将不同大小的嵌套XML元素提取到

Question

假设我们有一个任意的XML文档，如下所示

<?xml version="1.0" encoding="UTF-8"?>
<programs xmlns="http://something.org/schema/s/program">
   <program xmlns:xsd="http://www.w3.org/2001/XMLSchema"
            xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
            xsi:schemaLocation="http://something.org/schema/s/program  http://something.org/schema/s/program.xsd">
      <orgUnitId>Organization 1</orgUnitId>
      <requiredLevel>academic bachelor</requiredLevel>
      <requiredLevel>academic master</requiredLevel>
      <programDescriptionText xml:lang="nl">Here is some text; blablabla</programDescriptionText>
      <searchword xml:lang="nl">Scrum master</searchword>
   </program>
   <program xmlns:xsd="http://www.w3.org/2001/XMLSchema" 
            xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
            xsi:schemaLocation="http://something.org/schema/s/program  http://something.org/schema/s/program.xsd">
      <requiredLevel>bachelor</requiredLevel>
      <requiredLevel>academic master</requiredLevel>
      <requiredLevel>academic bachelor</requiredLevel>
      <orgUnitId>Organization 2</orgUnitId>
      <programDescriptionText xml:lang="nl">Text from another organization about some stuff.</programDescriptionText>
      <searchword xml:lang="nl">Excutives</searchword>
   </program>
   <program xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
      <orgUnitId>Organization 3</orgUnitId>
      <programDescriptionText xml:lang="nl">Also another huge text description from another organization.</programDescriptionText>
      <searchword xml:lang="nl">Negotiating</searchword>
      <searchword xml:lang="nl">Effective leadership</searchword>
      <searchword xml:lang="nl">negotiating techniques</searchword>
      <searchword xml:lang="nl">leadership</searchword>
      <searchword xml:lang="nl">strategic planning</searchword>
   </program>
</programs>

目前，我通过使用元素的绝对路径来looping遍历所需的元素，因为我无法使用ElementTree中的任何get或find方法。因此，我的代码如下所示：

import pandas as pd
import xml.etree.ElementTree as ET   
import numpy as np
import itertools

tree = ET.parse('data.xml')
root = tree.getroot()
root.tag

dfcols=['organization','description','level','keyword']
organization=[]
description=[]
level=[]
keyword=[]

for node in root:
    for child in 
       node.findall('.//{http://something.org/schema/s/program}orgUnitId'):
        organization.append(child.text) 
    for child in node.findall('.//{http://something.org/schema/s/program}programDescriptionText'):
        description.append(child.text) 
    for child in node.findall('.//{http://something.org/schema/s/program}requiredLevel'):
        level.append(child.text)
    for child in node.findall('.//{http://something.org/schema/s/program}searchword'):
        keyword.append(child.text)

当然，目标是创建一个数据帧。但是，由于XML文件中的每个节点都包含一个或多个元素，例如requiredLevel或searchword，因此在通过以下方式将数据强制转换到数据帧时，当前正在丢失数据：

df=pd.DataFrame(list(itertools.zip_longest(organization,
    description,level,searchword,
    fillvalue=np.nan)),columns=dfcols)

或者使用pd.Series给定的here或者另一个我似乎无法从here得到的解决方案

我最好的办法是根本不使用列表，因为它们似乎不能正确地索引数据。也就是说，我丢失了第2到第x个子节点的数据。但现在我被困住了，没有其他选择。你知道吗

我的最终结果应该是这样的：

organization    description  level                keyword
Organization 1  ....         academic bachelor,   Scrum master
                             academic master 
Organization 2  ....         bachelor,            Executives
                             academic master, 
                             academic bachelor    
Organization 3  ....                              Negotiating,
                                                  Effective leadership,
                                                  negotiating techniques,
                                                  ....

Answer 1

一个轻量级的xml_to_dict转换器可以找到here。可以通过this处理名称空间来改进它。你知道吗

def xml_to_dict(xml='', remove_namespace=True):
    """Converts an XML string into a dict

    Args:
        xml: The XML as string
        remove_namespace: True (default) if namespaces are to be removed

    Returns:
        The XML string as dict

    Examples:
        >>> xml_to_dict('<text><para>hello world</para></text>')
        {'text': {'para': 'hello world'}}

    """
    def _xml_remove_namespace(buf):
        # Reference: https://stackoverflow.com/a/25920989/1498199
        it = ElementTree.iterparse(buf)
        for _, el in it:
            if '}' in el.tag:
                el.tag = el.tag.split('}', 1)[1]
        return it.root

    def _xml_to_dict(t):
        # Reference: https://stackoverflow.com/a/10077069/1498199
        from collections import defaultdict

        d = {t.tag: {} if t.attrib else None}
        children = list(t)
        if children:
            dd = defaultdict(list)
            for dc in map(_xml_to_dict, children):
                for k, v in dc.items():
                    dd[k].append(v)
            d = {t.tag: {k: v[0] if len(v) == 1 else v for k, v in dd.items()}}

        if t.attrib:
            d[t.tag].update(('@' + k, v) for k, v in t.attrib.items())

        if t.text:
            text = t.text.strip()
            if children or t.attrib:
                if text:
                    d[t.tag]['#text'] = text
            else:
                d[t.tag] = text

        return d

    buffer = io.StringIO(xml.strip())
    if remove_namespace:
        root = _xml_remove_namespace(buffer)
    else:
        root = ElementTree.parse(buffer).getroot()

    return _xml_to_dict(root)

因此，让s成为保存xml的字符串。我们可以通过

d = xml_to_dict(s, remove_namespace=True)

现在的解决方案是直截了当的：

rows = []
for program in d['programs']['program']:
    cols = []
    cols.append(program['orgUnitId'])
    cols.append(program['programDescriptionText']['#text'])
    try:
        cols.append(','.join(program['requiredLevel']))
    except KeyError:
        cols.append('')

    try:
         searchwords = program['searchword']['#text']
    except TypeError:
         searchwords = []
         for searchword in program['searchword']:
            searchwords.append(searchword['#text'])
         searchwords = ','.join(searchwords)
    cols.append(searchwords)

    rows.append(cols)

df = pd.DataFrame(rows, columns=['organization', 'description', 'level', 'keyword'])