擅长:python、mysql、java
<p>如果您只想获取与条件匹配的URL,下面是代码:</p>
<pre><code>urls = df.groupby(level = 0).agg({'links': (lambda x: sum([(f in list(x.str.extract('(archive|xml)', expand=False))) for f in ['archive','xml']])==2)})['links']
print(urls)
Out[1]:
https://example.com False
https://example222.com False
https://example333.com True
Name: links, dtype: bool
print(list(urls[urls].index))
Out[2]:
['https://example333.com']
</code></pre>