Archive
XML to dict / XML to JSON
Problem
You have an XML file and you want to convert it to dict or JSON.
Well, if you have a dict, you can convert it to JSON with “json.dump()“, so the real question is: how to convert an XML file to a dictionary?
Solution
There is an excellent library for this purpose called xmltodict. Its usage is very simple:
import xmltodict
# It doesn't work with Python 3! Read on for the solution!
def convert(xml_file, xml_attribs=True):
with open(xml_file) as f:
d = xmltodict.parse(f, xml_attribs=xml_attribs)
return d
This worked well under Python 2.7 but I got an error under Python 3. I checked the project’s documentation and it claimed to be Python 3 compatible. What the hell?
The error message was this:
Traceback (most recent call last):
File "/home/jabba/Dropbox/python/lib/jabbapylib2/apps/xmltodict.py", line 247, in parse
parser.ParseFile(xml_input)
TypeError: read() did not return a bytes object (type=str)
During handling of the above exception, another exception occurred:
Traceback (most recent call last):
File "./xml2json.py", line 27, in <module>
print(convert(sys.argv[1]))
File "./xml2json.py", line 17, in convert
d = xmltodict.parse(f, xml_attribs=xml_attribs)
File "/home/jabba/Dropbox/python/lib/jabbapylib2/apps/xmltodict.py", line 249, in parse
parser.Parse(xml_input, True)
TypeError: '_io.TextIOWrapper' does not support the buffer interface
I even filed an issue ticket :)
After some debugging I found a hint here: you need to open the XML file in binary mode!
XML to dict (Python 2 & 3)
So the correct version that works with Python 3 too is this:
import xmltodict
def convert(xml_file, xml_attribs=True):
with open(xml_file, "rb") as f: # notice the "rb" mode
d = xmltodict.parse(f, xml_attribs=xml_attribs)
return d
XML to JSON (Python 2 & 3)
If you want JSON output:
import json
import xmltodict
def convert(xml_file, xml_attribs=True):
with open(xml_file, "rb") as f: # notice the "rb" mode
d = xmltodict.parse(f, xml_attribs=xml_attribs)
return json.dumps(d, indent=4)
Check Gmail for new messages
Problem
I want to check my new Gmail messages periodically. When I get a message from a specific sender (with a specific Subject), I want to trigger some action. How to do that?
Solution
Fortunately, there is an atom feed of unread Gmail messages at https://mail.google.com/mail/feed/atom. All you have to do it is visit this page, send your login credentials, fetch the feed and process it.
import urllib2
FEED_URL = 'https://mail.google.com/mail/feed/atom'
def get_unread_msgs(user, passwd):
auth_handler = urllib2.HTTPBasicAuthHandler()
auth_handler.add_password(
realm='New mail feed',
uri='https://mail.google.com',
user='{user}@gmail.com'.format(user=user),
passwd=passwd
)
opener = urllib2.build_opener(auth_handler)
urllib2.install_opener(opener)
feed = urllib2.urlopen(FEED_URL)
return feed.read()
##########
if __name__ == "__main__":
import getpass
user = raw_input('Username: ')
passwd = getpass.getpass('Password: ')
print get_unread_msgs(user, passwd)
For reading XML I use the untangle module:
import untangle # sudo pip install untangle
xml = get_unread_msgs(USER, PASSWORD)
o = untangle.parse(xml)
try:
for e in o.feed.entry:
title = e.title.cdata
print title
except IndexError:
pass # no new mail
Links
- How to auto log into gmail atom feed with Python? (I took the script from here)
- Read XML painlessly (the untangle module)
Serializations: data <-> XML, data <-> JSON, XML <->JSON
Problems
- Python data to XML and back
- Python data to JSON and back
- XML to JSON and back
Solution
import json
import xmlrpclib
from xml2json import Xml2Json
def data_to_xmlrpc(data):
# http://www.reddit.com/r/Python/comments/ole01/dictionary_to_xml_in_20_lines/
"""Return value: XML RPC string."""
return xmlrpclib.dumps((data,)) # arg. is tuple
def xmlrpc_to_data(xml):
# http://www.reddit.com/r/Python/comments/ole01/dictionary_to_xml_in_20_lines/
"""Return value: python data."""
return xmlrpclib.loads(xml)[0][0]
def data_to_json(data):
"""Return value: JSON string."""
data_string = json.dumps(data)
return data_string
def json_to_data(data_string):
"""Return value: python data."""
data = json.loads(data_string)
return data
def xml_to_json(xml):
"""Return value: JSON string."""
res = Xml2Json(xml).result
return json.dumps(res)
def json_to_xmlrpc(data_string):
"""Return value: XML RPC string."""
data = json.loads(data_string)
return data_to_xmlrpc(data)
def xmlrpc_to_json(xmlrpc):
"""Return value: JSON string."""
data = xmlrpc_to_data(xmlrpc)
return data_to_json(data)
The full source code (serialize.py) together with the imported xml2json.py are available here. This work is part of my jabbapylib library.
Examples
Unit tests are here, they show you how to use these functions. Examples with comments are here.
Read XML painlessly
Problem
I had an XML file (an RSS feed) from which I wanted to extract some data. I tried some XML libraries but I didn’t like any of them. Is there a simple, brain-friendly way for this? After all, it’s Python, so everything should be simple.
Solution
Yes, there is a simple library for reading XML called “untangle“, developed by Chris Stefanescu. It’s in PyPI, so installation is very easy:
sudo pip install untangle
For some examples, visit the project page.
Use Case
Let’s see a simple, real-world example. From the RSS feed of Planet Python, let’s extract the post titles and their URLs.
#!/usr/bin/env python
import untangle
#XML = 'examples/planet_python.xml' # can read a file too
XML = 'http://planet.python.org/rss20.xml'
o = untangle.parse(XML)
for item in o.rss.channel.item:
title = item.title.cdata
link = item.link.cdata
if link:
print title
print ' ', link
It couldn’t be any simpler :)
Limitations
According to Chris, untangle doesn’t support documents with namespaces (yet).
Related posts
Alternatives (update 20111031)
Here are some alternatives (thanks reddit).
- Python and XML (overview)
- lxml
- amara [official tutorial]
- xmltodict (converts XML to dict; added on 20141229)
lxml and amara are heavyweight solutions and are built upon C libraries so you may not be able to use them everywhere. untangle is a lightweight parser that can be a perfect choice to read a small and simple XML file.
Write XML to file
Problem
I wanted to create an XML file. The file was simple but I wanted to avoid producing it with “print” commands. Which API should be used for this purpose? The produced XML should be human readable, i.e. pretty printed (indented).
Solution
This post is based on the thread Best XML writing tool for Python.
(1) elementtree.SimpleXMLWriter (no indenting)
The SimpleXMLWriter module contains a simple helper class for applications that need to generate well-formed XML data. The interface is very simple:
#!/usr/bin/env python
from elementtree.SimpleXMLWriter import XMLWriter
import sys
w = XMLWriter(sys.stdout)
html = w.start("html")
w.start("head")
w.element("title", "my document")
w.element("meta", name="generator", value="my application 1.0")
w.end()
w.start("body")
w.element("h1", "this is a heading")
w.element("p", "this is a paragraph")
w.start("p")
w.data("this is ")
w.element("b", "bold")
w.data(" and ")
w.element("i", "italic")
w.data(".")
w.end("p")
w.close(html)
However, the output is not indented and as I saw, this feature is missing :( Here is the output of the code above:
<html><head><title>my document</title><meta name="generator" value="my application 1.0" /></head><body><h1>this is a heading</h1><p>this is a paragraph</p><p>this is <b>bold</b> and <i>italic</i>.</p></body></html>
If we prettify it, it will look like this:
<?xml version="1.0"?>
<html>
<head>
<title>my document</title>
<meta name="generator" value="my application 1.0"/>
</head>
<body>
<h1>this is a heading</h1>
<p>this is a paragraph</p>
<p>this is <b>bold</b> and <i>italic</i>.</p>
</body>
</html>
You can install elementtree from PyPI.
(2) lxml.etree (can do indenting)
This is what I chose for my project. This API is also very easy to use and it can do indenting. Documentation is here.
Example:
#!/usr/bin/env python
from lxml import etree as ET
root = ET.Element('background')
starttime = ET.SubElement(root, 'starttime')
hour = ET.SubElement(starttime, 'hour')
hour.text = '00'
minute = ET.SubElement(starttime, 'minute')
minute.text = '00'
second = ET.SubElement(starttime, 'second')
second.text = '01'
print ET.tostring(root, pretty_print=True, xml_declaration=True)
# write to file:
# tree = ET.ElementTree(root)
# tree.write('output.xml', pretty_print=True, xml_declaration=True)
Output:
<?xml version='1.0' encoding='ASCII'?>
<background>
<starttime>
<hour>00</hour>
<minute>00</minute>
<second>01</second>
</starttime>
</background>
Installation:
On PyPI, you can find lxml here. However, you will have to install some additional packages too:
sudo apt-get install libxml2-dev sudo apt-get install libxslt-dev # until Ubuntu 10.10: sudo apt-get install python2.6-dev # from Ubuntu 11.04: sudo apt-get install python2.7-dev # under Ubuntu 14.04 I needed this too: sudo apt-get install -y zlib1g-dev
Then, you can install the library with “sudo pip install lxml“.
Links
- Best XML writing tool for Python @ SO (Here you can read about several other XML writing APIs.)
- Pretty printing XML in python @ SO
Related posts
Python Adventures 