Try this: dd$i <- with(dd,interaction(z,x>0)) contrasts(dd$i,2) <- rbind( c(0,0),c(0,0),c(1,0),c(0,1) ) model.matrix( ~i, dd )[,-1]
-- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [email protected] 801.408.8111 > -----Original Message----- > From: [email protected] [mailto:r-help-boun...@r- > project.org] On Behalf Of Ben Bolker > Sent: Monday, November 02, 2009 8:41 PM > To: [email protected] > Subject: [R] design matrix construction question > > > with the following simple data frame > dd = structure(list(z = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L > ), .Label = c("a", "b"), class = "factor"), x = c(0.3, 0.2, 0.1, > 0, 0, 0, 0.2, 0.3)), .Names = c("z", "x"), row.names = c(NA, > -8L), class = "data.frame") > > I would like know if it's possible to use model.matrix() > to construct the following design matrix in some sensible way: > > za zb > 1 1 0 > 2 1 0 > 3 1 0 > 4 0 0 > 5 0 0 > 6 0 0 > 7 0 1 > 8 0 1 > > In other words, I want column 1 to be (z=="a" & x>0) and > column 2 to be (z=="b" & x>0). I can construct this matrix > using > > sweep(model.matrix(~z-1,dd),1,dd$x>0,"*") > > and then stick it into lm.fit -- but is there a more > elegant way to do this in general? I haven't found a formula combining > (z-1) and I(x>0) that works, although I can imagine there is one. > > thanks > Ben Bolker > > > -- > Ben Bolker > Associate professor, Biology Dep't, Univ. of Florida > [email protected] / www.zoology.ufl.edu/bolker > GPG key: www.zoology.ufl.edu/bolker/benbolker-publickey.asc ______________________________________________ [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.
