使用元素树python3解析XML文件

Question

我想解析下面给出的 XML 文件片段以提取视点标记及其属性名称。 我还想创建一个表来将提取的数据制成表格。

我的 XML 文件片段：

 <windows source-height='51'>
        <window class='dashboard' maximized='true' name='Figure 8-59'>
          <viewpoints>
            <viewpoint name='Good Filter Design'>
              <zoom type='entire-view' />
              <geo-search-visibility value='1' />
            </viewpoint>
            <viewpoint name='Poor Filter Design'>
              <zoom type='entire-view' />
              <geo-search-visibility value='1' />
            </viewpoint>
          </viewpoints>
          <active id='-1' />
        </window>
        <window class='dashboard' name='Figure 8-60 thought 8-65'>
          <viewpoints>
            <viewpoint name='Heat Map'>
              <zoom type='entire-view' />
              <geo-search-visibility value='1' />
            </viewpoint>
            <viewpoint name='Lightbulb'>
              <zoom type='entire-view' />
              <geo-search-visibility value='1' />
            </viewpoint>
            <viewpoint name='Sales Histogram'>
              <zoom type='entire-view' />
              <geo-search-visibility value='1' />
            </viewpoint>
          </viewpoints>
          <active id='-1' />
        </window>
</windows>

我想将"good filter design"和"poor filter design"提取并保留在一行中，并将其余三个视点名称作为第二行。

我的尝试：

root = getroot('example.xml')
for i in root.findall('windows/window/viewpoints/viewpoint'):
    print(i.get('name'))

Answer 1

使用 elementtree 应该很容易。 我不知道getroot()到底做了什么，但如果它真的返回 XML 文档的根元素，那么你不应该在findall参数中提到window ：

>>> from xml.etree import ElementTree as ET
>>> raw = '''your XML string'''
>>> root = ET.fromstring(raw)
>>> for v in root.findall('window/viewpoints'):
...     print([a.get('name') for a in v.findall('viewpoint')])
... 
['Good Filter Design', 'Poor Filter Design']
['Heat Map', 'Lightbulb', 'Sales Histogram']

demo

Answer 2

如果你可以很容易地使用beautifulsoup，那就是

from bs4 import BeautifulSoup
#xml = """your xml"""
soup = BeautifulSoup(xml, 'lxml')
names = [viewpt["name"] for viewpt in soup.find_all('viewpoint')]

这将给每个名为“viewpoint”的标签

如果你只想嵌套一个使用这个：

names = [viewpoint["name"]
        for windows in soup.find_all('windows')
            for window in windows.find_all("window")
                for viewpoints in window.find_all("viewpoints")
                    for viewpoint in viewpoints.find_all("viewpoint")]

在你的情况下，两者都会给出：

Out[18]: 
['Good Filter Design',
 'Poor Filter Design',
 'Heat Map',
 'Lightbulb',
 'Sales Histogram']