擅长:python、mysql、java
<p>试试这个:</p>
<pre><code>lst = [('word1', 'The word1 is in this sentence'), ('word2', 'The word2blawblaw is in this sentence'),
('word3', 'The word1 is in this sentence')]
for word, sentence in lst:
print(' '.join(i for i in sentence.split() if word not in i))
</code></pre>
<p>输出:</p>
<pre><code>The is in this sentence
The is in this sentence
The word1 is in this sentence
</code></pre>
<p>我故意把<code>word1</code>放在第三句</p>
<p><strong>说明:</strong></p>
<p>首先我们迭代<code>lst</code>,它在每次迭代中给我们一个元组,我们用<code>word, sentence</code>解压它</p>
<p>在分割<code>sentence</code>之后,它成为一个列表,即<code>['The', 'word1', 'is', 'in', 'this', 'sentence']</code>(在第一次迭代中)</p>
<p>然后我们再次遍历这个列表,它给出了单个单词,我们所要做的就是检查我们的<code>word</code>是否在这些单词中。如果它不在那里,那就是我们想要的</p>
<p>最后我们做<code>' '.join()</code>来造句</p>