Search and remove element with elementTree in Python
Question:
I have an XML document in which I want to search for some elements and if they match some criteria
I would like to delete them
However, I cannot seem to be able to access the parent of the element so that I can delete it
file = open('test.xml', "r")
elem = ElementTree.parse(file)
namespace = "{http://somens}"
props = elem.findall('.//{0}prop'.format(namespace))
for prop in props:
type = prop.attrib.get('type', None)
if type == 'json':
value = json.loads(prop.attrib['value'])
if value['name'] == 'Page1.Button1':
#here I need to access the parent of prop
# in order to delete the prop
Is there a way I can do this?
Thanks
Answers:
You can remove child elements with the according remove method. To remove an element you have to call its parents remove method. Unfortunately Element does not provide a reference to its parents, so it is up to you to keep track of parent/child relations (which speaks against your use of elem.findall())
A proposed solution could look like this:
root = elem.getroot()
for child in root:
if child.name != "prop":
continue
if True:# TODO: do your check here!
root.remove(child)
PS: don’t use prop.attrib.get(), use prop.get(), as explained here.
You could use xpath to select an Element’s parent.
file = open('test.xml', "r")
elem = ElementTree.parse(file)
namespace = "{http://somens}"
props = elem.findall('.//{0}prop'.format(namespace))
for prop in props:
type = prop.get('type', None)
if type == 'json':
value = json.loads(prop.attrib['value'])
if value['name'] == 'Page1.Button1':
# Get parent and remove this prop
parent = prop.find("..")
parent.remove(prop)
http://docs.python.org/2/library/xml.etree.elementtree.html#supported-xpath-syntax
Except if you try that it doesn’t work: http://elmpowered.skawaii.net/?p=74
So instead you have to:
file = open('test.xml', "r")
elem = ElementTree.parse(file)
namespace = "{http://somens}"
search = './/{0}prop'.format(namespace)
# Use xpath to get all parents of props
prop_parents = elem.findall(search + '/..')
for parent in prop_parents:
# Still have to find and iterate through child props
for prop in parent.findall(search):
type = prop.get('type', None)
if type == 'json':
value = json.loads(prop.attrib['value'])
if value['name'] == 'Page1.Button1':
parent.remove(prop)
It is two searches and a nested loop. The inner search is only on Elements known to contain props as first children, but that may not mean much depending on your schema.
Using the fact that every child must have a parent, I’m going to simplify @kitsu.eb’s example. f using the findall command to get the children and parents, their indices will be equivalent.
file = open('test.xml', "r")
elem = ElementTree.parse(file)
namespace = "{http://somens}"
search = './/{0}prop'.format(namespace)
# Use xpath to get all parents of props
prop_parents = elem.findall(search + '/..')
props = elem.findall('.//{0}prop'.format(namespace))
for prop in props:
type = prop.attrib.get('type', None)
if type == 'json':
value = json.loads(prop.attrib['value'])
if value['name'] == 'Page1.Button1':
#use the index of the current child to find
#its parent and remove the child
prop_parents[props.index[prop]].remove(prop)
I know this is an old thread but this kept popping up while I was trying to figure out a similar task. I did not like the accepted answer for two reasons:
1) It doesn’t handle multiple nested levels of tags.
2) It will break if multiple xml tags are deleted in the same level one-after-another. Since each element is an index of Element._children you shouldn’t delete while forward iterating.
I think a better more versatile solution is this:
import xml.etree.ElementTree as et
file = 'test.xml'
tree = et.parse(file)
root = tree.getroot()
def iterator(parents, nested=False):
for child in reversed(parents):
if nested:
if len(child) >= 1:
iterator(child)
if True: # Add your entire condition here
parents.remove(child)
iterator(root, nested=True)
For the OP, this should work – but I don’t have the data you’re working with to test if it’s perfect.
import xml.etree.ElementTree as et
file = 'test.xml'
tree = et.parse(file)
namespace = "{http://somens}"
props = tree.findall('.//{0}prop'.format(namespace))
def iterator(parents, nested=False):
for child in reversed(parents):
if nested:
if len(child) >= 1:
iterator(child)
if prop.attrib.get('type') == 'json':
value = json.loads(prop.attrib['value'])
if value['name'] == 'Page1.Button1':
parents.remove(child)
iterator(props, nested=True)
I like to use an XPath expression for this kind of filtering. Unless I know otherwise, such an expression must be applied at the root level, which means I can’t just get a parent and apply the same expression on that parent. However, it seems to me that there is a nice and flexible solution that should work with any supported XPath, as long as none of the sought nodes is the root. It goes something like this:
root = elem.getroot()
# Find all nodes matching the filter string (flt)
nodes = root.findall(flt)
while len(nodes):
# As long as there are nodes, there should be parents
# Get the first of all parents to the found nodes
parent = root.findall(flt+'/..')[0]
# Use this parent to remove the first node
parent.remove(nodes[0])
# Find all remaining nodes
nodes = root.findall(flt)
A solution using lxml module
from lxml import etree
root = ET.fromstring(xml_str)
for e in root.findall('.//{http://some.name.space}node'):
parent = e.getparent()
for child in parent.find('./{http://some.name.space}node'):
try:
parent.remove(child)
except ValueError:
pass
I also used XPath for this issue, but in a different way:
root = elem.getroot()
elementName = "YourElement"
#this will find all the parents of the elements with elementName
for elementParent in root.findall(".//{}/..".format(elementName)):
#this will find all the elements under the parent, and remove them
for element in elementParent.findall("{}".format(elementName)):
elementParent.remove(element)
I would like only to add a comment on the accepted answer, but my lack of reputation doesn’t allow me to. I wanted to add that it is important to add .findall("*")to the iterator to avoid issues, as stated in the documentation:
Note that concurrent modification while iterating can lead to problems, just like when iterating and modifying Python lists or dicts. Therefore, the example first collects all matching elements with root.findall(), and only then iterates over the list of matches.
Therefore, in the accepted answer the iteration should be for child in root.findal("*"):instead of for child in root:. Not doing so made my code skip some elements from the list.
I have an XML document in which I want to search for some elements and if they match some criteria
I would like to delete them
However, I cannot seem to be able to access the parent of the element so that I can delete it
file = open('test.xml', "r")
elem = ElementTree.parse(file)
namespace = "{http://somens}"
props = elem.findall('.//{0}prop'.format(namespace))
for prop in props:
type = prop.attrib.get('type', None)
if type == 'json':
value = json.loads(prop.attrib['value'])
if value['name'] == 'Page1.Button1':
#here I need to access the parent of prop
# in order to delete the prop
Is there a way I can do this?
Thanks
You can remove child elements with the according remove method. To remove an element you have to call its parents remove method. Unfortunately Element does not provide a reference to its parents, so it is up to you to keep track of parent/child relations (which speaks against your use of elem.findall())
A proposed solution could look like this:
root = elem.getroot()
for child in root:
if child.name != "prop":
continue
if True:# TODO: do your check here!
root.remove(child)
PS: don’t use prop.attrib.get(), use prop.get(), as explained here.
You could use xpath to select an Element’s parent.
file = open('test.xml', "r")
elem = ElementTree.parse(file)
namespace = "{http://somens}"
props = elem.findall('.//{0}prop'.format(namespace))
for prop in props:
type = prop.get('type', None)
if type == 'json':
value = json.loads(prop.attrib['value'])
if value['name'] == 'Page1.Button1':
# Get parent and remove this prop
parent = prop.find("..")
parent.remove(prop)
http://docs.python.org/2/library/xml.etree.elementtree.html#supported-xpath-syntax
Except if you try that it doesn’t work: http://elmpowered.skawaii.net/?p=74
So instead you have to:
file = open('test.xml', "r")
elem = ElementTree.parse(file)
namespace = "{http://somens}"
search = './/{0}prop'.format(namespace)
# Use xpath to get all parents of props
prop_parents = elem.findall(search + '/..')
for parent in prop_parents:
# Still have to find and iterate through child props
for prop in parent.findall(search):
type = prop.get('type', None)
if type == 'json':
value = json.loads(prop.attrib['value'])
if value['name'] == 'Page1.Button1':
parent.remove(prop)
It is two searches and a nested loop. The inner search is only on Elements known to contain props as first children, but that may not mean much depending on your schema.
Using the fact that every child must have a parent, I’m going to simplify @kitsu.eb’s example. f using the findall command to get the children and parents, their indices will be equivalent.
file = open('test.xml', "r")
elem = ElementTree.parse(file)
namespace = "{http://somens}"
search = './/{0}prop'.format(namespace)
# Use xpath to get all parents of props
prop_parents = elem.findall(search + '/..')
props = elem.findall('.//{0}prop'.format(namespace))
for prop in props:
type = prop.attrib.get('type', None)
if type == 'json':
value = json.loads(prop.attrib['value'])
if value['name'] == 'Page1.Button1':
#use the index of the current child to find
#its parent and remove the child
prop_parents[props.index[prop]].remove(prop)
I know this is an old thread but this kept popping up while I was trying to figure out a similar task. I did not like the accepted answer for two reasons:
1) It doesn’t handle multiple nested levels of tags.
2) It will break if multiple xml tags are deleted in the same level one-after-another. Since each element is an index of Element._children you shouldn’t delete while forward iterating.
I think a better more versatile solution is this:
import xml.etree.ElementTree as et
file = 'test.xml'
tree = et.parse(file)
root = tree.getroot()
def iterator(parents, nested=False):
for child in reversed(parents):
if nested:
if len(child) >= 1:
iterator(child)
if True: # Add your entire condition here
parents.remove(child)
iterator(root, nested=True)
For the OP, this should work – but I don’t have the data you’re working with to test if it’s perfect.
import xml.etree.ElementTree as et
file = 'test.xml'
tree = et.parse(file)
namespace = "{http://somens}"
props = tree.findall('.//{0}prop'.format(namespace))
def iterator(parents, nested=False):
for child in reversed(parents):
if nested:
if len(child) >= 1:
iterator(child)
if prop.attrib.get('type') == 'json':
value = json.loads(prop.attrib['value'])
if value['name'] == 'Page1.Button1':
parents.remove(child)
iterator(props, nested=True)
I like to use an XPath expression for this kind of filtering. Unless I know otherwise, such an expression must be applied at the root level, which means I can’t just get a parent and apply the same expression on that parent. However, it seems to me that there is a nice and flexible solution that should work with any supported XPath, as long as none of the sought nodes is the root. It goes something like this:
root = elem.getroot()
# Find all nodes matching the filter string (flt)
nodes = root.findall(flt)
while len(nodes):
# As long as there are nodes, there should be parents
# Get the first of all parents to the found nodes
parent = root.findall(flt+'/..')[0]
# Use this parent to remove the first node
parent.remove(nodes[0])
# Find all remaining nodes
nodes = root.findall(flt)
A solution using lxml module
from lxml import etree
root = ET.fromstring(xml_str)
for e in root.findall('.//{http://some.name.space}node'):
parent = e.getparent()
for child in parent.find('./{http://some.name.space}node'):
try:
parent.remove(child)
except ValueError:
pass
I also used XPath for this issue, but in a different way:
root = elem.getroot()
elementName = "YourElement"
#this will find all the parents of the elements with elementName
for elementParent in root.findall(".//{}/..".format(elementName)):
#this will find all the elements under the parent, and remove them
for element in elementParent.findall("{}".format(elementName)):
elementParent.remove(element)
I would like only to add a comment on the accepted answer, but my lack of reputation doesn’t allow me to. I wanted to add that it is important to add .findall("*")to the iterator to avoid issues, as stated in the documentation:
Note that concurrent modification while iterating can lead to problems, just like when iterating and modifying Python lists or dicts. Therefore, the example first collects all matching elements with root.findall(), and only then iterates over the list of matches.
Therefore, in the accepted answer the iteration should be for child in root.findal("*"):instead of for child in root:. Not doing so made my code skip some elements from the list.