0

Using ElementTree with Python 3.8, how can I convert the data into a Pandas dataframe?

Example XML:

<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<SystemSourceSet id="-1" name="UC33brAvg_FM31" weight="0.5">
  <!-- This model is an example and for review purposes only -->
  <!-- Reference: UC33brAvg_FM31 -->
  <!-- Description: UCERF 3.3 Branch Averaged Solution (FM31)-->
  <Settings>
    <DefaultMfds>
      <IncrementalMfd floats="false" m="6.5" rate="0.0" type="SINGLE" weight="1.0"/>
    </DefaultMfds>
  </Settings>
  <Source>
    <IncrementalMfd m="6.449" rate="3.3631184e-05" type="SINGLE"/>
    <Geometry depth="1.3" dip="50.0" indices="0:1" rake="-90.0" width="15.273"/>
  </Source>
  <Source>
    <IncrementalMfd m="6.638" rate="1.5340160e-05" type="SINGLE"/>
    <Geometry depth="1.3" dip="50.0" indices="0:2" rake="-90.0" width="15.273"/>
  </Source>
  <Source>
    <IncrementalMfd m="6.78" rate="1.0903030e-05" type="SINGLE"/>
    <Geometry depth="1.3" dip="50.0" indices="0:3" rake="-90.0" width="15.273"/>
  </Source>
  <Source>
    <IncrementalMfd m="6.893" rate="7.3397665e-06" type="SINGLE"/>
    <Geometry depth="1.3" dip="50.0" indices="0:4" rake="-90.0" width="15.273"/>
  </Source>

Expected Dataframe:

enter image description here

1 Answer 1

2

Navigate the tree manually and collect the data points you want to keep:

from xml.etree import ElementTree

root = ElementTree.parse('data.xml').getroot()
data = []
for node in root:
    if node.tag != 'Source':
        continue
    
    mfd = node.find('IncrementalMfd')
    geometry = node.find('Geometry')
    data.append({
        'indices': geometry.get('indices'),
        'IncrementalMfd m': mfd.get('m'),
        'rate': mfd.get('rate'),
        'type': mfd.get('type'),
        'Geometry depth': geometry.get('depth'),
        'dip': geometry.get('dip'),
        'rake': geometry.get('rake'),
        'width': geometry.get('width')
    })
    
df = pd.DataFrame(data)
Sign up to request clarification or add additional context in comments.

1 Comment

Very slick solution.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.