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.giraud...@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 à 16:42, Bert Gunter a écrit :
>> 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.giraud...@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@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.
>
>
--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Office: A 4.23
Email: pd....@cbs.dk Priv: pda...@gmail.com
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