假设我们有一个如下所示的任意XML文档
<?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>
当前，由于无法使用ElementTree中的任何looping或get方法，因此我使用它们的绝对路径来find。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框。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，所以当我通过以下任一方法将其投射到数据帧时，我当前正在丢失数据：
df=pd.DataFrame(list(itertools.zip_longest(organization,
description,level,searchword,
fillvalue=np.nan)),columns=dfcols)
或使用here给出的pd.Series或其他似乎无法从here适应的解决方案
我最好的选择是根本不使用列表，因为它们似乎无法正确索引数据。也就是说，我从第二个到第X个子节点丢失了数据。但是现在我被困住了，看不到其他选择。
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,
....
答案 0 :(得分：2)
请考虑使用逗号折叠的文本值构建字典列表。然后将列表传递到pandas.DataFrame构造函数中：
dicts = []
for node in root:
orgs = ", ".join([org.text for org in node.findall('.//{http://something.org/schema/s/program}orgUnitId')])
desc = ", ".join([desc.text for desc in node.findall('.//{http://something.org/schema/s/program}programDescriptionText')])
lvls = ", ".join([lvl.text for lvl in node.findall('.//{http://something.org/schema/s/program}requiredLevel')])
wrds = ", ".join([wrd.text for wrd in node.findall('.//{http://something.org/schema/s/program}searchword')])
dicts.append({'organization': orgs, 'description': desc, 'level': lvls, 'keyword': wrds})
final_df = pd.DataFrame(dicts, columns=['organization','description','level','keyword'])
输出
print(final_df)
# organization description level keyword
# 0 Organization 1 Here is some text; blablabla academic bachelor, academic master Scrum master
# 1 Organization 2 Text from another organization about some stuff. bachelor, academic master, academic bachelor Excutives
# 2 Organization 3 Also another huge text description from anothe... Negotiating, Effective leadership, negotiating...
答案 1 :(得分：1)
可以找到here的轻量级xml_to_dict转换器。 this可以改进它来处理名称空间。
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，假设s是保存您的xml的字符串。我们可以通过
d = xml_to_dict(s, remove_namespace=True)
现在解决方案很简单：
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'])