
> > >>> To me it seems like replacing 10 x 3 levels and then making
> > >> a factor
> > >>> of
> > >>> 1002 element vector x 10 is a "negligible" amount of operations
> > >>> needed.
> > >>>
> > >>> So, what's wrong with me? Any idea how to accelerate
> > >> significantly the
> > >>> transformation or (to go to the very beginning) to make
> > >> read.table use
> > >>> a fixed set of levels ("AA","AB", and "BB") and not to drop any
> > >>> (missing)
> > >>> level?
> > >>>
> > >>> R-devel_2006-08-26, Sun Solaris 10 OS - x86 64-bit
> > >>>
> > >>> The machine is with 32G RAM and AMD Opteron 285 (2.? GHz)
> > >> so it's not
> > >>> it.
> > >>>
> > >>> Thank you very much for the help,
> > >>>
> > >>> Latchezar Dimitrov,
> > >>> Analyst/Programmer IV,
> > >>> Wake Forest University School of Medicine, Winston-Salem, North
> > >>> Carolina, USA
> > >>>
> > >>> ______________________________________________
> > >>> R-help at stat.math.ethz.ch 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 stat.math.ethz.ch 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.
>
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
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