擅长:python、mysql、java
<p>您可以使用以下选项:</p>
<pre><code>d = {'S26': [['2016-03-18', '2016-03-28'], ['2016-03-18', '2016-03-28']], 'S24': [['2016-03-19', '2016-03-25'], ['2016-03-25', '2016-04-03']], 'S25': [['2016-03-18', '2016-03-25'], ['2016-03-18', '2016-03-25'], ['2016-03-20', '2016-03-25'], ['2016-03-20', '2016-03-25'], ['2016-03-25', '2016-04-03']]}
answer = {k:[list(el) for el in set([tuple(sublist) for sublist in v])]
for k, v in d.items()}
print(answer)
</code></pre>
<p><strong>输出</strong></p>
<pre><code>{'S24': [['2016-03-19', '2016-03-25'],
['2016-03-25', '2016-04-03']],
'S25': [['2016-03-20', '2016-03-25'],
['2016-03-18', '2016-03-25'],
['2016-03-25', '2016-04-03']],
'S26': [['2016-03-18', '2016-03-28']]}
</code></pre>
<p>这将迭代字典中的每个<code>(k, v)</code>元素对,并从列表值中删除重复的子列表。为此,我们使用列表理解将每个<code>sublist</code>转换为<code>tuple</code>,这样我们就可以创建一个<code>set</code>(因为列表是不可散列的)。你知道吗</p>