I have a huge xml file (1 Gig). I want to move some of the elements (entrys) to another file with the same header and specifications.
Let's say the original file contains this entry with tag <to_move>
:
<?xml version="1.0" encoding="ISO-8859-1"?>
<!DOCTYPE some SYSTEM "some.dtd">
<some>
...
<to_move date="somedate">
<child>some text</child>
...
...
</to_move>
...
</some>
I use lxml.etree.iterparse to iterate through the file. Works fine. When I find the element with tag <to_move>
, let's assume it is stored in the variable element
I do
new_file.write(etree.tostring(element))
But this results in
<?xml version="1.0" encoding="ISO-8859-1"?>
<!DOCTYPE some SYSTEM "some.dtd">
<some>
...
<to_move xmlns:="some" date="somedate"> # <---- Here is the problem. I don't want the namespace.
<child>some text</child>
...
...
</to_move>
...
</some>
So the question is: How to tell etree.tostring() not to write the xmlns:="some"
. Is this possible? I struggeled with the api-documentation of lxml.etree, but I couldn't find a satisfying answer.
This is what I found for etree.trostring
:
tostring(element_or_tree, encoding=None, method="xml",
xml_declaration=None, pretty_print=False, with_tail=True,
standalone=None, doctype=None, exclusive=False, with_comments=True)
Serialize an element to an encoded string representation of its XML tree.
To me every one of the parameters of tostring()
does not seem to help. Any suggestion or corrections?
I often grab a namespace to make an alias for it like this:
someXML = lxml.etree.XML(someString)
if ns is None:
ns = {"m": someXML.tag.split("}")[0][1:]}
someid = someXML.xpath('.//m:ImportantThing//m:ID', namespaces=ns)
You could do something similar to grab the namespace in order to make a regex that will clean it up after using tostring
.
Or you could clean up the input string. Find the first space, check if it is followed by xmlns, if yes, delete the whole xmlns bit up to the next space, if no delete the space. Repeat until there are no more spaces or xmlns declarations. But don't go past the first >
.