Yes, adding "-1" or "+0" both return a lm without an intercept term.
Michael On Tue, Nov 8, 2011 at 8:52 AM, Holger Taschenberger <holger.taschenber...@mpi-bpc.mpg.de> wrote: > Hi, > > I'm trying to compare two linear regressions. I'm using the > following approach: > ################## > xx<-1:100 > df1 <- data.frame(x = xx, y = xx * 2 + 30 + rnorm(n=length(xx),sd=10), g = 1) > df2 <- data.frame(x = xx, y = xx * 4 + 9 + rnorm(n=length(xx),sd=10), g = 2) > dta <- rbind(df1, df2) > dta$g <- factor(dta$g) > plot(df2$x,df2$y,type="l",col="red") > lines(df1$x,df1$y,col="blue") > summary(lm(formula = y ~ x + g + x:g, dta)) > ################## > I learned that the coefficients (g2 and x:g2) tell me about the > differences in intercept and slope and the corresponding p-values. > > Now I'm trying to do the same except that there should be no intercept > term: > ################## > xx<-1:100 > df1 <- data.frame(x = xx, y = xx * 2 + rnorm(n=length(xx),sd=10), g = 1) > df2 <- data.frame(x = xx, y = xx * 4 + rnorm(n=length(xx),sd=10), g = 2) > dta <- rbind(df1, df2) > dta$g <- factor(dta$g) > plot(df2$x,df2$y,type="l",col="red") > lines(df1$x,df1$y,col="blue") > summary(lm(formula = y ~ x - 1 + x:g, dta)) > ################## > I assume that the last line is the correct way to specify a linear model > without intercept. But I'm not certain about that. Can someone please > confirm? > > Thanks a lot, > Holger > > ______________________________________________ > 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.
