擅长:python、mysql、java
<p>您可以使用<code>zip</code>。</p>
<p><code>zip</code>从传递给它的iterables返回同一索引上的项。</p>
<pre><code>>>> from math import isnan
>>> list1 = [[1, 2, 'nan'], [3, 7, 8], [1, 1, 1], [10, -1,'nan']]
>>> list2 = [1, 2, 3, 4]
>>> out = [(x,y) for x,y in zip(list1,list2)
if not any(isnan(float(z)) for z in x)]
>>> out
[([3, 7, 8], 2), ([1, 1, 1], 3)]
</code></pre>
<p>现在解压缩<code>out</code>以获得所需的输出:</p>
<pre><code>>>> list1_clean, list2_clean = map(list, zip(*out))
>>> list1_clean
[[3, 7, 8], [1, 1, 1]]
>>> list2_clean
[2, 3]
</code></pre>
<p>有关<code>zip</code>的帮助:</p>
<pre><code>>>> print zip.__doc__
zip(seq1 [, seq2 [...]]) -> [(seq1[0], seq2[0] ...), (...)]
Return a list of tuples, where each tuple contains the i-th element
from each of the argument sequences. The returned list is truncated
in length to the length of the shortest argument sequence.
</code></pre>
<p>如果需要一个内存高效的解决方案,可以使用<code>itertools.izip</code>,因为它返回一个迭代器。</p>