That is curious, as I thought when I checked it that evalq did what I said
by default. Apparently not. However, to continue in the vein of complex
solutions for simple problems, either
explicitly using local() or specifying a local environment as an argument
evalq({a[is.na(a)]<-0;a},env=new.env())+b
does it. Either must be used to protect against argument evaluation, which
is the point of interest.
The solutions you suggested do this implicitly by creating local function
environments in which the assignment occurs of course.
This also means that my arcane explanation of evaluation is wrong: the
evaluator does not by default do the assignment in a local environment as I
stated, but follows the pointers (enclosures) -- of course. Local evaluation
must be explicitly forced.
-- Bert
> -----Original Message-----
> From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
> Sent: Tuesday, May 02, 2006 11:14 AM
> To: Berton Gunter
> Cc: John Kane; R R-help
> Subject: Re: [R] Adding elements in an array where I have
> missing data.
>
> But the evalq solution does change a.
>
> > a <- c(2, NA, 3)
> > b <- c(3,4, 5)
> > evalq({a[is.na(a)]<-0;a})+b
> [1] 5 4 8
> > a
> [1] 2 0 3
>
> If evalq were changed to local then it would not change a:
>
> > a <- c(2, NA, 3)
> > b <- c(3,4, 5)
> > local({a[is.na(a)]<-0;a})+b
> [1] 5 4 8
> > a
> [1] 2 NA 3
>
> Also the replace, ifelse and mapply solutions do not change a.
>
>
> On 5/2/06, Berton Gunter <[EMAIL PROTECTED]> wrote:
> > Below.
> >
> > > -----Original Message-----
> > > From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
> > > Sent: Tuesday, May 02, 2006 10:42 AM
> > > To: Berton Gunter
> > > Cc: John Kane; R R-help
> > > Subject: Re: [R] Adding elements in an array where I have
> > > missing data.
> > >
> > > On 5/2/06, Berton Gunter <[EMAIL PROTECTED]> wrote:
> > > > >
> > > > > Here are a few alternatives:
> > > > >
> > > > > replace(a, is.na(a), 0) + b
> > > > >
> > > > > ifelse(is.na(a), 0, a) + b
> > > > >
> > > > > mapply(sum, a, b, MoreArgs = list(na.rm = TRUE))
> > > > >
> > > >
> > > > Well, Gabor, if you want to get fancy...
> > > >
> > > > evalq({a[is.na(a)]<-0;a})+b
> > > >
> > >
> > > Note that the evalq can be omitted:
> > >
> > > { a[is.na] <- 0; a } + b
> > >
> >
> > No it can't. The idea is **not** to change the original a.
> >
> > -- Bert
> >
> >
>
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