get all parents of xml node using python

Question

for this xml

<Departments orgID="123" name="xmllist">
    <Department>
        <orgID>124</orgID>
        <name>A</name>
        <type>type a</type>
        <status>Active</status>
            <Department>
                <orgID>125</orgID>
                <name>B</name>
                <type>type b</type>
                <status>Active</status>
                <Department>
                    <orgID>126</orgID>
                    <name>C</name>
                    <type>type c</type>
                    <status>Active</status>
                </Department>
            </Department>
    </Department>
    <Department>
        <orgID>109449</orgID>
        <name>D</name>
        <type>type d</type>
        <status>Active</status>
    </Department>
</Departments>

How i can get all parents of a node using lxml etree in python.

Expected output : Input orgid=126 , it will return all the parents like ,

{'A':124,'B':125,'C':126}

falsetru · Accepted Answer · 2014-02-13 06:14:12Z

7

Using lxml and XPath:

>>> s = '''
... <Departments orgID="123" name="xmllist">
...     <Department>
...         <orgID>124</orgID>
...         <name>A</name>
...         <type>type a</type>
...         <status>Active</status>
...             <Department>
...                 <orgID>125</orgID>
...                 <name>B</name>
...                 <type>type b</type>
...                 <status>Active</status>
...                 <Department>
...                     <orgID>126</orgID>
...                     <name>C</name>
...                     <type>type c</type>
...                     <status>Active</status>
...                 </Department>
...             </Department>
...     </Department>
...     <Department>
...         <orgID>109449</orgID>
...         <name>D</name>
...         <type>type d</type>
...         <status>Active</status>
...     </Department>
... </Departments>
... '''

Using ancestor-or-self axis, you can find the node itself, parent, grandparent, ...

>>> import lxml.etree as ET
>>> root = ET.fromstring(s)
>>> for target in root.xpath('.//Department/orgID[text()="126"]'):
...     d = {
...         dept.find('name').text: int(dept.find('orgID').text)
...         for dept in target.xpath('ancestor-or-self::Department')
...     }
...     print(d)
...
{'A': 124, 'C': 126, 'B': 125}

answered Feb 13, 2014 at 6:14

falsetru

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4 Comments

Nishant Nawarkhede Over a year ago

Thanks , and what if i want include orgID=123 and name = xmllist in d ?

falsetru Over a year ago

@Nishant, for depts in target.xpath('ancestor-or-self::Departments'): d[depts.get('name')] = depts.get('orgID') before print statement.

Nishant Nawarkhede Over a year ago

Thanks , But output seems to be unordered is there is any way to make it ordered ? Here we are getting {'A': 124, 'C': 126, 'B': 125} can we get it like {'A': 124, 'B': 125 ,'C': 126} ??

falsetru Over a year ago

@Nishant, dict itself is unordered data structure. Use collection.OrderedDict if you want keep order. or list, ... if you don't need to use dict-like container.

shrewmouse · Accepted Answer · 2017-05-10 16:51:22Z

Use lxml's iterancestors() method.

from lxml import etree

doc = etree.fromstring(xml)
rval = {}
for org in doc.xpath('//orgID[text()="126"]'):
    for ancestor in org.iterancestors('Department'):
        id=ancestor.find('./orgID').text
        name=ancestor.find('./name').text
        rval[name]=id

print rval

output:

{'A': '124', 'C': '126', 'B': '125'}

If you're actually trying to preserve the order of the elements then you can't use a dict because you can't control the key order in a dict. You'll have to use an OrderedDict or just and array of tuples:

doc = etree.fromstring(xml)
a = []
for org in doc.xpath('//orgID[text()="126"]'):
    for ancestor in org.iterancestors():
        if ancestor.find('./orgID') is not None:
            id=ancestor.find('./orgID').text
            name=ancestor.find('./name').text
        elif ancestor.get('orgID'):
            id=ancestor.get('orgID')
            name=ancestor.get('name')
        else:
            continue

        print id,name
        a.append((name,id))

print "In order of discovery:\n    ", a 
print "From root to child\n    ", [x for x in reversed(a)]
print "dict keys are not sorted\n    ", dict(a)

Output:

126 C
125 B
124 A
123 xmllist
In order of discovery:
     [('C', '126'), ('B', '125'), ('A', '124'), ('xmllist', '123')]
From root to child
     [('xmllist', '123'), ('A', '124'), ('B', '125'), ('C', '126')]
dict keys are not sorted
     {'A': '124', 'xmllist': '123', 'C': '126', 'B': '125'}

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