RDFLIB

[Nisarg Patel]

import rdflib

g = rdflib.Graph()

g.parse("ExampleRdf1.rdf")  #I have uploaded RDF file.

qres = g.query(

    """

    select ?i ?type where {

  ?i a/(rdfs:subClassOf)* ?type

}""")

    

for row in qres:

    print(" %s is %s" % row  )

    

##output of this query:

 http://stackoverflow.com/q/20474862/1281433/Apurva is http://stackoverflow.com/q/20474862/1281433/Man

 http://stackoverflow.com/q/20474862/1281433/Apurva is http://stackoverflow.com/q/20474862/1281433/Mortal

 http://stackoverflow.com/q/20474862/1281433/Cathy is http://stackoverflow.com/q/20474862/1281433/Girl

 http://stackoverflow.com/q/20474862/1281433/Cathy is http://stackoverflow.com/q/20474862/1281433/Mad

 

 

 I want to remove this link(http://stackoverflow.com/q/20474862/1281433) from result. its Prefix.

 i need output like this:

 

 Apurva is Man

 Apurva is Mortal

 Cathy is Girl

 Cathy is Mad.

 

 

 Can anyone help me with this?

#ExampleRdf1.rdf file.

<?xml version="1.0" encoding="utf-8" ?>

<rdf:RDF xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"

         xmlns:rdfs="http://www.w3.org/2000/01/rdf-schema#">

  <rdf:Description rdf:about="http://stackoverflow.com/q/20474862/1281433/Apurva">

    <rdf:type>

      <rdf:Description rdf:about="http://stackoverflow.com/q/20474862/1281433/Man">

        <rdfs:subClassOf rdf:resource="http://stackoverflow.com/q/20474862/1281433/Mortal"/>

      </rdf:Description>

    </rdf:type>

  </rdf:Description>

  <rdf:Description rdf:about="http://stackoverflow.com/q/20474862/1281433/Cathy">

    <rdf:type>

      <rdf:Description rdf:about="http://stackoverflow.com/q/20474862/1281433/Girl">

        <rdfs:subClassOf rdf:resource="http://stackoverflow.com/q/20474862/1281433/Mad"/>

      </rdf:Description>

    </rdf:type>

  </rdf:Description>

</rdf:RDF>

[Boris Pelakh]

Easiest way is to strip the namespaces yourself:

import re

for row in qres:
    print(" %s is %s" % tuple(re.sub(r'^.*[/#]', '', r) for r in row))

If you want to use the namespace manager, you can do this similarly:

for row in qres:
    print(" %s is %s" % tuple(g.qname(r).split(':')[1] for r in row))

[Donny Winston]

Ideally, you'd associate a rdfs:label or something with each URI resource and print the labels instead, but you can also bind a namespace and format output according to that. In your case, perhaps

```python

import rdflib

g = rdflib.Graph()

g.parse("ex.rdf")
g.bind("so", "http://stackoverflow.com/q/20474862/1281433/")

qres = g.query(
    """
    select ?i ?type where {
  ?i a/(rdfs:subClassOf)* ?type
}""")
for row in qres:

    i, type_ = [e.n3(g.namespace_manager) for e in row]
    print(f"{i} is {type_}")

```

yields

```

so:Apurva is so:Man
so:Apurva is so:Mortal
so:Cathy is so:Girl
so:Cathy is so:Mad
```

[Wes Turner]

FWIW, in Python 3.9, there are new str.removeprefix() and str.removesuffix() methods; though you'd still be doing two containment tests whereas re.sub only does only containment test (but does require an 'import re')

https://www.python.org/dev/peps/pep-0616/

https://docs.python.org/3.9/library/stdtypes.html#str.removeprefix

--
http://github.com/RDFLib
---
You received this message because you are subscribed to the Google Groups "rdflib-dev" group.
To unsubscribe from this group and stop receiving emails from it, send an email to rdflib-dev+...@googlegroups.com.
To view this discussion on the web visit https://groups.google.com/d/msgid/rdflib-dev/0b0cd799-73d5-40e2-961a-c02ca8ad896eo%40googlegroups.com.

[Nicholas Car]

Hi all,

Let’s keep these “how to” question on stackoverflow, not here in the dev mailing list.

Nick

— 

Dr Nicholas Car

Data Systems Architect

SURROUND Australia

0477 560 177

nichol...@surroundaustralia.com

On 9 Jun 2020, at 9:10 am, Wes Turner <wes.t...@gmail.com> wrote:



To view this discussion on the web visit https://groups.google.com/d/msgid/rdflib-dev/CACfEFw83MFLMYJmBWind0CuHPar5YkV0t8aR%3DUUh3m3-xKAk8A%40mail.gmail.com.