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
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