How to use XPath in Python?


What are the libraries that support XPath? Is there a full implementation? How is the library used? Where is its website?

This question is tagged with python xml dom xpath python-2.x

~ Asked on 2008-08-12 11:28:36

12 Answers


libxml2 has a number of advantages:

  1. Compliance to the spec
  2. Active development and a community participation
  3. Speed. This is really a python wrapper around a C implementation.
  4. Ubiquity. The libxml2 library is pervasive and thus well tested.

Downsides include:

  1. Compliance to the spec. It's strict. Things like default namespace handling are easier in other libraries.
  2. Use of native code. This can be a pain depending on your how your application is distributed / deployed. RPMs are available that ease some of this pain.
  3. Manual resource handling. Note in the sample below the calls to freeDoc() and xpathFreeContext(). This is not very Pythonic.

If you are doing simple path selection, stick with ElementTree ( which is included in Python 2.5 ). If you need full spec compliance or raw speed and can cope with the distribution of native code, go with libxml2.

Sample of libxml2 XPath Use

import libxml2

doc = libxml2.parseFile("tst.xml")
ctxt = doc.xpathNewContext()
res = ctxt.xpathEval("//*")
if len(res) != 2:
    print "xpath query: wrong node set size"
if res[0].name != "doc" or res[1].name != "foo":
    print "xpath query: wrong node set value"

Sample of ElementTree XPath Use

from elementtree.ElementTree import ElementTree
mydoc = ElementTree(file='tst.xml')
for e in mydoc.findall('/foo/bar'):
    print e.get('title').text

~ Answered on 2008-08-26 13:06:39


The lxml package supports xpath. It seems to work pretty well, although I've had some trouble with the self:: axis. There's also Amara, but I haven't used it personally.

~ Answered on 2008-08-12 11:40:13


Sounds like an lxml advertisement in here. ;) ElementTree is included in the std library. Under 2.6 and below its xpath is pretty weak, but in 2.7+ much improved:

import xml.etree.ElementTree as ET
root = ET.parse(filename)
result = ''

for elem in root.findall('.//child/grandchild'):
    # How to make decisions based on attributes even in 2.6:
    if elem.attrib.get('name') == 'foo':
        result = elem.text

~ Answered on 2012-11-22 01:05:52


Use LXML. LXML uses the full power of libxml2 and libxslt, but wraps them in more "Pythonic" bindings than the Python bindings that are native to those libraries. As such, it gets the full XPath 1.0 implementation. Native ElemenTree supports a limited subset of XPath, although it may be good enough for your needs.

~ Answered on 2009-11-13 23:11:17


Another option is py-dom-xpath, it works seamlessly with minidom and is pure Python so works on appengine.

import xpath
xpath.find('//item', doc)

~ Answered on 2010-01-23 09:30:19


You can use:


from xml.dom.ext.reader import Sax2
from xml import xpath
doc = Sax2.FromXmlFile('foo.xml').documentElement
for url in xpath.Evaluate('//@Url', doc):
  print url.value


import libxml2
doc = libxml2.parseFile('foo.xml')
for url in doc.xpathEval('//@Url'):
  print url.content

~ Answered on 2010-08-23 13:00:01


The latest version of elementtree supports XPath pretty well. Not being an XPath expert I can't say for sure if the implementation is full but it has satisfied most of my needs when working in Python. I've also use lxml and PyXML and I find etree nice because it's a standard module.

NOTE: I've since found lxml and for me it's definitely the best XML lib out there for Python. It does XPath nicely as well (though again perhaps not a full implementation).

~ Answered on 2008-08-14 09:48:59


You can use the simple soupparser from lxml


from lxml.html.soupparser import fromstring

tree = fromstring("<a>Find me!</a>")
print tree.xpath("//a/text()")

~ Answered on 2015-11-15 05:31:21


If you want to have the power of XPATH combined with the ability to also use CSS at any point you can use parsel:

>>> from parsel import Selector
>>> sel = Selector(text=u"""<html>
            <h1>Hello, Parsel!</h1>
                <li><a href="">Link 1</a></li>
                <li><a href="">Link 2</a></li>
>>> sel.css('h1::text').extract_first()
'Hello, Parsel!'
>>> sel.xpath('//h1/text()').extract_first()
'Hello, Parsel!'

~ Answered on 2017-12-16 22:16:20


Another library is 4Suite:

I do not know how spec-compliant it is. But it has worked very well for my use. It looks abandoned.

~ Answered on 2010-08-23 12:57:47


PyXML works well.

You didn't say what platform you're using, however if you're on Ubuntu you can get it with sudo apt-get install python-xml. I'm sure other Linux distros have it as well.

If you're on a Mac, xpath is already installed but not immediately accessible. You can set PY_USE_XMLPLUS in your environment or do it the Python way before you import xml.xpath:

if sys.platform.startswith('darwin'):
    os.environ['PY_USE_XMLPLUS'] = '1'

In the worst case you may have to build it yourself. This package is no longer maintained but still builds fine and works with modern 2.x Pythons. Basic docs are here.

~ Answered on 2008-08-12 19:34:44


If you are going to need it for html:

import lxml.html as html
root  = html.fromstring(string)

~ Answered on 2019-05-29 13:48:30

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