You can use a combination of ElementTree's fromstring() method and the requests module's requests.get() to accomplish this.

https://docs.python.org/2/library/xml.etree.elementtree.html#parsing-xml

fromstring() parses XML from a string directly into an Element, which is the root element of the parsed tree.

Install the requests module:

pip install requests

Use the requests.get() to get your xml file from the url as a string. Pass that into the fromstring() function.

import xml.etree.cElementTree as ET
import requests
tree = ET.fromstring(requests.get('http://synd.cricbuzz.com/j2me/1.0/livematches.xml').text)
for child in tree:
   print("%s - %s"%(child.get('srs'),child.get('mchDesc')))

Results:

None - None
India tour of Sri Lanka, 2015 - Cricbuzz Cup - SL vs IND
Australia tour of Ireland, 2015 - IRE vs AUS
New Zealand tour of South Africa, 2015 - RSA vs NZ
Royal London One-Day Cup, 2015 - SUR vs KENT
Royal London One-Day Cup, 2015 - ESS vs YORKS

Inside the response element there's a row element, so your for loop should be in root[0] instead of root

Here's an example from your snippet, hope it helps you understand the issue

import xml.etree.ElementTree as ET
tree = ET.parse('rows.xml')
root = tree.getroot()

for _id in root[0].findall('row'):
    rank = _id.find('ethcty').text
    name = _id.find('cnt').text
    print(name, rank)

Also, findall should be the name of the node you want

As for loading directly from the url you should use the urllib as follows:

from urllib.request import urlopen
import xml.etree.ElementTree as ET

with urlopen('https://data.cityofnewyork.us/api/views/25th-nujf/rows.xml') as f:
    tree = ET.parse(f)
    root = tree.getroot()

    for _id in root[0].findall('row'):
        rank = _id.find('ethcty').text
        name = _id.find('cnt').text
        print(name, rank)

I edited the latter code because I forgot about the loading from the URL part of your question, i'm sorry about that

You can use urllib2 to download and parse the file in the same way. For e.g. the first few lines will be changed to:

import xml.etree.cElementTree as ET
import urllib2

for i in range(3):
    tree = ET.ElementTree(file=urllib2.urlopen('http://www.trion%i.com:6060/stat.xml' % i ))


    root = tree.getroot()
    root.tag, root.attrib

    # Rest of your code goes here....

From ElementTree docs:

We can import this data by reading from a file:

import xml.etree.ElementTree as ET

tree = ET.parse('country_data.xml')
root = tree.getroot()

Or directly from a string:

root = ET.fromstring(country_data_as_string)

and later in the same page, 20.5.1.4. Finding interesting elements:

for neighbor in root.iter('neighbor'):
    print(neighbor.attrib)

Which translate to:

import xml.etree.ElementTree as ET

root = ET.fromstring("""
<root>
<H D="14/11/2017">
<FC>
    <F LV="0">The quick</F>
    <F LV="1">brown</F>
    <F LV="2">fox</F>
</FC>
</H>
<H D="14/11/2017">
<FC>
    <F LV="0">The lazy</F>
    <F LV="1">fox</F>
</FC>
</H>
</root>""")
# root = tree.getroot()
for h in root.iter("H"):
    print (h.attrib["D"])
for f in root.iter("F"):
    print (f.attrib, f.text)

output:

14/11/2017
14/11/2017
{'LV': '0'} The quick
{'LV': '1'} brown
{'LV': '2'} fox
{'LV': '0'} The lazy
{'LV': '1'} fox

So I have ElementTree 1.2.6 on my box now, and ran the following code against the XML chunk you posted:

import elementtree.ElementTree as ET

tree = ET.parse("test.xml")
doc = tree.getroot()
thingy = doc.find('timeSeries')

print thingy.attrib

and got the following back:

{'name': 'NWIS Time Series Instantaneous Values'}

It appears to have found the timeSeries element without needing to use numerical indices.

What would be useful now is knowing what you mean when you say "it doesn't work." Since it works for me given the same input, it is unlikely that ElementTree is broken in some obvious way. Update your question with any error messages, backtraces, or anything you can provide to help us help you.

Before I try to answer, a tip. Your exception handler covers up the nature of the problem. Just let the original exception rise up and you'll have more information to share with people who are interested in helping you.

I like to use feedparser to parse Atom feeds. It does indeed give you dict-like objects. I submitted a patch to feedparser 4.1 to parse the GeoRSS elements into GeoJSON style dicts. See https://code.google.com/p/feedparser/issues/detail?id=62 and blog post at http://sgillies.net/blog/566/georss-patch-for-universal-feedparser/. You'd use it like this:

>>> import feedparser
>>> feed = feedparser.parse("http://earthquake.usgs.gov/earthquakes/catalogs/1hour-M1.xml")
>>> feed.entries[0]['where']
{'type': 'Point', 'coordinates': (-122.8282, 38.844700000000003)}

My patched version of 4.1 is in my Dropbox and you can get it using pip.

$ pip install http://dl.dropbox.com/u/10325831/feedparser-4.1-georss.tar.gz

Or just download and install with "python setup.py install".

You can parse the text as a string, which creates an Element, and create an ElementTree using that Element.

import xml.etree.ElementTree as ET
tree = ET.ElementTree(ET.fromstring(xmlstring))

I just came across this issue and the documentation, while complete, is not very straightforward on the difference in usage between the parse() and fromstring() methods.

The xml is gzip compressed - requests handles this automatically for you which you could use instead of urllib.

response = requests.get(url)tree = etree.fromstring(response.content)

http://stackoverflow.com/a/26435241 discusses solutions for doing it with urllib