This comes from the all.vars function and would indicate
a bug in that base R function.
> f = function(a) function() paste(a, a, sep="")
> all.vars(~ fo(o)())
character(0)
On Tue, Feb 3, 2009 at 8:24 AM, Wacek Kusnierczyk
<waclaw.marcin.kusnierc...@idi.ntnu.no> wrote:
> Peter Dalgaard wrote:
>> Gabor Grothendieck wrote:
>>> On Sat, Jan 31, 2009 at 6:01 PM, Wacek Kusnierczyk
>> th some additional boring pedantry wrt. ?gsubfn, which says:
>>>>
>>>> " If 'replacement' is a formula instead of a function then a one
>>>> line function is created whose body is the right hand side of the
>>>> formula and whose arguments are the left hand side separated by
>>>> '+' signs (or any other valid operator). The environment of the
>>>> function is the environment of the formula. If the arguments are
>>>> omitted then the free variables found on the right hand side are
>>>> used in the order encountered. "
>>>>
>>>> to my little mind, all of 'paste', 'rep', 'nchar', and 'x' in the
>>>> example above are *free variables* on the right of the formula. you
>>>
>>> The first three are functions, not variables.
>>
>> They are still free variables, subject to the same rules of variable
>> lookup. Wacek is right: The RHS is scanned recursively for objects of
>> mode "name" _except_ when they appear as function names (i.e. if
>> subexpression e is mode "call", then forget e[[1]] and look at the
>> arguments in as.list(e)[-1]. Not sure if this also happens if e[[1]]
>> is not a name, e.g. in f(a)(b), do you get both a and b or just b?)
>>
>
> an interesting point. the two calls to gsubfn below should, in this
> particular case, be equivalent:
>
> library(gsubfn)
>
> f = function(a) function(b) paste(a, b, sep="")
> gsubfn('o', ~ f('o')(o), 'foo')
> # "foooo"
> gsubfn('o', ~ f(o)('o'), 'foo')
> # the match seems to be ignored in the formula?
> the following fails, too:
>
> f = function(a) function() paste(a, a, sep="")
> gsubfn('o', ~ f(o)(), 'foo')
> # o won't capture the match
>
> this as well, though it's rather different:
>
> f = function() 'oo'
> gsubfn('o', ~ f(), 'foo')
> # really can't ignore the matched pattern if a formula is given?
>
>
> while an average statistician may never write such rubbish code, these
> are trivialized examples, and for a language advertised as one from the
> functional family this sort of code is not so unusual and it may be
> surprising that it fails.
Can you clarify this. In what way was the "match ignored"?
In the first case it added an o after each o. Were you expecting
something different?
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