"lim" is not the argument: "ylim" is. You put in a vector of length 2 comprising the min and max y you wish: consider this:
x = -5:5; y = x^2; z = rep(-5,11); layout(1:2) plot(x,y,type="b"); lines(x,z,col=2) plot(x,y,ylim = c(-8,max(y)+3),type="b"); lines(x,z,col=2) For your work, you'd need ylim = c(0, 1.03*max(y)) or something similar. Michael Weylandt On Thu, Aug 25, 2011 at 1:12 PM, Claudio Zanettini < claudio.zanett...@gmail.com> wrote: > Yes I tried but if I set the lim to 0 then it will not displayed the line > that is at -10, right? > > 2011/8/25 R. Michael Weylandt <michael.weyla...@gmail.com> > > Look at ylim, as an optional argument to plot. >> >> Michael >> >> On Thu, Aug 25, 2011 at 1:07 PM, Claudio Zanettini < >> claudio.zanett...@gmail.com> wrote: >> >>> Hello everyone, >>> I have a graph and a segment parallel to the x axis at y=-10, x=0, and >>> bars on it. >>> Now the question is, >>> Is there a way to leave the segment there but let the graph axis start >>> from >>> the origin? >>> In this way the segment will be out of the graph >>> >>> >>> Thanks >>> >>> [[alternative HTML version deleted]] >>> >>> ______________________________________________ >>> 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. >>> >> >> > [[alternative HTML version deleted]] ______________________________________________ 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.
