擅长:python、mysql、java
<p>我的方法是拆分字符串并使用<code>_car</code>前面的数字作为<code>key</code>进行比较。你知道吗</p>
<pre><code>>>> x = [
... '../../scene/temp_5a/458754/1_car.png',
... '../../scene/temp_5a/458754/2_car.png',
... '../../scene/temp_5a/458754/10_car.png',
... '../../scene/temp_5a/458754/15_car.png',
... '../../scene/temp_5a/458754/3_car.png']
>>>
>>> sorted(x,key=lambda i: int(i.split('/')[-1].split('_')[0]))
[[1, '../../scene/temp_5a/458754/1_car.png'], [2, '../../scene/temp_5a/458754/2_car.png'], [3, '../../scene/temp_5a/458754/3_car.png'], [10, '../../scene/temp_5a/458754/10_car.png'], [15, '../../scene/temp_5a/458754/15_car.png']]
</code></pre>