我想使用正则表达式匹配模式并提取模式的一部分。
我已经抓取了HTML数据,一个说明性代码段如下所示:
</script>
</li>
<li itemprop="itemListElement" itemscope="" itemtype="http://schema.org/ListItem">
<span class="hide" itemprop="position">1</span>
<div class="result-heading">
<a class="project-icon show-outline" href="/projects/quickfixj/" title="Find out more about QuickFIX/J - Open Source Java FIX Engine">
<img alt="QuickFIX/J - Open Source Java FIX Engine Icon" src="//a.fsdn.com/allura/p/quickfixj/icon?1533295730"/></a>
<div class="result-heading-texts">
<a href="/projects/quickfixj/" itemprop="url" title="Find out more
<a href="/projects/desmoj/" itemprop="url" title="Find out more about DESMO-J"><h2>DESMO-J</h2></a>
<div class="description">
<p class="description-inner">DESMO-<em>J</em> is a framework for
<a href="/projects/desmoj/files/stats/timeline" title="Downloads This Week">29 This Week</a>
</strong>
<strong>
find_all('a')
中更具代表性的子集突出显示问题:
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HTML当前存储为BeautifulSoup对象,即已通过以下方式传递:
html_soup= BeautifulSoup(response.text, 'html.parser')
我想在整个对象中搜索/projects/
的所有实例,并提取后续斜杠之间的字符串。例如:
from "/projects/quickfixj/" I would like to store "quickfixj".
我最初的想法是使用re.findall()
并尝试匹配(/projects/./)*
,但这不起作用。
非常感谢您的帮助。
答案 0 :(得分:1)
您已经走了一半
a='''</script>
</li>
<li itemprop="itemListElement" itemscope="" itemtype="http://schema.org/ListItem">
<span class="hide" itemprop="position">1</span>
<div class="result-heading">
<a class="project-icon show-outline" href="/projects/quickfixj/" title="Find out more about QuickFIX/J - Open Source Java FIX Engine">
<img alt="QuickFIX/J - Open Source Java FIX Engine Icon" src="//a.fsdn.com/allura/p/quickfixj/icon?1533295730"/></a>
<div class="result-heading-texts">
<a href="/projects/quickfixj/" itemprop="url" title="Find out more
<a href="/projects/desmoj/" itemprop="url" title="Find out more about DESMO-J"><h2>DESMO-J</h2></a>
<div class="description">
<p class="description-inner">DESMO-<em>J</em> is a framework for
<a href="/projects/desmoj/files/stats/timeline" title="Downloads This Week">29 This Week</a>
</strong>
<strong>'''
from bs4 import BeautifulSoup
soup = BeautifulSoup(a,"html.parser")
for i in soup.find_all('a'):
print(re.findall('/projects/(\w{1,})/',i.get('href')))
如果您需要独特的项目。将最后几行更改为
from bs4 import BeautifulSoup
soup = BeautifulSoup(a,"html.parser")
project_set=set()
for i in soup.find_all('a'):
project_set.add(*re.findall('/projects/(\w{1,})/',i.get('href')))
print(project_set) #{u'desmoj', u'quickfixj'}
答案 1 :(得分:0)
您可以提取所有链接,然后应用正则表达式:
from bs4 import BeautifulSoup
html = '''</script>
</li>
<li itemprop="itemListElement" itemscope="" itemtype="http://schema.org/ListItem">
<span class="hide" itemprop="position">1</span>
<div class="result-heading">
<a class="project-icon show-outline" href="/projects/quickfixj/" title="Find out more about QuickFIX/J - Open Source Java FIX Engine">
<img alt="QuickFIX/J - Open Source Java FIX Engine Icon" src="//a.fsdn.com/allura/p/quickfixj/icon?1533295730"/></a>
<div class="result-heading-texts">
<a href="/projects/quickfixj/" itemprop="url" title="Find out more
<a href="/projects/desmoj/" itemprop="url" title="Find out more about DESMO-J"><h2>DESMO-J</h2></a>
<div class="description">
<p class="description-inner">DESMO-<em>J</em> is a framework for
<a href="/projects/desmoj/files/stats/timeline" title="Downloads This Week">29 This Week</a>
</strong>
<strong>'''
html_soup = BeautifulSoup(html, 'html.parser')
links = [i.get('href') for i in html_soup.find_all('a', href=True)]
收益:
['/projects/quickfixj/', '/projects/quickfixj/', '/projects/desmoj/files/stats/timeline']
然后您可以应用正则表达式:
cleaned = [re.findall(r'(?<=projects\/)(.*?)\/', i)[0] for i in links]
收益:
['quickfixj', 'quickfixj', 'desmoj']
答案 2 :(得分:0)
这样的Regex应该可以解决问题(?<=\/projects\/).+?(?=\/)
可以这样工作
import re
regex = "(?<=\/projects\/).+?(?=\/)"
string = "<a href="/projects/quickfixj/" itemprop="url" title="Find out more...."
matches = re.findall(regex, string)
print(matches)
输出:["quickfixj"]