Thorn wrote:
Bert Gunter <gunter.berton <at> gene.com> writes:

Inline below.

-- No.

f <- function(x, y = x)x+y  ## lazy evaluation enables this

f(2)
[1] 4
f(2,3)
[1] 5

Ok, obviously I was not very precise about what I'd like to do. Sorry. I'm aware of this functionality. But I'd like to use the same idea for existing function. Take biplot as an example. There you have an argument called 'choices' controlling which components you'd like to plot. The standard labels of the axis are the names of the two chosen components, PC1 and PC2 say. If one wants to add the explained variance to the axis notation, one could provide the xlab argument:

biplot(pca-object, xlab=paste("PC1", explained.variance.by.pc1))

This works out well, and serves it purpose sufficiently. If it happens that you'd like to choose different components, you have to provide the choices argument. As a consequence you have to change your manually adjusted axis notation as well. Hence, I was wondering, if one could do it in a somewhat more automatic fashion:

biplot(pca-object, xlab=paste("PC", choices[1], explained.varaiance.by(choices
[1]))

I've to admit that this is an academic question, because at the time you call biplot, you know which components you want to plot. Thus, you could use the right values for choices and the labs. But I'm interested whether it can be done by telling R that it should evaluate choices[1] not at the prompt level, but at the moment it enters the function biplot, where choices will be known in any case (since it has default values).

Just to make it clear once again, I'm aware of the possibilities to get this specific task done. But I'm in particular interested, wheter it could be done using R's lazy evaluation mechanism.

I don't know if you can do that.  You'd need something like


f <- function(x, y) x+y

f(x, get("x", environment()))

but it doesn't work. The problem is you need access to the evaluation frame of f, which doesn't exist before you call the function, so you need to use lazy evaluation to find it after it has been created. I guessed that environment() would see it, but I was wrong: that is evaluated in the calling environment, so it sees that.

I'm not sure if there's some devious way to do this, but I'd certainly recommend against ever using it. To solve your original problem, just make a copy of the function and modify it.

Duncan Murdoch

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