Alex Rivera | Logout

ElementTree iterparse strategy

Asked 2012-10-09T04:51:47.420
27

I have to handle xml documents that are big enough (up to 1GB) and parse them with python. I am using the iterparse() function (SAX style parsing).

My concern is the following, imagine you have an xml like this

<?xml version="1.0" encoding="UTF-8" ?>
<families>
  <family>
    <name>Simpson</name>
    <members>
        <name>Homer</name>
        <name>Marge</name>
        <name>Bart</name>
    </members>
  </family>
  <family>
    <name>Griffin</name>
    <members>
        <name>Peter</name>
        <name>Brian</name>
        <name>Meg</name>
    </members>
  </family>
</families>

The problem is, of course to know when I am getting a family name (as Simpsons) and when I am getting the name of one of that family member (for example Homer)

What I have been doing so far is to use "switches" which will tell me if I am inside a "members" tag or not, the code will look like this

import xml.etree.cElementTree as ET

__author__ = 'moriano'

file_path = "test.xml"
context = ET.iterparse(file_path, events=("start", "end"))

# turn it into an iterator
context = iter(context)
on_members_tag = False
for event, elem in context:
    tag = elem.tag
    value = elem.text
    if value :
        value = value.encode('utf-8').strip()

    if event == 'start' :
        if tag == "members" :
            on_members_tag = True

        elif tag == 'name' :
            if on_members_tag :
                print "The member of the family is %s" % value
            else :
                print "The family is %s " % value

    if event =
Edit
Report

1 Answer

36

Here's one possible approach: we maintain a path list and peek backwards to find the parent node(s).

path = []
for event, elem in ET.iterparse(file_path, events=("start", "end")):
    if event == 'start':
        path.append(elem.tag)
    elif event == 'end':
        # process the tag
        if elem.tag == 'name':
            if 'members' in path:
                print 'member'
            else:
                print 'nonmember'
        path.pop()
answered 2012-10-09T06:24:35.747

Your Answer