Speeding up xpath

Question

I have a 1000 entry document whose format is something like:

<Example>
     <Entry>
          <n1></n1>
          <n2></n2>
      </Entry>
      <Entry>
          <n1></n1>
          <n2></n2>
      </Entry>
      <!--and so on-->

There are more than 1000 Entry nodes here. I am writing a Java program which basically gets all the node one by one and do some analyzing on each node. But the problem is that the retrieval time of the nodes increases with its no. For example it takes 78 millisecond to retrieve the first node 100 ms to retrieve the second and it keeps on increasing. And to retrieve the 999 node it takes more than 5 second. This is extremely slow. We would be plugging this code to XML files which have even more than 1000 entries. Some like millions. The total time to parse the whole document is more than 5 minutes.

I am using this simple code to traverse it. Here nxp is my own class which has all the methods to get nodes from xpath.

nxp.fromXpathToNode("/Example/Entry" + "[" + i  + "]", doc);

and doc is the document for the file. i is the no of node to retrieve.

Also when i try something like this

List<Node> nl = nxp.fromXpathToNodes("/Example/Entry",doc);  
      content = nl.get(i);

I face the same problem.

Anyone has any solution on how to speed up the tretirival of the nodes, so it takes the same amount of time to get the 1st node as well as the 1000 node from the XML file.

Here is the code for xpathtonode.

public Node fromXpathToNode(String expression, Node context)  
{  
    try  
    {  
        return (Node)this.getCachedExpression(expression).evaluate(context, XPathConstants.NODE);  
    }  
    catch (Exception cause)  
    {  
        throw new RuntimeException(cause);  
    }  
}

and here is the code for fromxpathtonodes.

public List<Node> fromXpathToNodes(String expression, Node context)  
{  
    List<Node> nodes = new ArrayList<Node>();  
    NodeList results = null;  
    
    try  
    {  
        results = (NodeList)this.getCachedExpression(expression).evaluate(context, XPathConstants.NODESET);  
          
        for (int index = 0; index < results.getLength(); index++)  
        {  
            nodes.add(results.item(index));  
        }  
    }  
    catch (Exception cause)  
    {  
        throw new RuntimeException(cause);  
    }  
    
    return nodes;  
}

and here is the starting

public class NativeXpathEngine implements XpathEngine  
{      
private final XPathFactory factory;  
  
private final XPath engine;  

/**
 * Cache for previously compiled XPath expressions. {@link XPathExpression#hashCode()}
 * is not reliable or consistent so use the textual representation instead.
 */  
private final Map<String, XPathExpression> cachedExpressions;  
  
public NativeXpathEngine()  
{
    super();  
    
    this.factory = XPathFactory.newInstance();  
    this.engine = factory.newXPath();  
    this.cachedExpressions = new HashMap<String, XPathExpression>();  
}

The code in `fromXpathToNode` and `fromXpathToNodes` would seem to be quite relevant here. Can you provide that code? — Gerco Dries, Mar 02 '10 at 17:32
I think he means that in this case, why not just iterate directly over the DOM node child lists. — PSpeed, Mar 02 '10 at 18:25

score 10 · Answer 1 · edited Dec 24 '10 at 18:04

Try VTD-XML. It uses less memory than DOM. It is easier to use than SAX and supports XPath. Here is some sample code to help you get started. It applies an XPath to get the Entry elements and then prints out the n1 and n2 child elements.

final VTDGen vg = new VTDGen();
vg.parseFile("/path/to/file.xml", false);

final VTDNav vn = vg.getNav();
final AutoPilot ap = new AutoPilot(vn);
ap.selectXPath("/Example/Entry");
int count = 1;
while (ap.evalXPath() != -1) {
    System.out.println("Inside Entry: " + count);

    //move to n1 child
    vn.toElement(VTDNav.FIRST_CHILD, "n1");
    System.out.println("\tn1: " + vn.toNormalizedString(vn.getText()));

    //move to n2 child
    vn.toElement(VTDNav.NEXT_SIBLING, "n2");
    System.out.println("\tn2: " + vn.toNormalizedString(vn.getText()));

    //move back to parent
    vn.toElement(VTDNav.PARENT);
    count++;
}

+1 for mentioning this awesome lib. I faced a similar problem by parsing some xPathExpressions which took near by 1min to complete. VTD-XML does the same job in 2seks. — onigunn, Nov 11 '10 at 08:52
The library mentioned is truly amazing. Depending on the use case, one has to check if shipping a code using this GPL'd library carries legal dependencies, since it builds on US patents 7133857, 7260652, and 7761459. — ikaerom, Feb 10 '17 at 15:04

score 7 · Answer 2 · edited May 23 '17 at 12:10

7

The correct solution is to detach the node right after you call item(i), like so:

Node node = results.item(index)
node.getParentNode().removeChild(node)
nodes.add(node)

See XPath.evaluate performance slows down (absurdly) over multiple calls

edited May 23 '17 at 12:10

Community

1
1

answered Feb 27 '13 at 00:55

Hani

276
3
4

score 4 · Answer 3 · answered Mar 12 '12 at 12:05

I had similar issue with the Xpath Evaluation , I tried using CachedXPathAPI’s which is faster by 100X than the XPathApi’s which was used earlier. more information about this Api is provided here : http://xml.apache.org/xalan-j/apidocs/org/apache/xpath/CachedXPathAPI.html

Hope it helps. Cheers, Madhusudhan

score 2 · Answer 4 · answered Mar 02 '10 at 19:54

If you need to parse huge but flat documents, SAX is a good alternative. It allows you to handle the XML as a stream instead of building a huge DOM. Your example could be parsed using a ContentHandler like this:

import org.xml.sax.Attributes;
import org.xml.sax.SAXException;
import org.xml.sax.ext.DefaultHandler2;

public class ExampleHandler extends DefaultHandler2 {

    private StringBuffer chars = new StringBuffer(1000);

    private MyEntry currentEntry;
    private MyEntryHandler myEntryHandler;

    ExampleHandler(MyEntryHandler myEntryHandler) {
        this.myEntryHandler = myEntryHandler;
    }

    @Override
    public void characters(char[] ch, int start, int length)
            throws SAXException {
        chars.append(ch);
    }

    @Override
    public void endElement(String uri, String localName, String qName)
            throws SAXException {
        if ("Entry".equals(localName)) {
            myEntryHandler.handle(currentEntry);
            currentEntry = null;
        }
        else if ("n1".equals(localName)) {
            currentEntry.setN1(chars.toString());
        }
        else if ("n2".equals(localName)) {
            currentEntry.setN2(chars.toString());
        }
    }


    @Override
    public void startElement(String uri, String localName, String qName,
            Attributes atts) throws SAXException {
        chars.setLength(0);
        if ("Entry".equals(localName)) {
            currentEntry = new MyEntry();
        }
    }
}

If the document has a deeper and more complex structure, you're going to need to use Stacks to keep track of the current path in the document. Then you should consider writing a general purpose ContentHandler to do the dirty work and use with your document type dependent handlers.

score 1 · Answer 5 · answered Mar 02 '10 at 17:33

1

What kind of parser are you using?

DOM pulls the whole document in memory - once you pull the whole document in memory then your operations can be fast but doing so in a web app or a for loop can have an impact.

SAX parser does on demand parsing and loads nodes as and when you request.

So try to use a parser implementation that suits your need.

answered Mar 02 '10 at 17:33

Kannan Ekanath

16,759
22
75
101

If he's planning to use this on a document with millions of entries, SAX is probably the better way to go. IMHO. – Reed Debaets Mar 02 '10 at 17:36
but why is it so slow. It should be fast for all the entry tags – jon Mar 02 '10 at 17:37

score 0 · Answer 6 · answered Mar 02 '10 at 17:43

0

Use the JAXEN library for xpaths: http://jaxen.codehaus.org/

answered Mar 02 '10 at 17:43

user284566

1
2

Speeding up xpath

6 Answers6

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