[R] Dataframe of factors transform speed?
Benilton Carvalho
bcarvalh at jhsph.edu
Fri Jul 20 06:25:03 CEST 2007
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. :-)
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).
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)
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
>
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