擅长:python、mysql、java
<p>您可以使用<code>any</code>或<code>all</code>来执行此操作</p>
<pre><code>>>> all(character.isalpha() for item in alist for character in item)
False
>>> any(character.isalpha() for item in alist for character in item)
True
>>> ## if you want the same output as using any
>>> not all(character.isalpha() for item in alist for character in item)
True
</code></pre>
<p>顺便说一下,<code>all</code>更快(至少在这个相对较小的情况下)</p>
^{pr2}$
<p>但如果你改变顺序,这就不成立了</p>
<pre><code>>>> timeit.timeit('not all(character.isalpha() for item in ["b1", "+", "1"] for character in item)', number=10000)
0.0663501940999538
>>> timeit.timeit('any(character.isalpha() for item in ["b1", "+", "1"] for character in item)', number=10000)
0.037434049209622344
</code></pre>
<p>所以选择其中一个很大程度上是个人偏好</p>