Thanks Uwe. I think your recipe is substantially sexier than mine.
However I will, I believe, ***NEVER*** understand how to put together
plotmath constructs. They seem to require some subset of:
* expression()
* bquote()
* substitute()
* parse()
* paste()
* as.expression()
* .(...)
(and quite possibly other things that I have forgotten) in a bewildering
variety of combinations with no perceptible rationale as what combination
is required in what circumstances.
It makes quantum mechanics look simple by comparison.
cheers,
Rolf
On 06/10/12 06:58, Uwe Ligges wrote:
On 05.10.2012 09:30, Rolf Turner wrote:
I want to do something like:
TH <- sprintf("%1.1f",c(0.3,0.5,0.7,0.9,1))
plot(1:10)
legend("bottomright",pch=1:5,legend=parse(text=paste("theta ==",TH)))
Notice that the final "1" comes out in the legend as just plain "1"
and NOT
as "1.0" although TH is
[1] "0.3" "0.5" "0.7" "0.9" "1.0"
I can get plotmath to preserve "1.0" as "1.0" and NOT convert it to "1"
if I use substitute, as in
text(2,5,labels=substitute(list(theta == a),list(a=TH[5])))
but substitute doesn't work appropriately with vectors.
Can anyone tell me how to get a "1.0" rather than "1" in the legend?
Ta.
cheers,
Rolf Turner
P.S. Just figured out a way using sapply():
leg <- sapply(TH,function(x){substitute(list(theta == a),list(a=x))})
plot(1:10)
legend("bottomright",pch=1:5,legend=parse(text=leg))
Note that the use of "parse" (pace Thomas Lumley! :-) ) is required ---
"legend=leg" does NOT work.
Getting here required an enormous amount of trial and error. And it
seems
pretty kludgy.
Is there a sexier way?
Not sure what is sexier here. I'd stay with
leg <- lapply(TH, function(x) bquote(list(theta == .(x))))
plot(1:10)
legend("bottomright", pch=1:5, legend=as.expression(leg))
Best,
Uwe Ligges
R. T.
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