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 co
This is more in comment to the answer by 'unutbu' in which a suggestion to cleanup namespace was desired without giving example. this might be what you are looking for...
from lxml import objectify
objectify.deannotate(root, cleanup_namespaces=True)
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 >
.
There is a way to remove namespaces with XSLT:
import io
import lxml.etree as ET
def remove_namespaces(doc):
# http://wiki.tei-c.org/index.php/Remove-Namespaces.xsl
xslt='''<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" indent="no"/>
<xsl:template match="/|comment()|processing-instruction()">
<xsl:copy>
<xsl:apply-templates/>
</xsl:copy>
</xsl:template>
<xsl:template match="*">
<xsl:element name="{local-name()}">
<xsl:apply-templates select="@*|node()"/>
</xsl:element>
</xsl:template>
<xsl:template match="@*">
<xsl:attribute name="{local-name()}">
<xsl:value-of select="."/>
</xsl:attribute>
</xsl:template>
</xsl:stylesheet>
'''
xslt_doc = ET.parse(io.BytesIO(xslt))
transform = ET.XSLT(xslt_doc)
doc = transform(doc)
return doc
doc = ET.parse('data.xml')
doc = remove_namespaces(doc)
print(ET.tostring(doc))
yields
<some>
<to_move date="somedate">
<child>some text</child>
</to_move>
</some>