Question

我想以.log应该是第一个文件，.gz文件应该是降序的方式对列表进行排序

my_list = [
     '/abc/a.log.1.gz',
     '/abc/a.log',
     '/abc/a.log.30.gz',
     '/abc/a.log.2.gz',
     '/abc/a.log.5.gz',
     '/abc/a.log.3.gz',
     '/abc/a.log.6.gz',
     '/abc/a.log.4.gz',
     '/abc/a.log.12.gz',
     '/abc/a.log.10.gz',
     '/abc/a.log.8.gz',
     '/abc/a.log.14.gz',
     '/abc/a.log.29.gz'
]

expected_result：

my_list = ['/abc/a.log',
        '/abc/a.log.30.gz',
        '/abc/a.log.29.gz',
        '/abc/a.log.29.gz',
        '/abc/a.log.14.gz',
        '/abc/a.log.12.gz',
        '/abc/a.log.10.gz',
        '/abc/a.log.8.gz',
        '/abc/a.log.6.gz',
        '/abc/a.log.5.gz',
        '/abc/a.log.4.gz',
        '/abc/a.log.3.gz',
        '/abc/a.log.2.gz'
        '/abc/a.log.1.gz']

reversed(mylist)也没有给我想要的解决方案。

Answer 1

将sorted与自定义key函数和reverse=True结合使用：

print(sorted(my_list, key=lambda x: (x.endswith('log'), x), reverse=True))
#['/abc/spa/a.log',
# '/abc/spa/a.log.30.gz',
# '/abc/spa/a.log.2.gz',
# '/abc/spa/a.log.1.gz']

基于更新的问题，似乎您正在尝试对文件名进行排序。我建议使用os.path来操纵这些字符串。

首先，您可以使用os.path.splitext拆分扩展名以在.log或.gz之间进行比较。然后再次剥离扩展名以获取文件号，并将其转换为整数。

例如：

import os

def get_sort_keys(filepath):
    split_file_path = os.path.splitext(filepath)
    sort_key = (split_file_path[1], *os.path.splitext(split_file_path[0]))
    return (sort_key[0], sort_key[1], int(sort_key[2].strip(".")) if sort_key[2] else 0)

print(sorted(my_list, key=get_sort_keys, reverse=True))
#['/abc/a.log',
# '/abc/a.log.30.gz',
# '/abc/a.log.29.gz',
# '/abc/a.log.14.gz',
# '/abc/a.log.12.gz',
# '/abc/a.log.10.gz',
# '/abc/a.log.8.gz',
# '/abc/a.log.6.gz',
# '/abc/a.log.5.gz',
# '/abc/a.log.4.gz',
# '/abc/a.log.3.gz',
# '/abc/a.log.2.gz',
# '/abc/a.log.1.gz']

在此版本中，我没有像以前一样显式检查endswith("log")，但是我依靠log扩展名将按字典顺序排在gz之后的事实。

Answer 2

如果有多个扩展名，则在字典中分配等级以具有更大的灵活性

my_list = ['/abc/spa/a.log.1.gz',
 '/abc/spa/a.log',
 '/abc/spa/a.log.30.tar',
 '/abc/spa/a.log.30.gz',
 '/abc/spa/a.log.2.gz']

rank={'log':1,'gz':2,'tar':3}    # 'tar' is optional here
sorted(my_list,key = lambda x : (-rank.get(x.rsplit('.')[-1],0), x),reverse=True)

这会给

['/abc/spa/a.log',
 '/abc/spa/a.log.30.gz',
 '/abc/spa/a.log.2.gz',
 '/abc/spa/a.log.1.gz',
 '/abc/spa/a.log.30.tar']

关于您的最新问题

my_list = [
     '/abc/a.log.1.gz',
     '/abc/a.log',
     '/abc/a.log.30.gz',
     '/abc/a.log.2.gz',
     '/abc/a.log.5.gz',
     '/abc/a.log.3.gz',
     '/abc/a.log.6.gz',
     '/abc/a.log.4.gz',
     '/abc/a.log.12.gz',
     '/abc/a.log.10.gz',
     '/abc/a.log.8.gz',
     '/abc/a.log.14.gz',
     '/abc/a.log.29.gz'
]
rank={'log':1,'gz':2,'tar':3}    # 'tar' is optional here
sorted(my_list,key = lambda x : (-rank.get(x.rsplit('.',1)[-1]), int(x.split('.')[-2]) if x.split('.')[-2].isdigit() else 0),reverse=True)

输出：

['/abc/a.log',
 '/abc/a.log.30.gz',
 '/abc/a.log.29.gz',
 '/abc/a.log.14.gz',
 '/abc/a.log.12.gz',
 '/abc/a.log.10.gz',
 '/abc/a.log.8.gz',
 '/abc/a.log.6.gz',
 '/abc/a.log.5.gz',
 '/abc/a.log.4.gz',
 '/abc/a.log.3.gz',
 '/abc/a.log.2.gz',
 '/abc/a.log.1.gz']

根据条件在python中反转列表

2 个答案: