Use Property Chains and Rolification Antoine Zimmermann’s answer is a very good start to this problem, and touches on the major point that you need to solve this sort of task: From x to each of x‘s brothers, there is a path of the form hasFather o hasFather-1. Now, that’s actually not true of just … Read more
Well, the obvious first start is to look at the classes and properties present in the data. Here is how to see what classes are being used: SELECT DISTINCT ?class WHERE { ?s a ?class . } LIMIT 25 OFFSET 0 (LIMIT and OFFSET are there for paging. It is worth getting used to these … Read more
There are two main options for retrieving the data you want. In some cases, it is possible to query the data using a publicly available SPARQL endpoint. This is probably the most convenient approach, and the one to take unless there’s some definite reason that you need the data locally. There are limitations to this … Read more
While there are some limitations in what property paths can do, depending on your exact requirements, you may be able to get what you need here. Consider this data: @prefix : <urn:ex:>. :a :relatedTo :b . :b :relatedTo :c . :c :relatedTo :d . :a :relatedTo :e . :e :relatedTo :f . :f :relatedTo :g … Read more
This is based on the same technique used to compute the position of an element in an RDF list using SPARQL that is described in: Is it possible to get the position of an element in an RDF Collection in SPARQL? If you have data like this: @prefix : <http://example.org> . :orgA :hasSuborganization :orgB, :orgC, … Read more
A Pure SPARQL 1.1 Solution I’ve extended the data to make the problem a little harder. Let’s add a duplicate element to the list, e.g., an additional :a at the end: @prefix : <http://example.org#> . :ls :list (:a :b :c :a) . Then we can use a query like this to extract each list node … Read more