Thank you so much for the help! I got it down to 1sec per run. -Chris
On Fri, Jul 29, 2011 at 1:12 PM, "Dénes TÓTH" <[email protected]> wrote: > > Hi, > > you can solve the task by simple matrix algebra. > Note that I find it really inconvenient to have a matrix with variables in > rows and cases in columns, so I transposed your predictors matrix. > > # your data > regress.y = rnorm(150) > predictors = matrix(rnorm(6000*150), ncol=150, nrow=6000) > tpreds <- t(predictors) > > # compute coefficients > coefs <- apply(tpreds,2,function(x) > solve(crossprod(x),crossprod(x,regress.y))) > > # compute residuals > resids <- regress.y - sweep(tpreds,2,coefs,"*") > > (Note that resids shall be transposed back if you really insist on your > original matrix format.) > > HTH, > Denes > > >> Hi, everyone. >> I need to run lm with the same response vector but with varying predictor >> vectors. (i.e. 1 response vector on each individual 6,000 predictor >> vectors) >> After looking through the R archive, I found roughly 3 methods that has >> been suggested. >> Unfortunately, I need to run this task multiple times(~ 5,000 times) and >> would like to find a faster way than the existing methods. >> All three methods I have bellow run on my machine timed with system.time >> 13~20 seconds. >> >> The opposite direction of 6,000 response vectors and 1 predictor vectors, >> that is supported with lm runs really fast ~0.5 seconds. >> They are pretty much performing the same number of lm fits, so I was >> wondering if there was a faster way, before I try to code this up in c++. >> >> thanks!! >> >> ## sample data ### >> regress.y = rnorm(150) >> predictors = matrix(rnorm(6000*150), ncol=150, nrow=6000) >> >> ## method 1 ## >> data.residuals = t(apply(predictors, 1, function(x)( lm(regress.y ~ -1 + >> as.vector(x))$residuals))) >> >> user system elapsed >> 15.076 0.048 15.128 >> >> ## method 2 ## >> data.residuals = matrix(rep(0, nrow(predictors) * ncol(predictors)), >> nrow=nrow(predictors), ncol=ncol(predictors) ) >> >> for( i in 1:nrow(predictors)){ >> pred = as.vector(predictors[i,]) >> data.residuals[i, ] = lm(regress.y ~ -1 + pred )$residuals >> } >> >> user system elapsed >> 13.841 0.012 13.854 >> >> ## method 3 ## >> library(nlme) >> >> all.data <- data.frame( y=rep(regress.y, nrow(predictors)), >> x=c(t(predictors)), g=gl(nrow(predictors), ncol(predictors)) ) >> all.fits <- lmList( y ~ x | g, data=all.data) >> data.residuals = matrix( residuals(all.fits), nrow=nrow(predictors), >> ncol=ncol(predictors)) >> >> user system elapsed >> 36.407 0.484 36.892 >> >> >> ## the opposite direction, supported by lm ### >> lm(t(predictors) ~ -1 + regress.y)$residuals >> >> user system elapsed >> 0.500 0.120 0.613 >> >> ______________________________________________ >> [email protected] 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. >> > > ______________________________________________ [email protected] 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.
