Parse XML file into Python object


I have an XML file which looks like this:

   <Name>some filename.mp3</Name>
   <Encoder>Gogo (after 3.0)</Encoder>
   <Mode>joint stereo</Mode>
   ..... and so forth ......

I want to read it into a python object, something like a list of dictionaries. Because the markup is absolutely fixed, I'm tempted to use regex (I'm quite good at using those). However, I thought I'll check if someone knows how to easily avoid regexes here. I don't have much experience with SAX or other parsing, though, but I'm willing to learn.

I'm looking forward to be shown how this is done quickly without regexes in Python. Thanks for your help!

4/3/2011 4:29:19 PM

Accepted Answer

My beloved SD Chargers hat is off to you if you think a regex is easier than this:

#!/usr/bin/env python
import xml.etree.cElementTree as et

   <Name>some filename.mp3</Name>
   <Encoder>Gogo (after 3.0)</Encoder>
   <Name>another filename.mp3</Name>

for el in tree.findall('file'):
    print '-------------------'
    for ch in el.getchildren():
        print '{:>15}: {:<30}'.format(ch.tag, ch.text) 

print "\nan alternate way:"  
el=tree.find('file[2]/Name')  # xpath
print '{:>15}: {:<30}'.format(el.tag, el.text)  


           Name: some filename.mp3             
        Encoder: Gogo (after 3.0)              
        Bitrate: 131                           
           Name: another filename.mp3          
        Encoder: iTunes                        
        Bitrate: 128                           

an alternate way:
           Name: another filename.mp3  

If your attraction to a regex is being terse, here is an equally incomprehensible bit of list comprehension to create a data structure:

[(ch.tag,ch.text) for e in tree.findall('file') for ch in e.getchildren()]

Which creates a list of tuples of the XML children of <file> in document order:

[('Name', 'some filename.mp3'), 
 ('Encoder', 'Gogo (after 3.0)'), 
 ('Bitrate', '131'), 
 ('Name', 'another filename.mp3'), 
 ('Encoder', 'iTunes'), 
 ('Bitrate', '128')]

With a few more lines and a little more thought, obviously, you can create any data structure that you want from XML with ElementTree. It is part of the Python distribution.


Code golf is on!

[{item.tag: item.text for item in ch} for ch in tree.findall('file')] 
[ {'Bitrate': '131', 
   'Name': 'some filename.mp3', 
   'Encoder': 'Gogo (after 3.0)'}, 
  {'Bitrate': '128', 
   'Name': 'another filename.mp3', 
   'Encoder': 'iTunes'}]

If your XML only has the file section, you can choose your golf. If your XML has other tags, other sections, you need to account for the section the children are in and you will need to use findall

There is a tutorial on ElementTree at

4/5/2011 5:36:31 AM

Use ElementTree. You don't need/want to muck about with a parse-only gadget like pyexpat ... you'd only end up re-inventing ElementTree partially and poorly.

Another possibility is lxml which is a third-party package which implements the ElementTree interface plus more.

Update Someone started playing code-golf; here's my entry, which actually creates the data structure you asked for:

# xs = """<encspot> etc etc </encspot"""
>>> import xml.etree.cElementTree as et
>>> from pprint import pprint as pp
>>> pp([dict((attr.tag, attr.text) for attr in el) for el in et.fromstring(xs)])
[{'Bitrate': '131',
  'Encoder': 'Gogo (after 3.0)',
  'Frame': 'no',
  'Frames': '6255',
  'Freq.': '44100',
  'Length': '00:02:43',
  'Mode': 'joint stereo',
  'Name': 'some filename.mp3',
  'Quality': 'good',
  'Size': '5,236,644'},
 {'Bitrate': '0', 'Name': 'foo.mp3'}]

You'd probably want to have a dict mapping "attribute" names to conversion functions:

converters = {
    'Frames': int,
    'Size': lambda x: int(x.replace(',', '')),
    # etc

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