擅长:python、mysql、java
<p>您可以使用<a href="https://docs.python.org/2/library/functions.html#filter" rel="nofollow noreferrer">filter</a>:</p>
<pre><code>>>> list(filter(lambda x:not any(w in x for w in wrong), old))
['Hi Whats with ', 'binga dingo']
</code></pre>
<p>或者,a <a href="https://docs.python.org/2/tutorial/datastructures.html#list-comprehensions" rel="nofollow noreferrer">list comprehension</a></p>
<pre><code>>>> [i for i in old if not any(x in i for x in wrong)]
['Hi Whats with ', 'binga dingo']
</code></pre>
<p>如果您对其中任何一个都不满意,请使用以下基于for循环的简单解决方案:</p>
<pre><code>>>> result = []
>>> for i in old:
... for x in wrong:
... if x in i:
... break
... else:
... result.append(i)
...
>>> result
['Hi Whats with ', 'binga dingo']
</code></pre>