在python中对列表中的文件进行排序

Question

glob.glob(file_naming)

这样做给了我两个文件:

['/srv/tmp/what/123_aaa_bbb_20190110043711.XML',
 '/srv/tmp/what/234_xxx_yyy_20190110043710.XML']

如何按最后一部分(20190110043711)的顺序对它们进行排序？我需要一个结果:

['/srv/tmp/what/123_aaa_bbb_20190110043710.XML',
 '/srv/tmp/what/234_xxx_yyy_20190110043711.XML']

Answer 1

用途sorted:

>>> sorted(l,key=lambda x: int(x.split('_')[-1]))
['/srv/tmp/what/234_xxx_yyy_20190110043710.XML', '/srv/tmp/what/123_aaa_bbb_20190110043711.XML']
>>>