Question

I’m using ElementTree findall() to find elements in my XML which have a certain tag. I want to turn the result into a list. At the moment, I’m iterating through the elements, picking out the .text for each element, and appending to the list. I’m sure there’s a more elegant way of doing this.

#!/usr/bin/python2.7
#
from xml.etree import ElementTree
import os
myXML = '''<root>
<project project_name="my_big_project">
<event name="my_first_event">
<location>London</location>
<location>Dublin</location>
<location>New York</location>
<month>January</month>
<year>2013</year>
</event>
</project>
</root>
'''

tree = ElementTree.fromstring(myXML)
for node in tree.findall('.//project'):
  for element in node.findall('event'):
    event_name=element.attrib.get('name')
    print event_name
    locations = []
    if element.find('location') is not None:
      for events in element.findall('location'):
        locations.append(events.text)
# Could I use something like this instead?
#      locations.append(''.join.text(*events) for events in element.findall('location'))

print locations

Outputs this (which is correct, but I’d like to assign the findall() results directly to a list, in text format, if possible;

my_first_event
['London', 'Dublin', 'New York']

Answer 1

You can try this – it uses a list comprehension to generate the list without having to create a blank one and then append.

if element.find('location') is not None:
  locations = [events.text for events in element.findall('location')]

With this, you can also get rid of the locations definition above, so your code would be:

tree = ElementTree.fromstring(myXML)
for node in tree.findall('.//project'):
  for element in node.findall('event'):
    event_name=element.attrib.get('name')
    print event_name
    if element.find('location') is not None:
      locations = [events.text for events in element.findall('location')]

print locations

One thing you will want to be wary of is what you are doing with locations – it won’t be defined if location doesn’t exist, so you will get a NameError if you try to print it and it doesn’t exist. If that is an issue, you can retain the locations = [] definition – if the matching element isn’t found, the result will just be an empty list.