The namespace of an XML document is significant. ElementTree requires tags to be fully qualified to find the right element. Here's an example of three elements with the same tag in different namespaces:

data = '''\
<root xmlns="xyz" xmlns:name="abc">
  <object name="one" />
  <name:object name="two" />
  <object xmlns="def" name="three" />
</root>
'''

Here's the elements that ElementTree sees:

>>> from xml.etree import ElementTree as et
>>> tree = et.fromstring(data)
>>> print(tree.findall('.//*'))
>>> et.dump(tree)
[<Element '{xyz}object' at 0x0000000003B07BD8>,
 <Element '{abc}object' at 0x0000000003B07C28>,
 <Element '{def}object' at 0x0000000003B07C78>]

So you have it right. Given the default namespace definition of:

<entry xmlns='http://www.w3.org/2005/Atom' ...

To access the 'title' tag, which uses the default namespace:

media['title'] = e.findall('{http://www.w3.org/2005/Atom}title')

to access the 'media:group' tag, refer to the media namespace definition:

<entry ... xmlns:media='http://search.yahoo.com/mrss/' ...

And use:

e.findall('{http://search.yahoo.com/mrss/}group')

Note the different ways a namespace can be specified:

<root xmlns="xyz" xmlns:name="abc">   # default namespace and
                                      # 'abc' namespace with id 'name'.
  <object name="one" />               # Uses default namespace 'xyz'.
  <name:object name="two" />          # uses 'abc' namespace (specified by id).
  <object xmlns="def" name="three" /> # change the default namespace to 'def'.
</root>

To read a specific tag from a specific namespace:

>>> print(tree.find('{abc}object').attrib['name'])
'two'

Note the namespace IDs are just shortcuts. Here's what happens when you dump the parsed XML tree. ElementTree doesn't bother to save the original namespace IDs and generates its own in the format ns#:

>>> et.dump(tree)
<ns0:root xmlns:ns0="xyz" xmlns:ns1="abc" xmlns:ns2="def">
  <ns0:object name="one" />
  <ns1:object name="two" />
  <ns2:object name="three" />
</ns0:root>

If you want specific shortcuts defined, use `register_namespace':

>>> et.register_namespace('','xyz') # default namespace
>>> et.register_namespace('name','abc')
>>> et.register_namespace('custom','def')
>>> et.dump(tree)
<root xmlns="xyz" xmlns:custom="def" xmlns:name="abc">
  <object name="one" />
  <name:object name="two" />
  <custom:object name="three" />
</root>

There is no Item_No_1 at the elements that are found by doc.findall('md/mi/ovalue/').

I think what you may try to do is get both lists

Copyitems = [e.text for e in doc.findall('md/mi/it')]
values = [e.text for e in doc.findall('md/mi/ovalue/v')]

Then find the index of the string 'Item_No_1' from items, and then index into values with that number.

Alternatively, zip the two lists together and check when you find one element.

Copyfor item,value in zip(doc.findall('md/mi/it'), doc.findall('md/mi/ovalue/v')):
    if item.text == 'Item_No_1':
        print(value.text)

There might be a better way, but those are the first ways that come to mind

Yes, in the package xml.etree you can find the built-in function related to XML. (also available for python2)

The one specifically you are looking for is findall.

For example:

import xml.etree.ElementTree as ET
tree = ET.fromstring(some_xml_data)
all_name_elements = tree.findall('.//name')

With:

In [1]: some_xml_data = "<help><person><name>dean</name></person></help>"

I get the following:

In [10]: tree.findall(".//name")
Out[10]: [<Element 'name' at 0x7ff921edd390>]

The first thing you should test is what happens if you simplify your XPath:

>>> print(root.find(".//field"))
None

So, what's going on? You don't have any elements of type field. You've got an explicit namespace, which means you have elements of type '{http://www.librarything.com/}field'. You can see this pretty easily:

>>> print(root.getchildren())
[<Element '{http://www.librarything.com/}item' at 0x1047580e8>]
>>> print(root.find(".//{http://www.librarything.com/}field"))
<Element '{http://www.librarything.com/}field' at 0x1047582c8>
>>> print(root.find(".//{http://www.librarything.com/}field[@type='5']"))
<Element '{http://www.librarything.com/}field' at 0x104758688>

If you want to know more, there are multiple questions on this site about how ETree deals with namespaces (from a quick search, 1 and 2 look relevant), and detailed information in the documentation; trying to explain it all in yet another answer would just lead to an inferior answer to the existing ones.

You can access the attribute value as this:

from elementtree.ElementTree import XML, SubElement, Element, tostring

text = """
<root>
    <phoneNumbers>
        <number topic="sys/phoneNumber/1" update="none" />
        <number topic="sys/phoneNumber/2" update="none" />
        <number topic="sys/phoneNumber/3" update="none" />
    </phoneNumbers>

    <gfenSMSnumbers>
        <number topic="sys2/SMSnumber/1" update="none" />
        <number topic="sys2/SMSnumber/2" update="none" />
    </gfenSMSnumbers>
</root>
"""

elem = XML(text)
for node in elem.find('phoneNumbers'):
    print node.attrib['topic']
    # Create sub elements
    if node.attrib['topic']=="sys/phoneNumber/1":
        tag = SubElement(node,'TagName')
        tag.attrib['attr'] = 'AttribValue'

print tostring(elem)

forget to say, if your ElementTree version is greater than 1.3, you can use XPath:

elem.find('.//number[@topic="sys/phoneNumber/1"]')

http://effbot.org/zone/element-xpath.htm

or you can use this simple one:

for node in elem.findall('.//number'):
    if node.attrib['topic']=="sys/phoneNumber/1":
        tag = SubElement(node,'TagName')
        tag.attrib['attr'] = 'AttribValue'

The previous posters have the right of it. The etree documentation can be found here:

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

And can help you out. Here's a code sample that might do the trick (partially taken from the above link):

Copyimport xml.etree.ElementTree as ET
tree = ET.parse('your_file.xml')
root = tree.getroot()

for group in root.findall('group'):
  title = group.find('title')
  titlephrase = title.find('phrase').text
  for doc in group.findall('document'):
    refid = doc.get('refid')

Or if you want the ID stored in the group tag, you'd use id = group.get('id') instead of searching for all the refids.

With lxml:

import lxml.etree
# xmlstr is your xml in a string
root = lxml.etree.fromstring(xmlstr)
textelem = root.find('result/field/value/text')
print textelem.text

Edit: But I imagine there could be more than one result...

import lxml.etree
# xmlstr is your xml in a string
root = lxml.etree.fromstring(xmlstr)
results = root.findall('result')
textnumbers = [r.find('field/value/text').text for r in results]