问题描述
我在写这个问题的答案时考虑了以下.
I thought about the following while writing an answer to this question.
假设我有一个像这样深度嵌套的 xml
文件(但嵌套更多且更长):
Suppose I have a deeply nested xml
file like this (but much more nested and much longer):
<section name="1">
<subsection name"foo">
<subsubsection name="bar">
<deeper name="hey">
<much_deeper name"yo">
<li>Some content</li>
</much_deeper>
</deeper>
</subsubsection>
</subsection>
</section>
<section name="2">
... and so forth
</section>
len(soup.find_all("section"))
的问题在于,在执行 find_all("section")
时,BS 一直在深入搜索一个标签我知道不会包含任何其他 section
标记.
The problem with len(soup.find_all("section"))
is that while doing find_all("section")
, BS keeps searching deep into a tag that I know won't contain any other section
tag.
那么,两个问题:
- 有没有办法让 BS 不递归搜索到已经找到的标签?
- 如果对 1 的回答是肯定的,是效率更高还是内部流程相同?
- Is there a way to make BS not search recursively into an already found tag?
- If the answer to 1 is yes, will it be more efficient or is it the same internal process?
推荐答案
BeautifulSoup
不能只提供它找到的标签的计数/数量.
BeautifulSoup
cannot give you just a count/number of tags it found.
不过,您可以改进的是:不要让 BeautifulSoup
通过传递 recursive=False
来搜索其他部分中的部分:
What you, though, can improve is: don't let BeautifulSoup
go searching sections inside other sections by passing recursive=False
:
len(soup.find_all("section", recursive=False))
除了改进之外,lxml
会更快地完成这项工作:
Aside from that improvement, lxml
would do the job faster:
tree.xpath('count(//section)')
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