擅长:python、mysql、java
<p>有一个来自<code>more_itertools</code>的<em>配方</em>,在这里很方便:</p>
<pre><code>import more_itertools as mit
d = {0:'I', 3: 'Mr.', 4: 'Akshay', 5: 'Nevrekar', 8: 'JK', 14: 'Soham', 15: 'Sharma'}
cons = [list(group) for group in mit.consecutive_groups(sorted(d.keys()))]
# [[0], [3, 4, 5], [8], [14, 15]]
res = [[d[y] for y in x] for x in cons if len(x)>1]
# [['Mr.', 'Akshay', 'Nevrekar'], ['Soham', 'Sharma']]
</code></pre>
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<p>如果只需要连续整数至少为<strong>3的组,则必须相应地修改不等式(<code>len(x) > 2</code>)。你知道吗</p>