[R] creating list with 200 identical objects

[Wacek Kusnierczyk]

Wacek Kusnierczyk wrote:
>
> consider:
>
>     setClass('foo',
>        representation=representation(content='environment'))
>     setMethod('initialize', 'foo', function(.Object) {
>         .Object at content = new.env()
>         .Object at content$name = 'foo'
>         .Object })
>
>     foos = rep(list(new('foo')), 2)
>     foos[[1]]@content$name = 'bar'
>     foos[[2]]@content$name
>     # 'bar'
>
> of course, because you have just one object twice on the list, its
> content is an environment, and with environments r does not pretend to
> be functional.  but:
>
>     foos = lapply(1:2, function(x) new('foo'))
>     foos[[1]]@content$name = 'bar'
>     foos[[2]]@content$name
>     # 'foo'
>   
> of course, because you have two distinct objects, and assignment to one
> of them has no effect on the other.  similarly if 'foo' is defined as
> follows:
>
>     setClass('foo',
>        representation=representation(content='environment'))
>     setMethod('initialize', 'foo', function(.Object) {
>        .Object at content$name = 'foo'
>        .Object })
>
> or as follows:
>
>     setClass('foo',
>        representation=representation(content='environment'),
>        prototype=list(content={
>            e=new.env()
>            e$name='foo'
>            e }))
>
>   

actually, not the last one;  i got caught by that redefining 'foo' with
setClass does not remove the initializer.  so here's another situation
(best to execute in a fresh session) you should beware:

    setClass('foo',
        representation=representation(content='environment'),
        prototype=list(content={
            e=new.env()
            e$name='foo'
            e }))

    foos = lapply(1:2, function(x) new('foo'))
    foos[[1]]@content$name = 'bar'
    foos[[2]]@content$name
    # "bar"

in this case, even if you're careful enough to create two objects, they
share the representation because they both get it from the same prototype.

vQ