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
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