[Rd] Is this a bug in `[`?
Kenny Bell
kmbell56 @ending from gm@il@com
Sun Aug 5 06:26:54 CEST 2018
This should more clearly illustrate the issue:
c(1, 2, 3, 4)[-seq_len(4)]
#> numeric(0)
c(1, 2, 3, 4)[-seq_len(3)]
#> [1] 4
c(1, 2, 3, 4)[-seq_len(2)]
#> [1] 3 4
c(1, 2, 3, 4)[-seq_len(1)]
#> [1] 2 3 4
c(1, 2, 3, 4)[-seq_len(0)]
#> numeric(0)
Created on 2018-08-05 by the reprex package (v0.2.0.9000).
On Sun, Aug 5, 2018 at 3:58 AM Rui Barradas <ruipbarradas using sapo.pt> wrote:
>
>
> Às 15:51 de 04/08/2018, Iñaki Úcar escreveu:
> > El sáb., 4 ago. 2018 a las 15:32, Rui Barradas
> > (<ruipbarradas using sapo.pt>) escribió:
> >>
> >> Hello,
> >>
> >> Maybe I am not understanding how negative indexing works but
> >>
> >> 1) This is right.
> >>
> >> (1:10)[-1]
> >> #[1] 2 3 4 5 6 7 8 9 10
> >>
> >> 2) Are these right? They are at least surprising to me.
> >>
> >> (1:10)[-0]
> >> #integer(0)
> >>
> >> (1:10)[-seq_len(0)]
> >> #integer(0)
> >>
> >>
> >> It was the last example that made me ask, seq_len(0) whould avoid an
> >> if/else or something similar.
> >
> > I think it's ok, because there is no negative zero integer, so -0 is 0.
>
> Ok, this makes sense, I should have thought about that.
>
> >
> > 1.0/-0L # Inf
> > 1.0/-0.0 # - Inf
> >
> > And the same can be said for integer(0), which is the result of
> > seq_len(0): there is no negative empty integer.
>
> I'm not completely convinced about this one, though.
> I would expect -seq_len(n) to remove the first n elements from the
> vector, therefore, when n == 0, it would remove none.
>
> And integer(0) is not the same as 0.
>
> (1:10)[-0] == (1:10)[0] == integer(0) # empty
>
> (1:10)[-seq_len(0)] == (1:10)[-integer(0)]
>
>
> And I have just reminded myself to run
>
> identical(-integer(0), integer(0))
>
> It returns TRUE so my intuition is wrong, R is right.
> End of story.
>
> Thanks for the help,
>
> Rui Barradas
>
> >
> > Iñaki
> >
> >>
> >>
> >> Thanks in advance,
> >>
> >> Rui Barradas
> >>
> >> ______________________________________________
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>
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