Look at: ?update
For example: lm.obj <- lm (y ~ x1 + ... + x300) lm.obj1 <- update(lm.obj, . ~ . - x1) lm.obj2 <- update(lm.obj1, . ~ . - x2) Ravi. ____________________________________________________________________ Ravi Varadhan, Ph.D. Assistant Professor, Division of Geriatric Medicine and Gerontology School of Medicine Johns Hopkins University Ph. (410) 502-2619 email: rvarad...@jhmi.edu ----- Original Message ----- From: ph84 <masterods...@gmx.de> Date: Thursday, March 12, 2009 3:28 pm Subject: [R] stats lm() function To: r-help@r-project.org > Hi, > > Im using the lm() function where the formula is quite big (300 arguments) > and the data is a frame of 3000 values. > > This is running in a loop where in each step the formula is reduced > by one > argument, and the lm command is called again (to check which > arguments are > useful) . > > This takes 1-2 minutes. > Is there a way to speed this up? > i checked the code of the lm function and its seems that its > preparing the > data and then calls lm.Fit(). i thought about just doing this praparing > stuff first and only call lm.fit() 300 times. > > > -- > View this message in context: > Sent from the R help mailing list archive at Nabble.com. > > ______________________________________________ > R-help@r-project.org mailing list > > PLEASE do read the posting guide > and provide commented, minimal, self-contained, reproducible code. ______________________________________________ R-help@r-project.org 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.
