Develop Reference

How to write ElementTree generated by iterparse into an xml file

Please, Note: Novice user of Python.
Hi,
I am working with more than 1Gb of XML file. Using Python2.7. Initially, I was using 'iter' to parse the XML. It worked fine with small files but with file such big I was getting a memory error. Then, I read the documentation and found out that iter load the whole file into memory at once and I should use iterparse. I used and able to load the xml file and make modification while I parse it.
The problem I am facing now is how to write this parsed element tree into a file. The methods I found on Google were suggesting 'write' method of ElementTree which was parsed using 'iter' but mine is parsed using iterparse.
Below is my code snippet. I had commented lines because inner logic of code is pretty big. The only part where I am struggling is writing the updated tree into 'output_pre' file.
The structure of my xml file is like this:
<users>
<user pin=''>
</user>
<user pin=''>
</user>
</users>
Code(inner logic has been removed):
----------------Parser---------------------------
import xml.etree.cElementTree as ET2
import xml.etree.ElementTree as ET
from xml.etree.ElementTree import Element
output_pre = open("pre_ouput.xml", 'w')
tree = ET2.iterparse("temp-output-preliminary.xml")
for event, elem in tree:
if elem.tag == "users":
pass
if elem.tag == "user":
userContent = list(elem)
#Number of children will help filter dummy users in user-state file.
numberOfChildren = len(userContent)
#assert numberOfChildren != 3
PIN = elem.get('pin')
assert PIN is not None
analysing += 1
logger.info ("Analysing user number: %d", analysing)
if numberOfChildren <= 2:
if numberOfChildren >=4:
if numberOfChildren == 3:
for e in ids:
node = ET2.Element("property", {eid: PROV_DATA})
elem.append(node)
container_id_set.add(e)
tree.write(output_pre, encoding='unicode')
output_pre.write("\n</perk-users")
output_pre.close()
Thanks!

How to get sub attribute value in Python

I have an xml file like this.
<approved-by approved="yes">
<person>
<name>XXX/name><signature>XXXXX</signature>
<location>XX</location><company>XXX</company><department>XX</department>
</person>
</approved-by>
<revision-history suppress="yes">
<rev-info>
<rev>PA1</rev>
<date><y>2013</y><m>01</m><d>22</d></date>
I need to retrieve values of 'rev' from all xmls and the value of approved. I want whtr the doc is approved or not. I have a script like this.
from xml.dom.minidom import parse, parseString
import os
import sys
def shahul(dir):
for r,d,f in os.walk(dir):
for files in f:
if files.endswith(".xml"):
dom=parse(os.path.join(r, files))
name = dom.getElementsByTagName('rev')
title = dom.getElementsByTagName('title')
approved=dom.getAttribute('approved')
print (files, title[0].firstChild.nodeValue,name[0].firstChild.nodeValue, approved, sep='\t')
shahul("location")
I am able to get the value under 'rev' but i am not able to get value of the attribute 'approved-by'. I understand my syntax is not right to get the value of approved, but i dont know it.
i need the following as output.
FILE_NAME, Title, PA1, yes
Please guide me.

Assuming that you have only one approved-by tag in the XML:
Change:
approved = dom.getAttribute('approved')
To:
approved_by = dom.getElementsByTagName('approved-by')
approved = approved_by[0].attributes['approved'].value

Retrieve XML parent and child attributes using Python and lxml

I'm trying to process an XML file using XPATH in Python / lxml.
I can pull out the values at a particular level of the tree using this code:
file_name = input('Enter the file name, including .xml extension: ') # User inputs file name
print('Parsing ' + file_name)
from lxml import etree
parser = etree.XMLParser()
tree = etree.parse(file_name, parser)
r = tree.xpath('/dataimport/programmelist/programme')
print (len(r))
with open(file_name+'.log', 'w', encoding='utf-8') as f:
for r in tree.xpath('/dataimport/programmelist/programme'):
progid = (r.get("id"))
print (progid)
It returns a list of values as expected. I also want to return the value of a 'child' (where it exists), but I can't work out how (I can only get it to work as a separate list, but I need to maintain the link between them).
Note: I will be writing the values out to a log file, but since I haven't been successful in getting everything out that I want, I haven't added the 'write out' code yet.
This is the structure of the XML:
<dataimport dtdversion="1.1">
<programmelist>
<programme id="eid-273168">
<imageref idref="img-1844575"/>
How can I get Python to return the id + idref?
The previous examples I have worked with had namespaces, but this file doesn't.

Since xpath() method returns tree, you can use xpath again to get idref list you want:
for r in tree.xpath('/dataimport/programmelist/programme')
progid = r.get("id")
ref_list = r.xpath('imageref/#idref')
print progid, ref_lis

Python: Update XML-file using ElementTree while conserving layout as much as possible

I have a document which uses an XML namespace for which I want to increase /group/house/dogs by one: (the file is called houses.xml)
<?xml version="1.0"?>
<group xmlns="http://dogs.house.local">
<house>
<id>2821</id>
<dogs>2</dogs>
</house>
</group>
My current result using the code below is: (the created file is called houses2.xml)
<ns0:group xmlns:ns0="http://dogs.house.local">
<ns0:house>
<ns0:id>2821</ns0:id>
<ns0:dogs>3</ns0:dogs>
</ns0:house>
</ns0:group>
I would like to fix two things (if it is possible using ElementTree. If it isn´t, I´d be greatful for a suggestion as to what I should use instead):
I want to keep the <?xml version="1.0"?> line.
I do not want to prefix all tags, I´d like to keep it as is.
In conclusion, I don´t want to mess with the document more than I absolutely have to.
My current code (which works except for the above mentioned flaws) generating the above result follows.
I have made a utility function which loads an XML file using ElementTree and returns the elementTree and the namespace (as I do not want to hard code the namespace, and am willing to take the risk it implies):
def elementTreeRootAndNamespace(xml_file):
from xml.etree import ElementTree
import re
element_tree = ElementTree.parse(xml_file)
# Search for a namespace on the root tag
namespace_search = re.search('^({\S+})', element_tree.getroot().tag)
# Keep the namespace empty if none exists, if a namespace exists set
# namespace to {namespacename}
namespace = ''
if namespace_search:
namespace = namespace_search.group(1)
return element_tree, namespace
This is my code to update the number of dogs and save it to the new file houses2.xml:
elementTree, namespace = elementTreeRootAndNamespace('houses.xml')
# Insert the namespace before each tag when when finding current number of dogs,
# as ElementTree requires the namespace to be prefixed within {...} when a
# namespace is used in the document.
dogs = elementTree.find('{ns}house/{ns}dogs'.format(ns = namespace))
# Increase the number of dogs by one
dogs.text = str(int(dogs.text) + 1)
# Write the result to the new file houses2.xml.
elementTree.write('houses2.xml')

An XML based solution to this problem is to write a helper class for ElementTree which:
Grabs the XML-declaration line before parsing as ElementTree at the time of writing is unable to write an XML-declaration line without also writing an encoding attribute(I checked the source).
Parses the input file once, grabs the namespace of the root element. Registers that namespace with ElementTree as having the empty string as prefix. When that is done the source file is parsed using ElementTree again, with that new setting.
It has one major drawback:
XML-comments are lost. Which I have learned is not acceptable for this situation(I initially didn´t think the input data had any comments, but it turns out it has).
My helper class with example:
from xml.etree import ElementTree as ET
import re
class ElementTreeHelper():
def __init__(self, xml_file_name):
xml_file = open(xml_file_name, "rb")
self.__parse_xml_declaration(xml_file)
self.element_tree = ET.parse(xml_file)
xml_file.seek(0)
root_tag_namespace = self.__root_tag_namespace(self.element_tree)
self.namespace = None
if root_tag_namespace is not None:
self.namespace = '{' + root_tag_namespace + '}'
# Register the root tag namespace as having an empty prefix, as
# this has to be done before parsing xml_file we re-parse.
ET.register_namespace('', root_tag_namespace)
self.element_tree = ET.parse(xml_file)
def find(self, xpath_query):
return self.element_tree.find(xpath_query)
def write(self, xml_file_name):
xml_file = open(xml_file_name, "wb")
if self.xml_declaration_line is not None:
xml_file.write(self.xml_declaration_line + '\n')
return self.element_tree.write(xml_file)
def __parse_xml_declaration(self, xml_file):
first_line = xml_file.readline().strip()
if first_line.startswith('<?xml') and first_line.endswith('?>'):
self.xml_declaration_line = first_line
else:
self.xml_declaration_line = None
xml_file.seek(0)
def __root_tag_namespace(self, element_tree):
namespace_search = re.search('^{(\S+)}', element_tree.getroot().tag)
if namespace_search is not None:
return namespace_search.group(1)
else:
return None
def __main():
el_tree_hlp = ElementTreeHelper('houses.xml')
dogs_tag = el_tree_hlp.element_tree.getroot().find(
'{ns}house/{ns}dogs'.format(
ns=el_tree_hlp.namespace))
one_dog_added = int(dogs_tag.text.strip()) + 1
dogs_tag.text = str(one_dog_added)
el_tree_hlp.write('hejsan.xml')
if __name__ == '__main__':
__main()
The output:
<?xml version="1.0"?>
<group xmlns="http://dogs.house.local">
<house>
<id>2821</id>
<dogs>3</dogs>
</house>
</group>
If someone has an improvement to this solution please don´t hesitate to grab the code and improve it.

Round-tripping, unfortunately, isn't a trivial problem. With XML, it's generally not possible to preserve the original document unless you use a special parser (like DecentXML but that's for Java).
Depending on your needs, you have the following options:
If you control the source and you can secure your code with unit tests, you can write your own, simple parser. This parser doesn't accept XML but only a limited subset. You can, for example, read the whole document as a string and then use Python's string operations to locate <dogs> and replace anything up to the next <. Hack? Yes.
You can filter the output. XML allows the string <ns0: only in one place, so you can search&replace it with < and then the same with <group xmlns:ns0=" → <group xmlns=". This is pretty safe unless you can have CDATA in your XML.
You can write your own, simple XML parser. Read the input as a string and then create Elements for each pair of <> plus their positions in the input. That allows you to take the input apart quickly but only works for small inputs.

when Save xml add default_namespace argument is easy to avoid ns0, on my code
key code: xmltree.write(xmlfiile,"utf-8",default_namespace=xmlnamespace)
if os.path.isfile(xmlfiile):
xmltree = ET.parse(xmlfiile)
root = xmltree.getroot()
xmlnamespace = root.tag.split('{')[1].split('}')[0] //get namespace
initwin=xmltree.find("./{"+ xmlnamespace +"}test")
initwin.find("./{"+ xmlnamespace +"}content").text = "aaa"
xmltree.write(xmlfiile,"utf-8",default_namespace=xmlnamespace)

etree from lxml provides this feature.
elementTree.write('houses2.xml',encoding = "UTF-8",xml_declaration = True) helps you in not omitting the declaration
While writing into the file it does not change the namespaces.
http://lxml.de/parsing.html is the link for its tutorial.
P.S : lxml should be installed separately.

Python Lxml (objectify): Checking whether a tag exists

I need to check whether a certain tag exists in an xml file.
For example, I want to see if the tag exists in this snippet:
<main>
<elem1/>
<elem2>Hi</elem2>
<elem3/>
...
</main>
Currently, I am using an ugly hack with error checking, like this:
try:
if root.elem1.tag:
foo = elem1
except AttributeError:
foo = "error finding elem1"
I also want to customize the string if it is unable to find the node (i.e. "unable to find -tagname-").
I have to check a long list of variables, and I don't want to repeat the code 100 times.
Any suggestions?
Edit:
Here is a snip of the actual xml file:
<main>
<asset name="Virtual Dvaered Unpresence">
<virtual/>
<presence>
<faction>Dvaered</faction>
<value>-1000.000000</value>
<range>0</range>
</presence>
</asset>
<asset name="Virtual Empire Small">
<virtual/>
<presence>
<faction>Empire</faction>
<value>100.000000</value>
<range>2</range>
</presence>
</asset>
</main>
I want to check whether the tag exists, and, if so, to get the contents.
Edit edit:
Ok, I am going to combine two of the answers, but I can only vote for one. Sorry.
Edit 3: Related question about XPath here: Python lxml (objectify): Xpath troubles

hasattr() works for this:
if hasattr(root, 'elem1'):
foo = root.elem1

Edit: updated answer for sample file.
I'm assuming you want to search each asset for certain tags. If so, the following worked for me:
import lxml.objectify
# Parse the file.
tree = lxml.objectify.parse('sample.xml')
root = tree.getroot()
# Which elements to find.
to_find = set(['presence/faction', 'presence/value', 'fake'])
# Go through each asset in the document.
for asset in root.findall('asset'):
# Check for each element.
for name in to_find:
node = asset.find(name)
if node is not None:
print 'Found %s, its value is %s' % (name, node)
else:
print 'Unable to find %s' % name
The output was:
Found presence/value, its value is -1000.0
Found presence/faction, its value is Dvaered
Unable to find fake
Found presence/value, its value is 100.0
Found presence/faction, its value is Empire
Unable to find fake

Assume you want to get elem2's value, you can use xpath to find it.
tree = etree.parse(StringIO(htmlString), etree.HTMLParser()).getroot()
youWantValue = tree.xpath('/main/elem2')[0].text

If your document tends to be relatively short you can iterate over all children of <main> looking for tags matching your set of variable names:
tree = lxml.etree.fromstring(DATA)
NAMES = set(['elem1', 'elem3'])
for node in tree.iterchildren():
if node.tag in NAMES:
print 'found', node.tag
Or you can search for each variable name one at a time:
for tag in ('elem1', 'elem3'):
if tree.find(tag) is not None:
print 'found', tag