[R] R versus SAS: lm performance
Arne.Muller@aventis.com
Arne.Muller at aventis.com
Mon May 10 16:36:58 CEST 2004
Hello,
A collegue of mine has compared the runtime of a linear model + anova in SAS and S+. He got the same results, but SAS took a bit more than a minute whereas S+ took 17 minutes. I've tried it in R (1.9.0) and it took 15 min. Neither machine run out of memory, and I assume that all machines have similar hardware, but the S+ and SAS machines are on windows whereas the R machine is Redhat Linux 7.2.
My question is if I'm doing something wrong (technically) calling the lm routine, or (if not), how I can optimize the call to lm or even using an alternative to lm. I'd like to run about 12,000 of these models in R (for a gene expression experiment - one model per gene, which would take far too long).
I've run the follwong code in R (and S+):
> options(contrasts=c('contr.helmert', 'contr.poly'))
The 1st colum is the value to be modeled, and the others are factors.
> names(df.gene1data) <- c("Va", "Ba", "Ti", "Do", "Ar", "Pr")
> df[c(1:2,1343:1344),]
Va Do Ti Ba Ar Pr
1 2.317804 000mM 24h NEW 1 1
2 2.495390 000mM 24h NEW 2 1
8315 2.979641 025mM 04h PRG 83 16
8415 4.505787 000mM 04h PRG 84 16
this is a dataframe with 1344 rows.
x <- Sys.time();
wlm <- lm(Va ~
Ba+Ti+Do+Pr+Ba:Ti+Ba:Do+Ba:Pr+Ti:Do+Ti:Pr+Do:Pr+Ba:Ti:Do+Ba:Ti:Pr+Ba:Do:Pr+Ti:Do:Pr+Ba:Ti:Do:Pr+(Ba:Ti:Do)/Ar, data=df, singular=T);
difftime(Sys.time(), x)
Time difference of 15.33333 mins
> anova(wlm)
Analysis of Variance Table
Response: Va
Df Sum Sq Mean Sq F value Pr(>F)
Ba 2 0.1 0.1 0.4262 0.653133
Ti 1 2.6 2.6 16.5055 5.306e-05 ***
Do 4 6.8 1.7 10.5468 2.431e-08 ***
Pr 15 5007.4 333.8 2081.8439 < 2.2e-16 ***
Ba:Ti 2 3.2 1.6 9.8510 5.904e-05 ***
Ba:Do 7 2.8 0.4 2.5054 0.014943 *
Ba:Pr 30 80.6 2.7 16.7585 < 2.2e-16 ***
Ti:Do 4 8.7 2.2 13.5982 9.537e-11 ***
Ti:Pr 15 2.4 0.2 1.0017 0.450876
Do:Pr 60 10.2 0.2 1.0594 0.358551
Ba:Ti:Do 7 1.4 0.2 1.2064 0.296415
Ba:Ti:Pr 30 5.6 0.2 1.1563 0.259184
Ba:Do:Pr 105 14.2 0.1 0.8445 0.862262
Ti:Do:Pr 60 14.8 0.2 1.5367 0.006713 **
Ba:Ti:Do:Pr 105 15.8 0.2 0.9382 0.653134
Ba:Ti:Do:Ar 56 26.4 0.5 2.9434 2.904e-11 ***
Residuals 840 134.7 0.2
The corresponding SAS program from my collegue is:
proc glm data = "the name of the data set";
class B T D A P;
model V = B T D P B*T B*D B*P T*D T*P D*P B*T*D B*T*P B*D*P T*D*P B*T*D*P A(B*T*D);
run;
Note, V = Va, B = Ba, T = Ti, D = Do, P = Pr, A = Ar of the R-example
kind regards + thanks a lot for your help,
Arne
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