Re: [Rd] Why does not the command 'length(a <- 1:5) <- 4' not work?

From: Thomas Lumley <tlumley_at_u.washington.edu>
Date: Tue 16 Jan 2007 - 19:26:27 GMT

On Tue, 16 Jan 2007, Charles Dupont wrote:

> I would assume that 'length(a <- 1:5) <- 4' should work because
> 'length(a <- 1:5)' does work.

Appearances can be deceptive.

length(a <- 1:5) evaluates a <- 1:5 and then passes its value to length(), in ordinary call-by-value semantics.

That can't be how complex assignment functions work, though.

Consider

length(names(a))<-5

and suppose names(a) is c("A","B"). This code doesn't just evaluate names(a) and pass the result on, since

    length(c("A","B"))<-5
would be silly (and invalid). It must be doing something more complicated to make sure that "a", buried deep inside the call, gets updated.

What actually happens is that

   f(x) <- y
is rewritten in terms of the assignment function "f<-" as

   x <- "f<-"(x, y)
so

   length(names(a))<-5
is

   names(a) <- "length<-"(names(a), 5)
which in turn is

   a <- "names<-"(a, "length<-"(names(a), 5))

This requires the assignment function "f<-" to exist, of course.

In your case

   length(a <- 1:5) <- 4
is rewritten as

   (a<-1:5) <- "length<-"(a<-1:5, 5)
and then as

   a <- "<-<-"(a, "length<-"(a<-1:5, 5)) which gives the error you report -- there is no function "<-<-".

You might think the last stage of rewriting could just be avoided, but

   (a<-1:5) <- "length<-"(a<-1:5, 5)
is invalid if not rewritten, since it tries to modify 1:5, which is not a variable.

If you want to compute

    a <- 1:5
    length(a)<-4
then just do it like that.

         -thomas



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https://stat.ethz.ch/mailman/listinfo/r-devel Received on Wed Jan 17 06:27:57 2007

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