How to combine do.call() plot() and expression()

Question

I am getting an error when I try and combine using expression with do.call and plot.

 x <- 1:10
 y <- x^1.5

I can get the plot I want by using only the plot function:

plot(y~x,xlab=expression(paste("Concentration (",mu,"M)")))

However, I would like to implement my plot using do.call. I have a really long list of parameters stored as a list, p. However, when I try and pass the list to do.call I get the following error:

p <- list(xlab=expression(paste("Concentration (",mu,"M)")))
do.call(plot,c(y~x,p))
Error in paste("Concentration (", mu, "M)") : 
  object 'mu' not found

I also tried defining the formula explicitly in the args passed to do.call. ie. do.call(plot,c(formula=y~x,p)). I do not understand why I am getting the error - especially because the following does not give an error:

do.call(plot,c(0,p))

(and gives the desired mu character in the xaxis).

+1 for your interesting observation about the behavior of `do.call(plot,c(0,p))`. — Josh O'Brien, Aug 16 '13 at 21:06
possible duplicate of [plot() and do.call(): How to pass expressions to plot title when '...' is used otherwise?](http://stackoverflow.com/questions/13982856/plot-and-do-call-how-to-pass-expressions-to-plot-title-when-is-used-o) — Metrics, Aug 16 '13 at 22:39

jdharrison · Answer 1 · 2013-08-17T00:59:39.770

14

You can use alist rather then list

p <- alist(xlab=expression(paste("Concentration (",mu,"M)")))
do.call(plot,c(y~x,p))

edited Aug 17 '13 at 00:59

answered Aug 16 '13 at 19:45

jdharrison

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Aaron left Stack Overflow · Accepted Answer · 2013-08-16T21:26:09.593

12

do.call evaluates the parameters before running the function; try wrapping the expression in quote:

p <- list(xlab=quote(expression(paste("Concentration (",mu,"M)"))))
do.call("plot", c(y~x, p))

edited Aug 16 '13 at 21:26

answered Aug 16 '13 at 19:46

Aaron left Stack Overflow

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1

Thanks! Do you know why `do.call(plot,c(0,p))` does not give an error? – dayne Aug 16 '13 at 19:49
1

Interesting find! Looks like my guess why it didn't work wasn't quite correct. One is using `graphics:::plot.default` and the other is using `graphics:::plot.formula`. I see that `graphics:::plot.formula` is doing some funky stuff with the `xlab` parameter using `enquote`; I'd have to look closer to understand why that causes it to fail. – Aaron left Stack Overflow Aug 16 '13 at 20:14
2

You should both stop using `sep` arguments in plotmath-`paste`. At best they are ignored, and at worst they get put at the end of the expression and not where you wanted them. – IRTFM Aug 16 '13 at 21:03
Nice catch @Dwin, `sep` removed. – Aaron left Stack Overflow Aug 16 '13 at 21:26
@DWin I also removed the `sep`. **BUT** I do not understand why it does not place the default space into the expression without the `sep` argument. – dayne Aug 16 '13 at 21:30
What "default space"? Spaces do not get interpreted as anything in R expressions. – IRTFM Aug 16 '13 at 21:54
Look at the documentation on `?plotmath`. There is no `sep` argument to that function. Try putting in sep="+" and see what happens. – IRTFM Aug 16 '13 at 22:06
1

@dayne Basically, as @DWin is pointing out, when it's inside of an `expression` object being interpreted by plotmath, `paste` does not refer at all to the familiar `base::paste`. Might be easier to grok this if you compare with something like `x[i]` which also has a very different meaning inside and outside of a plotmath call. – Josh O'Brien Aug 17 '13 at 00:05

score 7 · Answer 3 · answered Aug 16 '13 at 20:13

Setting quote=TRUE also works. It in effect prevents do.call() from evaluating the elements of args before it passes them to the function given by what.

x <- 1:10
y <- x^1.5
p <- list(xlab=expression(paste("Concentration (",mu,"M)",sep="")))

do.call(what = "plot", args = c(y ~ x, p), quote = TRUE)

How to combine do.call() plot() and expression()

3 Answers3

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