Python Beautiful Soup 最高效的标签查找方式

Question

我正在使用 python 和 BeautifulSoup 解析许多大型 XML 文件。我经常运行进入以下任务：

<Section1>
    <Report>
        <Matrix>...</Matrix>
        <Matrix>...</Matrix>
        <Matrix>...</Matrix>
        <Matrix>...</Matrix>
    </Report>
</Section1>

我正在尝试收集并遍历所有矩阵。我使用如下代码：

res = urlopen(url)
html = res.read()
soup = BeautifulSoup(html, 'xml')
matrices = soup.find("Section1").find_all("Matrix")
#Then I handle each matrix

为什么我不能使用这样的选择器？

matrices = soup.find("Section1 Matrix")

有更快的方法吗？有时我正在访问嵌套在 XML 中更远的节点，我需要确保它们是后代，但不一定是其他几个节点的直接子节点。提供的示例是一种简化。任何帮助将不胜感激。

Answer 1

BeautifulSoup "supports CSS selectors" 您需要将选择器传递给 .select 方法

In [1]: from bs4 import BeautifulSoup as BS

In [2]: soup = BS("""<Section1>
   ...:     <Report>
   ...:         <Matrix>...</Matrix>
   ...:         <Matrix>...</Matrix>
   ...:         <Matrix>...</Matrix>
   ...:         <Matrix>...</Matrix>
   ...:     </Report>
   ...: </Section1>""", "xml")

In [3]: soup.select("Section1 Matrix")
Out[3]: 
[<Matrix>...</Matrix>,
 <Matrix>...</Matrix>,
 <Matrix>...</Matrix>,
 <Matrix>...</Matrix>]

如果您想要获取文档中的所有 Matrix 个节点；您可以使用 CSSSelector class from lxml.cssselect¹.

In [3]: from lxml.etree import fromstring

In [4]: xml_doc = '''<Section1>
   ...:     <Report>
   ...:         <Matrix>...</Matrix>
   ...:         <Matrix>...</Matrix>
   ...:         <Matrix>...</Matrix>
   ...:         <Matrix>...</Matrix>
   ...:     </Report>
   ...: </Section1>'''

In [5]: tree = fromstring(xml_doc)

In [6]: matrix = [el for el in sel(tree)]

In [7]: matrix
Out[7]: 
[<Element Matrix at 0x7f84b5b8f388>,
 <Element Matrix at 0x7f84b5b8fc48>,
 <Element Matrix at 0x7f84b5b8fd88>,
 <Element Matrix at 0x7f84b5b8fdc8>]

¹ 如果 cssselect 尚未与 pip 一起安装，则需要安装它：pip install cssselect

Python Beautiful Soup 最高效的标签查找方式

Python Beautiful Soup Most Efficient Way to Find Tags

python

xml

beautifulsoup

css-selectors