Prof Brian Ripley skrev: > -0.5*(A+B) is not a contrast, which is the seat of your puzzlement. > > All you can get from y ~ x is an intercept (a column of ones) and a > single 'contrast' column for 'x'. > > If you use y ~ 0+x you can get two columns for 'x', but R does not > give you an option of what columns in the case: see the source of > contrasts(). So you would need to replace contrasts(), which I think > will be hard as model.matrix.default will look in the 'stats' > namespace. It would probably be easier to create the model matrix > yourself. > Or accept the default and do the parameter transformations yourself.
l <- lm(y~x) T <- rbind( c(-1,-.5), c(0,1)) c2 <- T%*%coef(l) V2 <- T%*%vcov(l) %*% t(T) cbind(coef=c(c2), s.e.=sqrt(diag(V2))) > On Tue, 9 Sep 2008, Kenneth Knoblauch wrote: > >> Hi, >> >> I'm trying to redefine the contrasts for a linear model. >> With a 2 level factor, x, with levels A and B, a two level >> factor outputs A and B - A from an lm fit, say >> lm(y ~ x). I would like to set the contrasts so that >> the coefficients output are -0.5 (A + B) and B - A, >> but I can't get the sign correct for the first coefficient >> (Intercept). >> >> Here is a toy example, >> >> set.seed(12161952) >> y <- rnorm(10) >> x <- factor(rep(letters[1:2], each = 5)) >> ## so A and B = >> tapply(y, x, mean) >> >> a b >> -0.7198888 0.8323837 >> >> ## and with treatment contrasts >> coef(lm(y ~ x)) ## A and B - A >> >> (Intercept) xb >> -0.7198888 1.5522724 >> >> Then, I try to redefine the contrasts >> >> ### would like contrasts: -0.5 (A + B) and B - A >> D1 <- matrix( c(-0.5, -0.5, >> -1, 1), >> 2, 2, byrow = TRUE) >> C1 <- solve(D1) >> Cnt <- C1[, -1] >> contrasts(x) <- Cnt >> coef(lm(y ~ x)) >> >> (Intercept) x1 >> 0.05624745 1.55227241 >> >> but note that the desired value is >> -0.5 * sum(tapply(y, x, mean)) >> >> [1] -0.05624745 >> >> I note that the first column of C1 is -1's not +1's >> and that working by hand, if I tamper with the model matrix >> >> mm <- model.matrix(y ~ x) >> mm[, 1] <- -1 >> >> mm >> (Intercept) x1 >> 1 -1 -0.5 >> 2 -1 -0.5 >> 3 -1 -0.5 >> 4 -1 -0.5 >> 5 -1 -0.5 >> 6 -1 0.5 >> 7 -1 0.5 >> 8 -1 0.5 >> 9 -1 0.5 >> 10 -1 0.5 >> attr(,"assign") >> [1] 0 1 >> attr(,"contrasts") >> attr(,"contrasts")$x >> [,1] >> a -0.5 >> b 0.5 >> >> solve(t(mm) %*% mm) %*% t(mm) %*% y ##Yes, I know. Use QR >> [,1] >> (Intercept) -0.05624745 >> x1 1.55227241 >> >> gives the correct sign. >> >> So, I guess my question reduces to how one would set the >> contrasts for the model.matrix to be correct >> for this to work out correctly? >> >> Thank you. >> >> Ken >> >> >> -- >> Ken Knoblauch >> Inserm U846 >> Institut Cellule Souche et Cerveau >> Département Neurosciences Intégratives >> 18 avenue du Doyen Lépine >> 69500 Bron >> France >> tel: +33 (0)4 72 91 34 77 >> fax: +33 (0)4 72 91 34 61 >> portable: +33 (0)6 84 10 64 10 >> http://www.sbri.fr >> >> ______________________________________________ >> 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. > > ------------------------------------------------------------------------ > > ______________________________________________ > 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. > -- O__ ---- Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~~~~~~~~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 ______________________________________________ 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.
