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 at sapo.pt
<mailto:ruipbarradas at sapo.pt>> wrote:
?s 15:51 de 04/08/2018, I?aki ?car escreveu:
> El s?b., 4 ago. 2018 a las 15:32, Rui Barradas
> (<ruipbarradas at sapo.pt <mailto:ruipbarradas at sapo.pt>>) escribi?:
>>
>> 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