?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>) 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
>
> I?aki
>
>>
>>
>> Thanks in advance,
>>
>> Rui Barradas
>>
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