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How do I remove a comment outside of the root element of an XML document using python lxml

How do you remove comments above or below the root node of an xml document using python's lxml module? I want to remove only one comment above the root node, NOT all comments in the entire document. For instance, given the following xml document
<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<!-- This comment needs to be removed -->
<root>
<!-- This comment needs to STAY -->
<a/>
</root>
I want to output
<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<root>
<!-- This comment needs to STAY -->
<a/>
</root>
The usual way to remove an element would be to do element.getparent().remove(element), but this doesn't work for the root element since getparent returns None. I also tried the suggestions from this stackoverflow answer, but the first answer (using a parser that remove comments) removes all comments from the document including the ones I want to keep, and the second answer (adding a dummy opening and closing tag around the document) doesn't work if the document has a directive above the root element.
I can get access to the comment above the root element using the following code, but how do I remove it from the document?
from lxml import etree as ET
tree = ET.parse("./sample_file.xml")
root = tree.getroot()
comment = root.getprevious()
# What do I do with comment now??
I've tried doing the following, but none of them worked:
comment.getparent().remove(comment) says AttributeError: 'NoneType' object has no attribute 'remove'
del comment does nothing
comment.clear() does nothing
comment.text = "" renders an empty comment <!---->
root.remove(comment) says ValueError: Element is not a child of this node.
tree.remove(comment) says AttributeError: 'lxml.etree._ElementTree' object has no attribute 'remove'
tree[:] = [root] says TypeError: 'lxml.etree._ElementTree' object does not support item assignment
Initialize a new tree with tree = ET.ElementTree(root). Serializing this new tree still has the comments somehow.

You could just build another tree by using fromstring() and passing in the root element.
from lxml import etree
tree = etree.parse("sample_file.xml")
new_tree = etree.fromstring(etree.tostring(tree.getroot()))
print(etree.tostring(new_tree, xml_declaration=True, encoding="UTF-8", standalone=True).decode())
printed output...
<?xml version='1.0' encoding='UTF-8' standalone='yes'?>
<root>
<!-- This comment needs to STAY -->
<a/>
</root>
Note: This will also remove any processing instructions before root, so another option is to append the comment to root before removing...
from lxml import etree
tree = etree.parse("sample_file.xml")
root = tree.getroot()
for comment_to_delete in root.xpath("preceding::comment()"):
root.append(comment_to_delete)
root.remove(comment_to_delete)
print(etree.tostring(tree, xml_declaration=True, encoding="UTF-8", standalone=True).decode())
This produces the same output as above, but will retain any processing instructions that occur before root.

You can parse a XML file with comments with the xmlPullParser:
If your input file looks like:
<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<!-- This comment needs to be removed -->
<root>
<!-- This comment needs to STAY -->
<a/>
<b>Text</b>
</root>
Parse the file and write it to a new one:
import xml.etree.ElementTree as ET
import re
# Write XML declaration line into neu file without comment 1
def write_delte_xml(input):
with open('Cleaned.xml', 'a') as my_file:
my_file.write(f'{input}')
with open('Remove_Comment.xml', 'r', encoding='utf-8') as xml:
feedstring = xml.readlines()
parser = ET.XMLPullParser(['start','end', 'comment'])
for line in enumerate(feedstring):
if line[0] == 0 and line[1].startswith('<?'):
write_delte_xml(line[1])
parser.feed(line[1])
for event, elem in parser.read_events():
if event == "comment" and line[0] != 1:
write_delte_xml(line[1])
#print(line[1])
if event == "start" and r'\>' not in line[1]:
write_delte_xml(f"{line[1]}")
#print("start",f"{line[1]},Element: {elem}")
if event == "end":
write_delte_xml(f"{line[1]}")
#print(f"END: {line[1]}")
# Clean douplicates
xml_list = []
with open('Cleaned.xml', 'rb') as xml:
lines = xml.readlines()
for line in lines:
if line not in xml_list:
xml_list.append(line)
with open('Cleaned_final.xml', 'wb') as my_file:
for line in xml_list:
my_file.write(line)
print('Cleaned.xml')
Output:
<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<root>
<!-- This comment needs to STAY -->
<a/>
<b>Text</b>
</root>

Removing Elements from a KML (Python)

I generated a KML file using Python's SimpleKML library and the following script, the output of which is also shown below:
import simplekml
kml = simplekml.Kml()
ground = kml.newgroundoverlay(name='Aerial Extent')
ground.icon.href = 'C:\\Users\\mdl518\\Desktop\\aerial_image.png'
ground.latlonbox.north = 46.55537
ground.latlonbox.south = 46.53134
ground.latlonbox.east = 48.60005
ground.latlonbox.west = 48.57678
ground.latlonbox.rotation = 0.090320
kml.save(".//aerial_extent.kml")
The output KML:
<?xml version="1.0" encoding="UTF-8"?>
<kml xmlns="http://www.opengis.net/kml/2.2" xmlns:gx="http://www.google.com/kml/ext/2.2">
<Document id="1">
<GroundOverlay id="2">
<name>Aerial Extent</name>
<Icon id="3">
<href>C:\\Users\\mdl518\\Desktop\\aerial_image.png</href>
</Icon>
<LatLonBox>
<north>46.55537</north>
<south>46.53134</south>
<east>48.60005</east>
<west>48.57678</west>
<rotation>0.090320</rotation>
</LatLonBox>
</GroundOverlay>
</Document>
However, I am trying to remove the "Document" tag from this KML since it is a default element generated with SimpleKML, while keeping the child elements (e.g. GroundOverlay). Additionally, is there a way to remove the "id" attributes associated with specific elements (i.e. for the GroundOverlay, Icon elements)? I am exploring the usage of ElementTree/lxml to enable this, but these seem to be more specific to XML files as opposed to KMLs. Here's what I'm trying to use to modify the KML, but it is unable to remove the Document element:
from lxml import etree
tree = etree.fromstring(open("C:\\Users\\mdl518\\Desktop\\aerial_extent.kml").read())
for item in tree.xpath("//Document[#id='1']"):
item.getparent().remove(item)
print(etree.tostring(tree, pretty_print=True))
Here is the final desired output XML:
<?xml version="1.0" encoding="UTF-8"?>
<kml xmlns="http://www.opengis.net/kml/2.2" xmlns:gx="http://www.google.com/kml/ext/2.2">
<GroundOverlay>
<name>Aerial Extent</name>
<Icon>
<href>C:\\Users\\mdl518\\Desktop\\aerial_image.png</href>
</Icon>
<LatLonBox>
<north>46.55537</north>
<south>46.53134</south>
<east>48.60005</east>
<west>48.57678</west>
<rotation>0.090320</rotation>
</LatLonBox>
</GroundOverlay>
</kml>
Any insights are most appreciated!

You are getting tripped up on the dreaded namespaces...
Try using something like this:
ns = {'kml': 'http://www.opengis.net/kml/2.2'}
for item in tree.xpath("//kml:Document[#id='1']",namespaces=ns):
item.getparent().remove(item)
Edit:
To remove just the parent and retain all its descendants, try the following:
retain = doc.xpath("//kml:Document[#id='1']/kml:GroundOverlay",namespaces=ns)[0]
for item in doc.xpath("//kml:Document[#id='1']",namespaces=ns):
anchor = item.getparent()
anchor.remove(item)
anchor.insert(1,retain)
print(etree.tostring(doc, pretty_print=True).decode())
This should get you the desired output.

How to save certain sections in a line to a new .txt file?

So I have the following .txt file of data, where the data highlighted with yellow needs to be saved to a new txt file:
I managed to print certain sections in Python, but that's about it:
with open('Podatki-zima-MEDVES.txt', mode='r+t') as file:
for line in file:
print(line[18:39])
Resulting in:
EntryDate="20101126"
EntryDate="20101126"
EntryDate="20101126"
EntryDate="20101126"
EntryDate="20101127"
EntryDate="20101128"
EntryDate="20101128"
EntryDate="20101128"
EntryDate="20101128"
I know it's a very basic question, but for someone experienced this wouldn't take a minute.
Thanks

It looks like you're trying to parse xml data.
There is a standard library package that can do this. The documentation is pretty good and it includes a tutorial. Take a look at The ElementTree XML API.
In you case the code would look something like:
data = """
<data>
<ROW EntryData="20101126" SnowDepth="4"/>
<ROW EntryData="20101127" SnowDepth="8"/>
</data>"""
import xml.etree.ElementTree as ET
root = ET.fromstring(data)
for child in root:
entries = child.attrib
print(entries["EntryData"], entries["SnowDepth"])
This gives the output you're looking for:
20101126 4
20101127 8

As an alternative to using Element Tree you could use an Expat parser for your Structured Markup data.
You first need to specify document type and wrap a top level element around your data as follows:
<?xml version="1.0"?>
<podatki>
<ROW RowState="5" EntryDate="20101126" Entry="" SnowDepth="4" />
<ROW RowState="13" EntryDate="20101126" Entry="Prvi sneg to zimo" SnowDepth="10" />
</podatki>
Then you could use an expat parser.
import xml.parsers.expat
def podatki(name, attrs):
if name == "ROW":
print(f'EntryDate={attrs["EntryDate"]},',
f'SnowDepth={attrs["SnowDepth"]}')
parser = xml.parsers.expat.ParserCreate()
parser.StartElementHandler = podatki
with open('podatki.xml', 'rb') as input_file:
parser.ParseFile(input_file)
The result should be
EntryDate=20101126, SnowDepth=4
EntryDate=20101126, SnowDepth=10

How to force ElementTree to keep xmlns attribute within its original element?

I have an input XML file:
<?xml version='1.0' encoding='utf-8'?>
<configuration>
<runtime name="test" version="1.2" xmlns:ns0="urn:schemas-microsoft-com:asm.v1">
<ns0:assemblyBinding>
<ns0:dependentAssembly />
</ns0:assemblyBinding>
</runtime>
</configuration>
...and Python script:
import xml.etree.ElementTree as ET
file_xml = 'test.xml'
tree = ET.parse(file_xml)
root = tree.getroot()
print (root.tag)
print (root.attrib)
element_runtime = root.find('.//runtime')
print (element_runtime.tag)
print (element_runtime.attrib)
tree.write(file_xml, xml_declaration=True, encoding='utf-8', method="xml")
...which gives the following output:
>test.py
configuration
{}
runtime
{'name': 'test', 'version': '1.2'}
...and has an undesirable side-effect of modifying XML into:
<?xml version='1.0' encoding='utf-8'?>
<configuration xmlns:ns0="urn:schemas-microsoft-com:asm.v1">
<runtime name="test" version="1.2">
<ns0:assemblyBinding>
<ns0:dependentAssembly />
</ns0:assemblyBinding>
</runtime>
</configuration>
My original script modifies XML so I do have to call tree.write and save edited file. But the problem is that ElementTree parser moves xmlns attribute from runtime element up to the root element configuration which is not desirable in my case.
I can't remove xmlns attribute from the root element (remove it from the dictionary of its attributes) as it is not listed in a list of its attributes (unlike the attributes listed for runtime element).
Why does xmlns attribute never gets listed within the list of attributes for any element?
How to force ElementTree to keep xmlns attribute within its original element?
I am using Python 3.5.1 on Windows.

xml.etree.ElementTree pulls all namespaces into the first element as it internally doesn't track on which element the namespace was declared originally.
If you don't want that, you'll have to write your own serialisation logic.
The better alternative would be to use lxml instead of xml.etree, because it preserves the location where a namespace prefix is declared.

Following #mata advice, here I give an answer with an example with code and xml file attached.
The xml input is as shown in the picture (original and modified)
The python codes check the NtnlCcy Name and if it is "EUR", convert the Price to USD (by multiplying EURUSD: = 1.2) and change the NtnlCcy Name to "USD".
The python code is as follows:
from lxml import etree
pathToXMLfile = r"C:\Xiang\codes\Python\afmreports\test_original.xml"
tree = etree.parse(pathToXMLfile)
root = tree.getroot()
EURUSD = 1.2
for Rchild in root:
print ("Root child: ", Rchild.tag, ". \n")
if Rchild.tag.endswith("Pyld"):
for PyldChild in Rchild:
print ("Pyld Child: ", PyldChild.tag, ". \n")
Doc = Rchild.find('{001.003}Document')
FinInstrNodes = Doc.findall('{001.003}FinInstr')
for FinInstrNode in FinInstrNodes:
FinCcyNode = FinInstrNode.find('{001.003}NtnlCcy')
FinPriceNode = FinInstrNode.find('{001.003}Price')
FinCcyNodeText = ""
if FinCcyNode is not None:
CcyNodeText = FinCcyNode.text
if CcyNodeText == "EUR":
PriceText = FinPriceNode.text
Price = float(PriceText)
FinPriceNode.text = str(Price * EURUSD)
FinCcyNode.text = "USD"
tree.write(r"C:\Xiang\codes\Python\afmreports\test_modified.xml", encoding="utf-8", xml_declaration=True)
print("\n the program runs to the end! \n")
As we compare the original and modified xml files, the namespace remains unchanged, the whole structure of the xml remains unchanged, only some NtnlCcy and Price Nodes have been changed, as desired.
The only minor difference we do not want is the first line. In the original xml file, it is <?xml version="1.0" encoding="UTF-8"?>, while in the modified xml file, it is <?xml version='1.0' encoding='UTF-8'?>. The quotation sign changes from double quotation to single quotation. But we think this minor difference should not matter.
The original file context will be attached for your easy test:
<?xml version="1.0" encoding="UTF-8"?>
<BizData xmlns="001.001">
<Hdr>
<AppHdr xmlns="001.002">
<Fr>
<Id>XXX01</Id>
</Fr>
<To>
<Id>XXX02</Id>
</To>
<CreDt>2019-10-25T15:38:30</CreDt>
</AppHdr>
</Hdr>
<Pyld>
<Document xmlns="001.003">
<FinInstr>
<Id>NLENX240</Id>
<FullNm>AO.AAI</FullNm>
<NtnlCcy>EUR</NtnlCcy>
<Price>9</Price>
</FinInstr>
<FinInstr>
<Id>NLENX681</Id>
<FullNm>AO.ABN</FullNm>
<NtnlCcy>USD</NtnlCcy>
<Price>10</Price>
</FinInstr>
<FinInstr>
<Id>NLENX320</Id>
<FullNm>AO.ING</FullNm>
<NtnlCcy>EUR</NtnlCcy>
<Price>11</Price>
</FinInstr>
</Document>
</Pyld>

XPath with LXML Element

I am trying to parse an XML document using lxml etree. The XML doc I am parsing looks like this:
<?xml version="1.0" encoding="UTF-8"?>
<metadata xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://www.openarchives.org/OAI/2.0/">\t
<codeBook version="2.5" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="ddi:codebook:2_5" xsi:schemaLocation="ddi:codebook:2_5 http://www.ddialliance.org/Specification/DDI-Codebook/2.5/XMLSchema/codebook.xsd">
<docDscr>
<citation>
<titlStmt>
<titl>Test Title</titl>
</titlStmt>
<prodStmt>
<prodDate/>
</prodStmt>
</citation>
</docDscr>
<stdyDscr>
<citation>
<titlStmt>
<titl>Test Title 2</titl>
<IDNo agency="UKDA">101</IDNo>
</titlStmt>
<rspStmt>
<AuthEnty>TestAuthEntry</AuthEnty>
</rspStmt>
<prodStmt>
<copyright>Yes</copyright>
</prodStmt>
<distStmt/>
<verStmt>
<version date="">1</version>
</verStmt>
</citation>
<stdyInfo>
<subject>
<keyword>2009</keyword>
<keyword>2010</keyword>
<topcClas>CLASS</topcClas>
<topcClas>ffdsf</topcClas>
</subject>
<abstract>This is an abstract piece of text.</abstract>
<sumDscr>
<timePrd event="single">2020</timePrd>
<nation>UK</nation>
<anlyUnit>Test</anlyUnit>
<universe>test</universe>
<universe>hello</universe>
<dataKind>fdsfdsf</dataKind>
</sumDscr>
</stdyInfo>
<method>
<dataColl>
<timeMeth>test timemeth</timeMeth>
<dataCollector>test data collector</dataCollector>
<sampProc>test sampprocess</sampProc>
<deviat>test deviat</deviat>
<collMode>test collMode</collMode>
<sources/>
</dataColl>
</method>
<dataAccs>
<setAvail>
<accsPlac>Test accsPlac</accsPlac>
</setAvail>
<useStmt>
<restrctn>NONE</restrctn>
</useStmt>
</dataAccs>
<othrStdyMat>
<relPubl>122</relPubl>
<relPubl>12332</relPubl>
</othrStdyMat>
</stdyDscr>
</codeBook>
</metadata>
I wrote the following code to try and process it:
from lxml import etree
import pdb
f = open('/vagrant/out2.xml', 'r')
xml_str = f.read()
xml_doc = etree.fromstring(xml_str)
f.close()
From what I understand from the lxml xpath docs, I should be able to get the text from a specific element as follows:
xml_doc.xpath('/metadata/codeBook/docDscr/citation/titlStmt/titl/text()')
However, when I run this it returns an empty array.
The only xpath I can get to return something is using a wildcard:
xml_doc.xpath('*')
Which returns [<Element {ddi:codebook:2_5}codeBook at 0x7f8da8a413f8>].
I've read through the docs and I'm not understanding what is going wrong with this. Any help is appreciated.

You need to take the default namespace into account so instead of
xml_doc.xpath('/metadata/codeBook/docDscr/citation/titlStmt/titl/text()')
use
xml_doc.xpath.xpath(
'/oai:metadata/ddi:codeBook/ddi:docDscr/ddi:citation/ddi:titlStmt/ddi:titl/text()',
namespaces={
'oai': 'http://www.openarchives.org/OAI/2.0/',
'ddi': 'ddi:codebook:2_5'
}
)