擅长:python、mysql、java
<p>单线解决方案可能如下所示:</p>
<pre class="lang-py prettyprint-override"><code>list_in = {'key0': {'a': ['A', 'B', 'C'], 'b': ['A', 'B', 'C']}, 'key1': {'c': ['A', 'B', 'C'], 'd': ['A', 'B', 'C']}, 'key2': {'e': 0, 'f': 0}}
nth = 1
list_out = {k: {k2: v2 if not isinstance(v2,list) else v2[nth] \
for k2,v2 in v.items() } \
for k,v in list_in.items() }
print list_out
</code></pre>
<pre><code>{'key2': {'e': 0, 'f': 0}, 'key1': {'c': 'B', 'd': 'B'}, 'key0': {'a': 'B', 'b': 'B'}}
</code></pre>
<p>注意,我用字符串替换了键和列表的值</p>