[Rd] bug and fix : using panel.first with plot() in do.call() (PR#458)

[ripley@stats.ox.ac.uk]

On Fri, 25 Feb 2000 c.farrow@compserv.gla.ac.uk wrote:

> The following works as expected
>         plot(1:5, panel.first=grid(2,2))
> and if
>         my.panel<-function() grid(2,2)
> then
>         plot(1:5, panel.first=my.panel() )
> is also OK
> 
> but, 
>         do.call("plot", list(x=1:5, panel.first=grid(2,2))
>         do.call("plot", list(x=1:5, panel.first=my.panel)) 
>         do.plot("plot", list(x=1:5, panel.first=my.panel() ))
> do not draw the grid because the panel.first expression is evaluated
> before the plot gets drawn.

Yes, and a solution is

         do.call("plot", list(x=1:5, panel.first=quote(grid(2,2))))

as when you do this sort of thing you can't rely on lazy evaluation.
replacing `quote' by `delay' will also work.  The help page says
they should be unevaluated expressions.

> A solution is to modify plot.default such that the line
> 
> panel.first
> is replaced by
> 
> if(is.function(panel.first)) {
>         panel.first()
> }else {
>         panel.first
> }
> This allows the previous behaviour, but also allows a function name to
> be passed to plot using do.call()
> now all the following produce the expected graph
> plot(1:5, panel.first=grid(2,2))
> plot(1:5, panel.first=my.panel())
> plot(1:5,panel.first=my.panel)
> 
> do.call("plot", list(x=1:5, panel.first=my.panel ))
> 
> Similarly the line 
> 
> panel.last 
> should be replaced by 
> if(is.function(panel.last)) {
>         panel.last()
> } else {
> panel.last
> }

Um, ?plot.default say panel.first and panel.last should be *expressions*
*to be evaluated*, so is this not working as documented?


-- 
Brian D. Ripley,                  ripley@stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272860 (secr)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595


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