21

I'm new to xml parsing and Python so bear with me. I'm using lxml to parse a wiki dump, but I just want for each page, its title and text.

For now I've got this:

from xml.etree import ElementTree as etree

def parser(file_name):
    document = etree.parse(file_name)
    titles = document.findall('.//title')
    print titles

At the moment titles isn't returning anything. I've looked at previous answers like this one: ElementTree findall() returning empty list and the lxml documentation, but most things seemed to be tailored towards parsing HTML.

This is a section of my XML:

<mediawiki xmlns="http://www.mediawiki.org/xml/export-0.7/"     xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.mediawiki.org/xml/export-0.7/ http://www.mediawiki.org/xml/export-0.7.xsd" version="0.7" xml:lang="en">
  <siteinfo>
  <sitename>Wikipedia</sitename>
<base>http://en.wikipedia.org/wiki/Main_Page</base>
<generator>MediaWiki 1.20wmf9</generator>
<case>first-letter</case>
<namespaces>
  <namespace key="-2" case="first-letter">Media</namespace>
  <namespace key="-1" case="first-letter">Special</namespace>
  <namespace key="0" case="first-letter" />
  <namespace key="1" case="first-letter">Talk</namespace>
  <namespace key="2" case="first-letter">User</namespace>
  <namespace key="3" case="first-letter">User talk</namespace>
  <namespace key="4" case="first-letter">Wikipedia</namespace>
  <namespace key="5" case="first-letter">Wikipedia talk</namespace>
  <namespace key="6" case="first-letter">File</namespace>
  <namespace key="7" case="first-letter">File talk</namespace>
  <namespace key="8" case="first-letter">MediaWiki</namespace>
  <namespace key="9" case="first-letter">MediaWiki talk</namespace>
  <namespace key="10" case="first-letter">Template</namespace>
  <namespace key="11" case="first-letter">Template talk</namespace>
  <namespace key="12" case="first-letter">Help</namespace>
  <namespace key="13" case="first-letter">Help talk</namespace>
  <namespace key="14" case="first-letter">Category</namespace>
  <namespace key="15" case="first-letter">Category talk</namespace>
  <namespace key="100" case="first-letter">Portal</namespace>
  <namespace key="101" case="first-letter">Portal talk</namespace>
  <namespace key="108" case="first-letter">Book</namespace>
  <namespace key="109" case="first-letter">Book talk</namespace>
</namespaces>
  </siteinfo>
  <page>
    <title>Aratrum</title>
    <ns>0</ns>
    <id>65741</id>
    <revision>
  <id>349931990</id>
  <parentid>225434394</parentid>
  <timestamp>2010-03-15T02:55:02Z</timestamp>
  <contributor>
    <ip>143.105.193.119</ip>
  </contributor>
  <comment>/* Sources */</comment>
  <sha1>2zkdnl9nsd1fbopv0fpwu2j5gdf0haw</sha1>
  <text xml:space="preserve" bytes="1436">'''Aratrum''' is the Latin word for  [[plough]], and &quot;arotron&quot; (αροτρον) is the [[Greek language|Greek]] word. The   [[Ancient Greece|Greeks]] appear to have had diverse kinds of plough from the earliest  historical records. [[Hesiod]] advised the farmer to have always two ploughs, so that if  one broke the other might be ready for use. These ploughs should be of two kinds, the one  called &quot;autoguos&quot; (αυτογυος, &quot;self-limbed&quot;), in which the plough-tail  was of the same piece of timber as the share-beam and the pole; and the other called  &quot;pekton&quot; (πηκτον, &quot;fixed&quot;), because in it, three parts, which were of  three kinds of timber, were adjusted to one another, and fastened together by nails.

The ''autoguos'' plough was made from a [[sapling]] with two branches growing from its   trunk in opposite directions. In ploughing, the trunk served as the pole, one of the two     branches stood upwards and became the tail, and the other penetrated the ground and,    sometimes shod with bronze or iron, acted as the [[ploughshare]]. 

==Sources==
Based on an article from ''A Dictionary of Greek and Roman Antiquities,'' John Murray,     London, 1875.
ἄρατρον

==External links==
*[http://penelope.uchicago.edu/Thayer/E/Roman/Texts/secondary/SMIGRA*/Aratrum.html Smith's     Dictionary article], with diagrams, further details, sources.
[[Category:Agricultural machinery]]
[[Category:Ancient Greece]]
[[Category:Animal equipment]]</text>
</revision>
</page>

I've also tried iterparse and then printing the tag of the element it finds:

for e in etree.iterparse(file_name):
    print e.tag

but it complains about the e not having a tag attribute.

EDIT: screenshot

Community
  • 1
  • 1
liloka
  • 1,016
  • 4
  • 14
  • 29

2 Answers2

37

The problem is that you are not taking XML namespaces into account. The XML document (and all the elements in it) is in the http://www.mediawiki.org/xml/export-0.7/ namespace. To make it work, you need to change

titles = document.findall('.//title')

to

titles = document.findall('.//{http://www.mediawiki.org/xml/export-0.7/}title')

The namespace can also be provided via the namespaces parameter, which is a prefix:URI dictionary:

NSMAP = {'mw':'http://www.mediawiki.org/xml/export-0.7/'}
titles = document.findall('.//mw:title', namespaces=NSMAP)

See the Parsing XML with Namespaces section in the ElementTree documentation for more information.

A third option (added in Python 3.8) is to use a namespace wildcard:

titles = document.findall('.//{*}title')

The trouble with iterparse() is caused by the fact that this function provides (event, element) tuples (not just elements). In order to get the tag name, change

for e in etree.iterparse(file_name):
    print(e.tag)

to this:

for ev, el in etree.iterparse(file_name):
    print(el.tag)
mzjn
  • 48,958
  • 13
  • 128
  • 248
  • Isn't there another method which takes namespaces as another parameter for lxml? Either way that works! So thank you! I'm just going to add getting the text of title for other people viewing this question. – liloka Dec 07 '13 at 23:09
  • 3
    This is exactly what I was looking for! Thank you! – Apolymoxic Aug 18 '16 at 18:44
0

First, you need to locate the parent element, page. I don't know how many layers is this nested, but once you find it, you can immmidiately obtain the title tag:

>>> page_tag = ET.fromstring(xdata)
>>> title_tag = page_tag.find('title')
>>> title_tag.text
'Aratrum'

With more information flooded in, you can do this:

def parser(file_name):
    document = etree.parse(file_name)
    titles = []
    for page_tag in document.findall('page'):
        titles.append(page_tag.find('title').text)
    return titles

Hope this helps!

aIKid
  • 26,968
  • 4
  • 39
  • 65
  • Forgive me, but what exactly is xdata? I've tried the root of the parsed xml file, and the parsed document directly but I'm just getting errors. – liloka Dec 07 '13 at 00:07
  • Oh, xdata is a part of your xml. The whole `page` tag. If it still have a parent other than root, obtain it from the parent. If it's not, you can search for `root.findall('page')` straight away. – aIKid Dec 07 '13 at 00:07
  • ... and how do I create this object? >.o – liloka Dec 07 '13 at 00:22
  • Can you post your whole xml in a pastebin or somewhere? I'm sorry, but i'm going out, and i can't answer in 5-6 hours. – aIKid Dec 07 '13 at 00:38
  • It was a bit big, so I had to cut it down, but I just took out a few pages. http://pastebin.com/gUi226Bj Thank you! – liloka Dec 07 '13 at 11:17
  • Hum.. Is there's nothing else above `mediawiki`? It's quite important to get everything from the top to the title part. – aIKid Dec 07 '13 at 13:53
  • Nope, nothing above that. The only thing that is missing is more pages. – liloka Dec 07 '13 at 14:05
  • Updated the answer. Please try it. – aIKid Dec 07 '13 at 14:13
  • Titles is still returning empty. When I print the document all I get is: so that might be why it's not getting anything... – liloka Dec 07 '13 at 14:32
  • Oh? Literally nothing? Try changing the append to `page_tag.find('title').text`, as in my updated answer. – aIKid Dec 07 '13 at 15:51
  • Literally nothing. I've included a screenshot to show you. I moved return titles to outside of the for loop; which just returns []. – liloka Dec 07 '13 at 17:33