擅长:python、mysql、java
<p>如果,比如说</p>
<pre><code>x = [['these are some words'], ['these are some more words'], ['these are even more words'], ['these are the last words']]
</code></pre>
<p>那么</p>
^{pr2}$
<p>会给你</p>
<pre><code>[['these', 'are', 'some', 'words'], ['these', 'are', 'some', 'more', 'words'], ['these', 'are', 'even', 'more', 'words'], ['these', 'are', 'the', 'last', 'words']]
</code></pre>
<p>如你所愿。在</p>
<p>但是,如果你原来的表情</p>
<pre><code>contentLists = [content[i:i+1] for i in range(0, len(content), 1)]
</code></pre>
<p>产生我称之为<code>x</code>这里,为什么首先构建一个长度为1的子列表的列表是毫无意义的?!在</p>
<p>看起来你想直接:</p>
<pre><code>y = [item.split() for item in content]
</code></pre>
<p>而不是产生<code>contentLists</code>,也就是<code>x</code>,然后从中<code>y</code>,不是吗?在</p>