擅长:python、mysql、java
<p>试试这个</p>
<pre><code>remove = ['', '', '']
print [data for data in s if all(x not in remove for x in data)]
</code></pre>
<p>remove列表包含与要删除的列表完全相同的列表,<code>s</code>是包含所有子列表的完整列表</p>
<p>仅供参考<code>s</code></p>
<pre><code>s=[['', '', '', ''],
['2.3', '82.2', '0.6', '1.5'],
['3.6', '92.9', '0.5', '2.1'],
['6.3', '82.9', '0.7', '2.1'],
['7.0', '70.8', '0.5', '1.8'],
['7.7', '56.3', '0.4', '2.0'],
['8.3', '97.0', '0.8', '1.8'],
['10.4', '67.0', '0.6', '1.5'],
['11.8', '89.3', '0.7', '1.4'],
['13.0', '75.8', '0.8', '1.3'],
['14.1', '77.1', '0.6', '1.7'],
['15.8', '74.6', '0.6', '1.8'],
['16.9', '69.0', '0.4', '2.5'],
['18.4', '89.9', '0.6', '2.4'],
['20.3', '93.5', '0.9', '2.3'],
['21.4', '80.9', '0.6', '1.9'],
['21.9', '81.6', '0.9', '2.2'],
['23.5', '65.0', '0.6', '2.5'],
['24.4', '78.4', '0.4', '1.8'],
['27.2', '81.5', '0.7', '2.3'],
['28.8', '73.4', '0.4', '1.7']]
</code></pre>
<p>干杯</p>