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.gz',
     '/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'
]

预期结果：

my_list = ['/abc/a.log',
        '/abc/a.log.gz',
        '/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']

我的解决方案：导入操作系统

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))

获取错误：

ValueError: invalid literal for int() with base 10: 'log'

Answer 1

您可以将sorted与自定义函数一起使用，该函数可以执行一些“试一试”检查。

def try_convert(x):
    y = x.rsplit('.', 2)[-2]
    return ('log' not in x, int(y) if y.isdigit() else float('inf'), x)

sorted(my_list, key=try_convert, reverse=True)

['/abc/a.log.gz',
 '/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']

该函数确保不带整数部分的文件名最后排序（如果按降序排序，则为第一个）。此外，所有“ .log”文件都位于第一位。

Answer 2

您正在尝试将字符串'log'传递到int()中。

然后无法将其转换为int并引发ValueError: invalid literal for int() with base 10: 'log'

这是在return (sort_key[0], sort_key[1], int(sort_key[2].strip(".")) if sort_key[2] else 0)

尝试在转换中使用try catch块

Answer 3

或使用：

>>> sorted(my_list,key=lambda x: int(x.split('.')[2]) if x.split('.')[2].isdigit() else 31,reverse=True)
['/abc/a.log.gz', '/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']
>>>

更新的问题：

>>> sorted(my_list,key=lambda x: int(x.split('.')[-2]) if x.split('.')[-2].isdigit() else 31,reverse=True)
['/abc/a.log', '/abc/a.log.gz', '/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python中按扩展名对文件路径进行排序

3 个答案: