Constructing titles from list of expressions

Hello!

I have the following problem:
I have a function to construct three surfaceplots with a marker for an optimum,
each of the plots has as title paste("Estimated ",pred.var.lab," for
",var.lab[1]," vs. ",var.lab[2],sep="") with different var.lab[1,2] each time.
My problem is now that I need to allow for plotmath expressions in the
variables pred.var.lab and var.lab[1,2] and get these evaluated correctly.

I tested the usage of parse(), paste(), substitute() in different combinations,
but could not find a version, which worked. The problem seems to be the order
of evaluation...

Let's work on this from the inside out.

Suppose you want the label "Estimated alpha for beta vs gamma^3" but with
alpha, beta, and gamma^3 replaced by the appropriate expressions.

    plot(1:10,main=quote("Estimated "*alpha*" for "*beta*" vs."*gamma^3))

Now, suppose you might want to change the expressions.  You need to get
them into the formula. This is what substitute() is good for.

  substitute("Estimated "*a*" for "*b*" vs."*c,
	list(a=quote(alpha),b=quote(beta),c=quote(gamma^3)))
gives
  "Estimated " * alpha * " for " * beta * " vs." * gamma^3

so the plot command can be
 plot(1:10,main=substitute("Estimated "*a*" for "*b*" vs."*c,
        list(a=quote(alpha),b=quote(beta),c=quote(gamma^3))))

Now in fact you have the expressions a, b and c in variables

expr1<-quote(alpha)
expr2<-quote(beta)
expr3<-quote(gamma^3)
plot(1:10,main=substitute("Estimated "*a*" for "*b*" vs."*c,
	list(a=expr1,b=expr2,c=expr3)))

In your example you wanted b and c to be elements of a vector.  They
actually have to be elements of a list

c(quote(beta),quote(gamma^3))

[[1]]
beta

[[2]]
gamma^3

So
 expr1<-quote(alpha)
 expr2<-c( quote(beta), quote(gamma^3))
 plot(1:10,main=substitute("Estimated "*a*" for "*b*" vs."*c,
        list(a=expr1,b=expr2[[1]],c=expr2[[2]])))

Finally, you could even make the expressions from text strings if you want
to

str1<-"alpha"
str2<-c("beta","gamma^3")

expr1<-parse(text=str1)[[1]]
expr2<-parse(text=str2)
 plot(1:10,main=substitute("Estimated "*a*" for "*b*" vs."*c,
        list(a=expr1,b=expr2[[1]],c=expr2[[2]])))


Finally, what's the difference between using quote() as I did and
expression() as you did? Not a lot.  quote() produces a call and
expression() produces an expression (roughly a list of calls). In this
context it doesn't matter. The documentation for plotmath() talks about
expressions, but most of the examples use substitute() in a way that
produces a call.

 	-thomas



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