擅长:python、mysql、java
<p>如果在列表理解中找到任何筛选器,则可以通过删除url在其中使用<code>any()</code>:</p>
<pre><code>from itertools import chain
p = [['www.sample.de/fl/autor/xxx',
'www.sample.de/fl/autor/xxx',
'www.sample.de/fl/autor/xxx',],
['www.temp.de/thema/xxx',
'www.temp.de/thema/xxx',]
]
filters = ['/autor/', '/thema/' ]
p = [x for x in chain.from_iterable(p) if not any(f in x for f in filters)]
# []
</code></pre>