Firstly, thank you so much for helping me discover new knowledge (for
myself) through this beautifully crafted SPARQL. This makes a lot more
sense that you have used the DISTINCT keyword to actually eliminate any
duplicates, but will this query ensure the truth is captured in one table
in the same way as the individual two tables did, because I want to make
sure I can use this dataset to count indegrees (high influencers) properly.
It is impossible to survey all the rows to ensure the knowledge is true,
but I am asking anyway.
On Sat, Dec 21, 2013 at 7:00 PM, Joshua TAYLOR <joshuaaa...@gmail.com>wrote:
>
> On Sat, Dec 21, 2013 at 1:04 PM, Ali Gajani <aligaj...@gmail.com> wrote:
>
>> SELECT *
>> WHERE {
>> ?p a
>> <http://dbpedia.org/ontology/Person> .
>> { ?p <http://dbpedia.org/ontology/influenced> ?influenced. }
>> UNION
>> { ?p <http://dbpedia.org/ontology/influencedBy> ?influencedBy.}
>> }
>>
>
> With a query like this, I'd expect that both properties would be used to
> find a match for ?influencer and ?influenced. E.g.,
>
> select distinct ?influencer ?influencee where {
> { ?influencer dbpedia-owl:influenced ?influencee }
> UNION
> { ?influencee dbpedia-owl:influencedBy ?influencer }
> }
> limit 50
>
> Of course, this can be done even more simply with property paths:
>
> select distinct ?influencer ?influencee where {
> ?influencer dbpedia-owl:influenced|^dbpedia-owl:influencedBy ?influencee
> }
> limit 50
>
> //JT
>
>
> --
> Joshua Taylor, http://www.cs.rpi.edu/~tayloj/
>
--
Ali Gajani
Founder at Mr. Geek
www.mrgeek.me
www.aligajani.com
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