R Legend变量替代

Question

I always desire to have my R code as flexible as possible; at present I have three (potentially more) curves to compare based on a parameter delta, but I don't want to hardcode the values of delta anywhere (or even how many values if I can avoid it).

I am trying to make a legend that involves both Greek and a variable substitution for the delta values, so each legend entry is of the form like 'delta = 0.01', where delta is Greek and 0.01 is determined by variable. Many different combinations of paste , substitute , bquote and expression have been tried, but always end up with some verbatim code leftover in the finished legend, OR fail to put 'delta' into symbolic form.

delta <- c(0.01,0.05,0.1)
plot(type="n", x=1:5, y=1:5) #the curves themselves are irrelevant
legend_text <- vector(length=length(delta)) #I don't think lists work either
for(i in 1:length(delta)){
  legend_text[i] <- substitute(paste(delta,"=",D),list(D=delta[i]) )
}
legend(x="topleft", fill=rainbow(length(delta)), legend=legend_text)

Since legend=substitute(paste(delta,"=",D),list(D=delta[1]) works for a single entry, I've also tried doing a 'semi-hardcoded' version, fixing the length of delta:

legend(x="topleft", fill=rainbow(length(delta)),
       legend=c(substitute(paste(delta,"=",A), list(A=delta[1])),
                substitute(paste(delta,"=",B), list(B=delta[2])),
                substitute(paste(delta,"=",C), list(C=delta[3])) )
      )

but this has the same issues as before.

Is there a way I can do this, or do I need to change the code by hand with each update of delta?

Answer 1

Try using lapply() with as.expression() to generate your legend labels. Also use bquote to create your individual expressions

legend_text <- as.expression(lapply(delta, function(d) {
    bquote(delta==.(d))
} ))

Note that with plotmath you need == to get an equals sign. Also no need for paste() since nothing is really a string here.