擅长:python、mysql、java
<p>你可以这样做:</p>
<pre><code>matrix = [
[[0], 'whatever0'],
[[1], 'whatever1'],
[[1, 0], 'whatever10'],
[[1, 1], 'whatever11'],
[[1, 2], 'whatever12'],
[[1, 3], 'whatever13'],
[[1, 4], 'whatever14'],
]
result = next((e for e in reversed(matrix) if len(e[0]) == 1), None)
print(result)
</code></pre>
<p><strong>输出</strong></p>
<pre><code>[[1], 'whatever1']
</code></pre>
<p><strong>解释</strong></p>
<p>首先使用<a href="https://docs.python.org/3/library/functions.html#reversed" rel="nofollow noreferrer">reversed</a>向后迭代列表,只返回具有<code>len(e[0]) == 1</code>的元素。然后用<a href="https://docs.python.org/3/library/functions.html#next" rel="nofollow noreferrer">next</a>只得到第一个结果。你知道吗</p>