R listers: I have been foiled by plotmath! (in R 2.01,Windows 2000) The task: Plot a normal density and label the ticks as mu - 3 sigma, mu - 2 sigma, ...., mu + 3 sigma, where the mu's and sigmas appear as Greek symbols, of course. The following code does this: x<-seq(-3,to=3,by=.01) y<-dnorm(x) plot(x,y,type='h',col='lightblue',axes=FALSE) lines(x,y,col='darkblue') axis(2) for(i in seq(-3,to=3)) axis(1,at=i, lab=switch(sign(i)+2, eval(substitute(expression(mu-j*sigma),list(j=-i))), expression(mu), eval(substitute(expression(mu+j*sigma),list(j=i))))) box() However, I think the code in the for loop is ugly and probably means that I'm doing it wrong. In particular: 1) Is there a neat way to use one axis() call and a vector (of expressions?) for the lab=argument? 2) The plotmath Help state that expressions can be used for axis labels, so I would have expected the above to work without the eval()call -- but it does not. Would someone kindly explain to me why not -- i.e., what I have misunderstood. That is, to be clear, why does the following not work: for(i in seq(-3,to=3)) axis(1,at=i, lab=switch(sign(i)+2, substitute(expression(mu-j*sigma),list(j=i)), expression(mu), substitute(expression(mu+j*sigma),list(j=i)))) Any further ideas,insights, or pointers to reference materials would also be appreciated. Many thanks. -- Bert Gunter Genentech Non-Clinical Statistics South San Francisco, CA "The business of the statistician is to catalyze the scientific learning process." - George E. P. Box
Bert, This works fine and and seems a little simpler: x<-seq(-3,to=3,by=.01) y<-dnorm(x) plot(x,y,type='h',col='lightblue',axes=FALSE) lines(x,y,col='darkblue') axis(2) ll <- expression(mu-3*sigma, mu-2*sigma, mu-sigma, mu, mu+sigma, mu+2*sigma, mu+3*sigma) axis(1, at=-3:3, lab=ll) box() Cheers, Andy __________________________________ Andy Jaworski 518-1-01 Process Laboratory 3M Corporate Research Laboratory ----- E-mail: apjaworski at mmm.com Tel: (651) 733-6092 Fax: (651) 736-3122 Berton Gunter <gunter.berton at ge ne.com> To Sent by: "'R-help'" r-help-bounces at st <r-help at stat.math.ethz.ch> at.math.ethz.ch cc Subject 03/18/2005 03:41 [R] plotmath question PM R listers: I have been foiled by plotmath! (in R 2.01,Windows 2000) The task: Plot a normal density and label the ticks as mu - 3 sigma, mu - 2 sigma, ...., mu + 3 sigma, where the mu's and sigmas appear as Greek symbols, of course. The following code does this: x<-seq(-3,to=3,by=.01) y<-dnorm(x) plot(x,y,type='h',col='lightblue',axes=FALSE) lines(x,y,col='darkblue') axis(2) for(i in seq(-3,to=3)) axis(1,at=i, lab=switch(sign(i)+2, eval(substitute(expression(mu-j*sigma),list(j=-i))), expression(mu), eval(substitute(expression(mu+j*sigma),list(j=i))))) box() However, I think the code in the for loop is ugly and probably means that I'm doing it wrong. In particular: 1) Is there a neat way to use one axis() call and a vector (of expressions?) for the lab=argument? 2) The plotmath Help state that expressions can be used for axis labels, so I would have expected the above to work without the eval()call -- but it does not. Would someone kindly explain to me why not -- i.e., what I have misunderstood. That is, to be clear, why does the following not work: for(i in seq(-3,to=3)) axis(1,at=i, lab=switch(sign(i)+2, substitute(expression(mu-j*sigma),list(j=i)), expression(mu), substitute(expression(mu+j*sigma),list(j=i)))) Any further ideas,insights, or pointers to reference materials would also be appreciated. Many thanks. -- Bert Gunter Genentech Non-Clinical Statistics South San Francisco, CA "The business of the statistician is to catalyze the scientific learning process." - George E. P. Box ______________________________________________ R-help at stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
On Friday 18 March 2005 15:41, Berton Gunter wrote:> R listers: > > I have been foiled by plotmath! > > (in R 2.01,Windows 2000) > > The task: Plot a normal density and label the ticks as mu - 3 sigma, mu - 2 > sigma, ...., mu + 3 sigma, where the mu's and sigmas appear as Greek > symbols, of course. > > The following code does this: > > x<-seq(-3,to=3,by=.01) > y<-dnorm(x) > plot(x,y,type='h',col='lightblue',axes=FALSE) > lines(x,y,col='darkblue') > axis(2) > for(i in seq(-3,to=3)) > axis(1,at=i, lab=switch(sign(i)+2, > eval(substitute(expression(mu-j*sigma),list(j=-i))), > expression(mu), > eval(substitute(expression(mu+j*sigma),list(j=i))))) > box() > > However, I think the code in the for loop is ugly and probably means that > I'm doing it wrong. In particular: > > 1) Is there a neat way to use one axis() call and a vector (of > expressions?) for the lab=argument?Yes, expression objects can be vectors. e.g.: ## use switch as above for better formatting lab = do.call(expression, lapply(-3:3, function(i) { bquote(mu + .(i) * sigma) } )) axis(1, at = -3:3, lab = lab)> 2) The plotmath Help state that expressions can be used for axis labels, so > I would have expected the above to work without the eval()call -- but it > does not. Would someone kindly explain to me why not -- i.e., what I have > misunderstood. That is, to be clear, why does the following not work: > > for(i in seq(-3,to=3)) > axis(1,at=i, lab=switch(sign(i)+2, > substitute(expression(mu-j*sigma),list(j=i)), > expression(mu), > substitute(expression(mu+j*sigma),list(j=i))))> is.expression(substitute(expression(mu-j*sigma),list(j=1)))[1] FALSE> is.expression(eval(substitute(expression(mu-j*sigma),list(j=1))))[1] TRUE ?substitute says Value: The 'mode' of the result is generally '"call"' ... which evidently have to be evaluated. Hth, Deepayan
On Fri, 18 Mar 2005, Berton Gunter wrote:> > 2) The plotmath Help state that expressions can be used for axis labels, so > I would have expected the above to work without the eval()call -- but it > does not. Would someone kindly explain to me why not -- i.e., what I have > misunderstood. That is, to be clear, why does the following not work:Because substitute() doesn't evaluate its argument: the result is not an expression but a call to the expression() function. An example where it is clearer what is going on substitute(log(x),list(x=1)) doesn't return a number, even though log() returns a number. It returns a call to log() that still has to be evaluated -thomas> for(i in seq(-3,to=3)) > axis(1,at=i, lab=switch(sign(i)+2, > substitute(expression(mu-j*sigma),list(j=i)), > expression(mu), > substitute(expression(mu+j*sigma),list(j=i)))) >