问题描述
我正在处理一个非常大的数据集(大约 500Mio-Triples),该数据集存储在 graphDB Free 中并在我的本地开发人员机器上运行。
我想用 RDF4J 对数据集进行一些操作,并且必须或多或少地选择整个数据集。为了进行测试,我只选择所需的元组。代码在第一个百万元组上运行良好,之后它变得非常慢,因为 graphDB 继续分配更多的 RAM。
是否有可能对非常大的数据集进行 SELECT-Query 并批量获取它们?
基本上我只想通过一些选定的三元组“迭代”,所以应该不需要使用来自 graphDB 的那么多 RAM。我可以看到在查询完成之前我已经在 RDF4J 中获取了数据,因为它仅在大约 1.4 Mio 读取元组时崩溃(HeapSpaceError)。不幸的是,graphDB 并没有释放已经读取的元组的内存。我错过了什么吗?
非常感谢您的帮助。
ps。我已经将graphDB的可用heapSpace设置为20GB。
RDF4J (Java) 代码如下所示:
package ch.test;
import org.eclipse.rdf4j.query.*;
import org.eclipse.rdf4j.repository.RepositoryConnection;
import org.eclipse.rdf4j.repository.http.HTTPRepository;
import java.io.File;
import java.io.IOException;
import java.nio.charset.StandardCharsets;
import java.nio.file.Files;
import java.nio.file.Paths;
public class RDF2RDF {
public static void main(String[] args) {
System.out.println("Running RDF2RDF");
HTTPRepository sourceRepo = new HTTPRepository("http://localhost:7200/repositories/datatraining");
try {
String path = new File("").getAbsolutePath();
String sparqlCommand= Files.readString(Paths.get(path + "/src/main/resources/sparql/select.sparql"),StandardCharsets.ISO_8859_1);
int chunkSize = 10000;
int positionInChunk = 0;
long loadedTuples = 0;
RepositoryConnection sourceConnection = sourceRepo.getConnection();
TupleQuery query = sourceConnection.prepareTupleQuery(sparqlCommand);
try (TupleQueryResult result = query.evaluate()) {
for (BindingSet solution:result) {
loadedTuples++;
positionInChunk++;
if (positionInChunk >= chunkSize) {
System.out.println("Got " + loadedTuples + " Tuples");
positionInChunk = 0;
}
}
}
} catch (IOException err) {
err.printStackTrace();
}
}
}
select.sparql:
PREFIX XXX_meta_schema: <http://schema.XXX.ch/meta/>
PREFIX XXX_post_schema: <http://schema.XXX.ch/post/>
PREFIX XXX_post_tech_schema: <http://schema.XXX.ch/post/tech/>
PREFIX XXX_geo_schema: <http://schema.XXX.ch/geo/>
PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
PREFIX XXX_raw_schema: <http://schema.XXX.ch/raw/>
PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
SELECT * WHERE {
BIND(<http://data.XXX.ch/raw/Table/XXX.csv> as ?table).
?row XXX_raw_schema:isDefinedBy ?table.
?cellStreetAdress XXX_raw_schema:isDefinedBy ?row;
XXX_raw_schema:ofColumn <http://data.XXX.ch/raw/Column/Objektadresse>;
rdf:value ?valueStreetAdress.
?cellOrt mobi_raw_schema:isDefinedBy ?row;
XXX_raw_schema:ofColumn <http://XXX.mobi.ch/raw/Column/Ort>;
rdf:value ?valueOrt.
?cellPlz mobi_raw_schema:isDefinedBy ?row;
XXX_raw_schema:ofColumn <http://XXX.mobi.ch/raw/Column/PLZ>;
rdf:value ?valuePLZ.
BIND (URI(concat("http://data.XXX.ch/post/tech/Adress/",MD5(STR(?cellStreetAdress)))) as ?iri_tech_Adress).
}
我的解决方案: 使用首先获取所有“行”的子选择语句。
PREFIX mobi_post_schema: <http://schema.mobi.ch/post/>
PREFIX mobi_post_tech_schema: <http://schema.mobi.ch/post/tech/>
PREFIX mobi_geo_schema: <http://schema.mobi.ch/geo/>
PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
PREFIX mobi_raw_schema: <http://schema.mobi.ch/raw/>
PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
SELECT * WHERE {
{
SELECT ?row WHERE
{
BIND(<http://data.mobi.ch/raw/Table/Gebaeudeobjekte_August2020_ARA_Post.csv> as ?table).
?row mobi_raw_schema:isDefinedBy ?table.
}
}
?cellStreetAdress mobi_raw_schema:isDefinedBy ?row;
mobi_raw_schema:ofColumn <http://data.mobi.ch/raw/Column/Objektadresse>;
rdf:value ?valueStreetAdress.
?cellOrt mobi_raw_schema:isDefinedBy ?row;
mobi_raw_schema:ofColumn <http://data.mobi.ch/raw/Column/Ort>;
rdf:value ?valueOrt.
?cellPlz mobi_raw_schema:isDefinedBy ?row;
mobi_raw_schema:ofColumn <http://data.mobi.ch/raw/Column/PLZ>;
rdf:value ?valuePLZ.
BIND (URI(concat("http://data.mobi.ch/post/tech/Adress/",MD5(STR(?cellStreetAdress)))) as ?iri_tech_Adress).
}
解决方法
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