使用Perl访问RDF/XML/OWL文件节点

Question

我有一个RDF/XML数据，我想解析并访问该节点。它看起来像这样：

<!-- http://purl.obolibrary.org/obo/VO_0000185 -->
    <owl:Class rdf:about="&obo;VO_0000185">
        <rdfs:label>Influenza virus gene</rdfs:label>
        <rdfs:subClassOf rdf:resource="&obo;VO_0000156"/>
        <obo:IAO_0000117>YH</obo:IAO_0000117>
    </owl:Class>

    <!-- http://purl.obolibrary.org/obo/VO_0000186 -->
    <owl:Class rdf:about="&obo;VO_0000186">
        <rdfs:label>RNA vaccine</rdfs:label>
        <owl:equivalentClass>
            <owl:Class>
                <owl:intersectionOf rdf:parseType="Collection">
                    <rdf:Description rdf:about="&obo;VO_0000001"/>
                    <owl:Restriction>
                        <owl:onProperty rdf:resource="&obo;BFO_0000161"/>
                        <owl:someValuesFrom rdf:resource="&obo;VO_0000728"/>
                    </owl:Restriction>
                </owl:intersectionOf>
            </owl:Class>
        </owl:equivalentClass>
        <rdfs:subClassOf rdf:resource="&obo;VO_0000001"/>
        <obo:IAO_0000116>Using RNA may eliminate the problem of having to tailor a vaccine for each individual patient with their specific immunity. The advantage of RNA is that it can be used for all immunity types and can be taken from a single cell. DNA vaccines need to produce RNA which then prompts the manufacture of proteins. However, RNA vaccine eliminates the step from DNA to RNA.</obo:IAO_0000116>
        <obo:IAO_0000115>A vaccine that uses RNA(s) derived from a pathogen organism.</obo:IAO_0000115>
        <obo:IAO_0000117>YH</obo:IAO_0000117>
    </owl:Class>

完整的RDF/XML文件可以在这里找到。

我想做的是做以下事情：

1. 查找包含条目<rdfs:subClassOf rdf:resource="&obo;VO_0000001"/>的区块
2. 访问<rdfs:label>...</rdfs:label>定义的文字术语

因此，在上面的示例中，代码将通过第二个块并输出："RNA疫苗"。

我目前被以下代码卡住了。在我做不到的地方访问节点。正确的方法是什么？使用XML:：LibXML以外的解决方案欢迎。

#!/usr/bin/perl -w
use strict;
use Data::Dumper;
use Carp;
use File::Basename;
use XML::LibXML 1.70;
my $filename = "VO.owl";
# Obtained from http://svn.code.sf.net/p/vaccineontology/code/trunk/src/ontology/VO.owl
my $parser = XML::LibXML->new();
my $doc = $parser->parse_file( $filename );
foreach my $chunk ($doc->findnodes('/owl:Class')) {
        my ($label) = $chunk->findnodes('./rdfs:label');
        my ($subclass) = $chunk->findnodes('./rdfs:subClassOf');
        print $label->to_literal;
        print $subclass->to_literal;
}

Answer 1

将RDF当作XML进行解析是愚蠢的。完全相同的数据可以以多种不同的方式出现。例如，以下所有RDF文件都携带相同的数据。任何一致的RDF实现都必须以相同的方式处理它们。。。

<!-- example 1 -->
<rdf:RDF xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#">
  <rdf:Description rdf:about="#me">
    <rdf:type rdf:resource="http://xmlns.com/foaf/0.1/Person" />
    <foaf:name>Toby Inkster</foaf:name>
  </rdf:Description>
</rdf:RDF>
<!-- example 2 -->
<rdf:RDF
    xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"
    xmlns:foaf="http://xmlns.com/foaf/0.1/">
  <foaf:Person rdf:about="#me">
    <foaf:name>Toby Inkster</foaf:name>
  </foaf:Person>
</rdf:RDF>
<!-- example 3 -->
<rdf:RDF
    xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"
    xmlns:foaf="http://xmlns.com/foaf/0.1/">
  <foaf:Person rdf:about="#me" foaf:name="Toby Inkster" />
</rdf:RDF>
<!-- example 4 -->
<rdf:RDF
    xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"
    xmlns:foaf="">
  <rdf:Description rdf:about="#me"
    rdf:type="http://xmlns.com/foaf/0.1/Person"
    foaf:name="Toby Inkster" />
</rdf:RDF>
<!-- example 5 -->
<rdf:RDF xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#">
  <rdf:Description rdf:ID="me">
    <rdf:type>
      <rdf:Description rdf:about="http://xmlns.com/foaf/0.1/Person" />
    </rdf:type>
    <foaf:name>Toby Inkster</foaf:name>
  </rdf:Description>
</rdf:RDF>
<!-- example 6 -->
<foaf:Person
    xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"
    xmlns:foaf="http://xmlns.com/foaf/0.1/"
    rdf:about="#me"
    foaf:name="Toby Inkster" />

我也可以很容易地列出六种其他变体，但我仅限于此。这个RDF文件只包含两个语句——I’m a Person；我的名字是"Toby Inkster"——OP的数据包含50000多条语句。

这只是RDF的XML序列化；还有其他序列化。

如果你尝试用XPath处理所有这些，你很可能会变成一个被锁在某个塔里的疯子，在睡梦中喃喃自语地谈论三元组；三元组。。。

幸运的是，格雷格·威廉姆斯为你打下了心理健康的子弹。RDF：：Trine和RDF：：Query不仅是Perl最好的RDF框架；它们是所有编程语言中最好的。

以下是如何使用RDF：：Trine和RDF：：Query：来实现OP的任务

#!/usr/bin/env perl
use v5.12;
use RDF::Trine;
use RDF::Query;
my $model = 'RDF::Trine::Model'->new(
    'RDF::Trine::Store::DBI'->new(
        'vo',
        'dbi:SQLite:dbname=/tmp/vo.sqlite',
        '',  # no username
        '',  # no password
    ),
);
'RDF::Trine::Parser::RDFXML'->new->parse_url_into_model(
    'http://svn.code.sf.net/p/vaccineontology/code/trunk/src/ontology/VO.owl',
    $model,
) unless $model->size > 0;
my $query = RDF::Query->new(<<'SPARQL');
PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
SELECT ?super_label ?sub_label
WHERE {
    ?sub rdfs:subClassOf ?super .
    ?sub rdfs:label ?sub_label .
    ?super rdfs:label ?super_label .
}
LIMIT 5
SPARQL
print $query->execute($model)->as_string;

样本输出：

+----------------------------+----------------------------------+
| super_label                | sub_label                        |
+----------------------------+----------------------------------+
| "Aves vaccine"             | "Ducks vaccine"                  |
| "route of administration"  | "intravaginal route"             |
| "Shigella gene"            | "aroA from Shigella"             |
| "Papillomavirus vaccine"   | "Bovine papillomavirus vaccine"  |
| "virus protein"            | "Feline leukemia virus protein"  |
+----------------------------+----------------------------------+

UPDATE：这里有一个SPARQL查询，可以插入上面的脚本来检索您想要的数据：

PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
PREFIX obo:  <http://purl.obolibrary.org/obo/>
SELECT ?subclass ?label
WHERE {
    ?subclass
        rdfs:subClassOf obo:VO_0000001 ;
        rdfs:label ?label .
}

Answer 2

/owl:Class不是XML文档中的根元素。您必须将根元素包含到XPath:/rdf:RDF/owl:Class中。或者，如果您想获得所有出现的内容，无论XML树的深度如何，都可以使用双斜杠表示法：//owl:Class。