Patrick Giraudoux
2019-Oct-20 18:01 UTC
[R] using a variable and a superscript in a legend
Now, we have two solutions working. This is great since I did not find any example of the kind searching r-help archives and google... Thanks ! Le 20/10/2019 ? 19:31, Peter Dalgaard a ?crit?:> It's tricky, but I think what you want is > > legend(list(x=0,y=100), > legend=as.expression(list( > "Sans renard", > bquote(.(densren) * " ind."/"km"^2) > )), > lty=c(1,2),col=c("black","red"),bty="n") > > Generally, if you want a vector of unevaluated expressions, you need an object of mode "expression", but you cannot create it directly with expression() because then the bquote() is left unevaluated: > >> expression("Sans renard",bquote(.(densren) * " ind."/"km"^2)) > expression("Sans renard", bquote(.(densren) * " ind."/"km"^2)) > > Putting the bquote on the outside _looks_ like it might work: > >> bquote(expression("Sans renard",.(densren) * " ind."/"km"^2)) > expression("Sans renard", 1.25 * " ind."/"km"^2) > > but that is not an "expression" object, but a call to expression() (!). Try it and see. > > Evaluating the call does actually work (notice that the printed value is exactly the same, but the object is not): > >> eval(bquote(expression("Sans renard",.(densren) * " ind."/"km"^2))) > expression("Sans renard", 1.25 * " ind."/"km"^2) > > but I think I prefer the as.expression(list(....)) construction. > > An alternative tack is this: > >> e <- expression(0,0) >> e[[1]] <- "sans renard" >> e[[2]] <- bquote(.(densren) * " ind."/"km"^2) >> plot(1:100,1:100,type="n") >> legend(list(x=0,y=100),legend=e, lty=c(1,2),col=c("black","red"),bty="n") > > >> On 20 Oct 2019, at 18:02 , Patrick Giraudoux <patrick.giraudoux at univ-fcomte.fr> wrote: >> >> Thanks Bert and Peter, >> >> Yes Bert, I was aware of the legend() function syntax, and just quoting the legend argument within the function. >> >> However, Bert and Peter, I do not understand why it works with your absolutely reproducible examples and not in the slightly (not so slightly apparently) different context where I used it... >> >> densren=1.25 >> plot(1:100,1:100,type="n") >> legend(list(x=0,y=100),legend=c("Sans renard",bquote(.(densren) (ind./km)^2)),lty=c(1,2),col=c("black","red"),bty="n") >> >> densren=1.25 >> plot(1:100,1:100,type="n") >> legend(list(x=0,y=100),legend=c("Sans renard",bquote(.(densren) * " ind."/"km"^2)),lty=c(1,2),col=c("black","red"),bty="n" >> >> Probably because the result of bquote() is concatenated in a character vector, but how to deal with this ? >> >> Best, >> >> Patrick >> >> >> >> Le 20/10/2019 ? 16:42, Bert Gunter a ?crit : >>> Assuming you are using base graphics, your syntax for adding the legend appears to be wrong. >>> legend() is a separate function, not a parameter of plot.default afaics. >>> >>> The following works for me: >>> >>>> densren <- 1.25 >>>> plot(1:10) >>>> legend (x="center", legend =bquote(.(densren) (ind./km)^2)) >>> See ?legend >>> >>> Bert Gunter >>> >>> "The trouble with having an open mind is that people keep coming along and sticking things into it." >>> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) >>> >>> >>> On Sun, Oct 20, 2019 at 5:30 AM Patrick Giraudoux <patrick.giraudoux at univ-fcomte.fr> wrote: >>> Dear listers, >>> >>> I am trying to pass an expression inlcuding a variable and a >>> superpscript to a legend. What I want to obtain is e.g. with densren = 1.25 >>> >>> 1.25 ind./km^2 >>> >>> I have tried many variants of the following: >>> >>> legend=bquote(.(densren) (ind./km)^2) >>> >>> but if not errors, do obtain >>> >>> 1.25 (ind./km^2) >>> >>> hence not what I want (no parenthesis, 2 in superscript...) >>> >>> Any idea about a correct syntax to get what I need ? >>> >>> Best, >>> >>> Patrick >>> >>> >>> [[alternative HTML version deleted]] >>> >>> ______________________________________________ >>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see >>> 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. >>
To continue down this rabbit hole ... Actually, both solutions are the same; Peter's is just more general than mine, as it works more conveniently for more lines in the legend. However, note that:> class(c("Sans renard", bquote(.(densren) (ind./km)^2)))[1] "list" # by coercion so it does not seem necessary to explicitly call list(). That is: plot(1:100,1:100,type="n") legend(list(x=0,y=100), legend = as.expression(c("Sans renard", bquote(.(densren) (ind./km)^2))),lty=c(1,2),col=c("black","red"),bty="n") appears to suffice. I would appreciate correction if I'm wrong about this. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Sun, Oct 20, 2019 at 11:01 AM Patrick Giraudoux < patrick.giraudoux at univ-fcomte.fr> wrote:> Now, we have two solutions working. This is great since I did not find > any example of the kind searching r-help archives and google... > Thanks ! > > Le 20/10/2019 ? 19:31, Peter Dalgaard a ?crit : > > It's tricky, but I think what you want is > > > > legend(list(x=0,y=100), > > legend=as.expression(list( > > "Sans renard", > > bquote(.(densren) * " ind."/"km"^2) > > )), > > lty=c(1,2),col=c("black","red"),bty="n") > > > > Generally, if you want a vector of unevaluated expressions, you need an > object of mode "expression", but you cannot create it directly with > expression() because then the bquote() is left unevaluated: > > > >> expression("Sans renard",bquote(.(densren) * " ind."/"km"^2)) > > expression("Sans renard", bquote(.(densren) * " ind."/"km"^2)) > > > > Putting the bquote on the outside _looks_ like it might work: > > > >> bquote(expression("Sans renard",.(densren) * " ind."/"km"^2)) > > expression("Sans renard", 1.25 * " ind."/"km"^2) > > > > but that is not an "expression" object, but a call to expression() (!). > Try it and see. > > > > Evaluating the call does actually work (notice that the printed value is > exactly the same, but the object is not): > > > >> eval(bquote(expression("Sans renard",.(densren) * " ind."/"km"^2))) > > expression("Sans renard", 1.25 * " ind."/"km"^2) > > > > but I think I prefer the as.expression(list(....)) construction. > > > > An alternative tack is this: > > > >> e <- expression(0,0) > >> e[[1]] <- "sans renard" > >> e[[2]] <- bquote(.(densren) * " ind."/"km"^2) > >> plot(1:100,1:100,type="n") > >> legend(list(x=0,y=100),legend=e, > lty=c(1,2),col=c("black","red"),bty="n") > > > > > >> On 20 Oct 2019, at 18:02 , Patrick Giraudoux < > patrick.giraudoux at univ-fcomte.fr> wrote: > >> > >> Thanks Bert and Peter, > >> > >> Yes Bert, I was aware of the legend() function syntax, and just quoting > the legend argument within the function. > >> > >> However, Bert and Peter, I do not understand why it works with your > absolutely reproducible examples and not in the slightly (not so slightly > apparently) different context where I used it... > >> > >> densren=1.25 > >> plot(1:100,1:100,type="n") > >> legend(list(x=0,y=100),legend=c("Sans renard",bquote(.(densren) > (ind./km)^2)),lty=c(1,2),col=c("black","red"),bty="n") > >> > >> densren=1.25 > >> plot(1:100,1:100,type="n") > >> legend(list(x=0,y=100),legend=c("Sans renard",bquote(.(densren) * " > ind."/"km"^2)),lty=c(1,2),col=c("black","red"),bty="n" > >> > >> Probably because the result of bquote() is concatenated in a character > vector, but how to deal with this ? > >> > >> Best, > >> > >> Patrick > >> > >> > >> > >> Le 20/10/2019 ? 16:42, Bert Gunter a ?crit : > >>> Assuming you are using base graphics, your syntax for adding the > legend appears to be wrong. > >>> legend() is a separate function, not a parameter of plot.default > afaics. > >>> > >>> The following works for me: > >>> > >>>> densren <- 1.25 > >>>> plot(1:10) > >>>> legend (x="center", legend =bquote(.(densren) (ind./km)^2)) > >>> See ?legend > >>> > >>> Bert Gunter > >>> > >>> "The trouble with having an open mind is that people keep coming along > and sticking things into it." > >>> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) > >>> > >>> > >>> On Sun, Oct 20, 2019 at 5:30 AM Patrick Giraudoux < > patrick.giraudoux at univ-fcomte.fr> wrote: > >>> Dear listers, > >>> > >>> I am trying to pass an expression inlcuding a variable and a > >>> superpscript to a legend. What I want to obtain is e.g. with densren > 1.25 > >>> > >>> 1.25 ind./km^2 > >>> > >>> I have tried many variants of the following: > >>> > >>> legend=bquote(.(densren) (ind./km)^2) > >>> > >>> but if not errors, do obtain > >>> > >>> 1.25 (ind./km^2) > >>> > >>> hence not what I want (no parenthesis, 2 in superscript...) > >>> > >>> Any idea about a correct syntax to get what I need ? > >>> > >>> Best, > >>> > >>> Patrick > >>> > >>> > >>> [[alternative HTML version deleted]] > >>> > >>> ______________________________________________ > >>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see > >>> 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]]
You're right. I was worried that c() would create a character vector and deparse the unevaluated call in the process, but apparently it is an implicit as.character _inside_ legend that is doing us in. (I can't offhand see where it is happening, but there might be scope for improvement if legend() would just accept a list object and treat the elements separately). -pd> On 20 Oct 2019, at 20:28 , Bert Gunter <bgunter.4567 at gmail.com> wrote: > > However, note that: > > > class(c("Sans renard", bquote(.(densren) (ind./km)^2))) > [1] "list" # by coercion > > so it does not seem necessary to explicitly call list(). That is: > > plot(1:100,1:100,type="n") > legend(list(x=0,y=100), legend = as.expression(c("Sans renard", bquote(.(densren) (ind./km)^2))),lty=c(1,2),col=c("black","red"),bty="n") > > appears to suffice. I would appreciate correction if I'm wrong about this. >-- Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Office: A 4.23 Email: pd.mes at cbs.dk Priv: PDalgd at gmail.com