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Problem To Transform An Owl File Related To Cell Line Ontology Using An Xslt

I am trying to parse the owl file called "CLO_v2.0.27.owl" available at the bottom of this webpage http://bioportal.bioontology.org/ontologies/45376.

I want to extract name of cell lines that are subClassOf CLO_0000019 and that are embedded in this kind of element:

<owl:Class rdf:about="&obo;CLO_0008089">
    <rdfs:label>NCI-H460</rdfs:label>
    <rdfs:subClassOf rdf:resource="&obo;CLO_0000019"/>
    <rdfs:seeAlso>ATCC: HTB-177</rdfs:seeAlso>
</owl:Class>

In order to do that I have this XSLT : 

<xsl:stylesheet version="1.0" xmlns:xsl="&lt;a href=" http:="" www.w3.org="" 1999="" XSL="" Transform"="" rel="nofollow">http://www.w3.org/1999/XSL/Transform" 
xmlns:owl="http://www.w3.org/2002/07/owl#" 
xmlns:rdfs="http://www.w3.org/2000/01/rdf-schema#" 
xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"
xmlns:dc="http://purl.org/dc/elements/1.1/">

<xsl:output method="text"/>
<xsl:template match="/">
    <xsl:apply-templates select="//owl:Class[rdfs:subClassOf[@rdf:resource = '*CLO_0000019']]"/>
</xsl:template>

<xsl:template match="owl:Class">
    <xsl:value-of select="./rdfs:label"/>
        <xsl:if test="not(position()=last())">
            <xsl:text>
</xsl:text>
        </xsl:if>
</xsl:template>

</xsl:stylesheet>

But unfortunatelly my textfile is empty. And if I change a single line (see below) I get a text file filled out but with items I am not interested in.

<xsl:apply-templates select="//owl:Class[rdfs:subClassOf[@rdf:resource]]" />

So if there is XSLT masters that could correct my newbie mistake, I would appreciate very much.

I am using the evaluation version the software XMLspy

xml
• 4,481 views • link
updated by Egon Willighagen Egon Willighagen • written by Fred Fleche Fred Fleche

BTW: the scope of rdfs:seeAlso is a resource, not a literal. So (I know that's not your fault) but the statement 'rdfs:seeAlso' in your example seems wrong to me.

• 0 views • link
written by Pierre Lindenbaum Pierre Lindenbaum

3 answers

Apologies for not being able to resist to give the SPARQL version, as done in R with rrdf...

library(rrdf)

clo = load.rdf(file="CLO.owl")
# never mind the warnings
summarize.rdf(clo)

sparql = "
  PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
  PREFIX obo: <http://purl.obolibrary.org/obo/>

  SELECT ?line ?label WHERE {
    ?line rdfs:subClassOf obo:CLO_0000019 ;
          rdfs:label ?label .
  }
"

lines = sparql.rdf(clo, sparql)

Output looks like:

> lines[1:3,]
      line              label         
 [1,] "obo:CLO_0002076" "C1.18.4"     
 [2,] "obo:CLO_0006202" "IGF104/90"   
 [3,] "obo:CLO_0002345" "CDR2"
• 0 views • link
written by Egon Willighagen Egon Willighagen

Egon, your answer is very welcome. Thanks a lot.

• 0 views • link
written by Fred Fleche Fred Fleche

Is it what you need ? I'm not sure about what you want for your output. Your select statement was wrong and AFAIK, there is no wildcards in XSLT.

<xsl:stylesheet version="1.0" xmlns:xsl="&lt;a href=" http:="" www.w3.org="" 1999="" XSL="" Transform"="" rel="nofollow">http://www.w3.org/1999/XSL/Transform" 
    xmlns:owl="http://www.w3.org/2002/07/owl#" 
    xmlns:rdfs="http://www.w3.org/2000/01/rdf-schema#" 
    xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"
    xmlns:dc="http://purl.org/dc/elements/1.1/">

<xsl:output method="text"/>
<xsl:template match="/">
    <xsl:for-each select="//owl:Class[rdfs:subClassOf/@rdf:resource='&lt;a href=" http:="" purl.obolibrary.org="" obo="" CLO_0000019'"="" rel="nofollow">http://purl.obolibrary.org/obo/CLO_0000019']">
        <xsl:value-of select="@rdf:about"/>
        <xsl:text>  </xsl:text>
        <xsl:value-of select="rdfs:label"/>
        <xsl:text>  </xsl:text>
        <xsl:value-of select="rdfs:seeAlso"/>
        <xsl:text>
</xsl:text>
    </xsl:for-each>
</xsl:template>

</xsl:stylesheet>
• 0 views • link
written by Pierre Lindenbaum Pierre Lindenbaum

I am wondering why you prefer to use XSLT to get the subclasses? Wouldn't it be easier and more reliable to load the file and query it (using Jena or the OWLAPI), or even try and see if the resource is available via a SPARQL endpoint? This would for example allow you to load resources imported by the CLO.

• 0 views • link
written by Melanie Courtot • Melanie Courtot

Hello Melanie. I use XSLT because it is very convenient to use it. Even if I am a newbie and I got a bug. By the way since you submited a comment and not an answer could you please cut/paste it in the comment section just below my answer. Thanks in advance.

• 0 views • link
written by Fred Fleche Fred Fleche

Mélanie, I do understand that SPARQL is the solution of choice to handle a RDF graph. But here the statements were all identically structured , so I think XSLT is a solution of choice here, especially to transform the XML into something else.

• 0 views • link
written by Pierre Lindenbaum Pierre Lindenbaum

Well my answer was you should try and avoid XSLT, as this will most probably get you in trouble in the long run :) For example, you are using [?] as a template, but it would be perfectly valid OWL to write [?] [?] [?]

The CLO developers use Protege to edit their file, as AFAIK there is no way to predict which serialization will happen.

• 0 views • link
written by Melanie Courtot • Melanie Courtot

You may also want to consider that CLO_0000019 is a defined class, which means a reasoner should be run before trying to get subclasses if one wishes to get asserted and inferred subclasses. Using XSLT is your choice of course, I was just trying to point to some limitations and other options available.

• 0 views • link
written by Melanie Courtot • Melanie Courtot

We all agree. That's why I said "here, the statements were all identically structured" :-)

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written by Pierre Lindenbaum Pierre Lindenbaum

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