using a variable and a superscript in a legend

Great !? You have helped to solve a problem on which I was sweating 
(sporadically, however) since months...

Thanks,

Best,


Le 20/10/2019 ? 18:29, Bert Gunter a ?crit?:

The legend must be "an expression vector."
c("Sans renard",bquote(.(densren) (ind./km)^2))?? is not because the 
first element is a character string.

This works:

plot(1:100,1:100,type="n")
? ?legend(list(x=0,y=100),legend=c(expression("Sans 
renard"),bquote(.(densren) 
(ind./km)^2)),lty=c(1,2),col=c("black","red"),bty="n")

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 9:02 AM Patrick Giraudoux 
<patrick.giraudoux at univ-fcomte.fr 
<mailto: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
    <mailto: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]]