擅长:python、mysql、java
<p>将<code>d35</code>字典转换为已排序的列表并逐步执行:</p>
<pre><code>In [4]: d35 = {"10": "A",
...: "19": "B",
...: "25": "C",
...: "29": "D",
...: "32": "E",
...: "36": "F",
...: }
In [5]: sorted(d35.items())
Out[5]: [('10', 'A'), ('19', 'B'), ('25', 'C'), ('29', 'D'), ('32', 'E'), ('36', 'F')]
In [7]: time = 15
In [11]: for max_time, group_name in sorted(d35.items()):
...: if int(max_time) >= time:
...: break
...:
In [12]: max_time
Out[12]: '19'
In [13]: group_name
Out[13]: 'B'
</code></pre>
<p>修改你的方法会得到这个结果。我在for循环中添加了else来处理任何组都没有覆盖的时间。在</p>
^{pr2}$