python版本号排序

1. 方法一（纯数字或带字母），使用distutils下自带的工具

from distutils.version import LooseVersion from distutils.version import StrictVersion

2. 方法二（纯数字），版本号类似：

['1.0.0', '1.0.2', '1.0.12', '1.1.2', '1.3.3']

import os import sys basedir = os.path.abspath(os.path.dirname(__file__)) lib_path = os.path.dirname(os.path.dirname(basedir)) sys.path.append(lib_path) from demo import logger if __name__ == '__main__': order = sys.argv[1] logger.info('Order is [ %s ].' % order) version_list = ["1.1.2", "1.0.0", "1.3.3", "1.0.12", "1.0.2"] if order == 'asc': version_list.sort(key=lambda s: map(int, s.split('.'))) if order == 'desc': version_list.sort(key=lambda s: map(int, s.split('.')), reverse=True) logger.info('sort versions is [ %s ].' % version_list)

(venv) C:\Users\jalen\demo>python dashboard\test\commontest.py asc
2020-08-11 02:44:33,391 commontest.py:12:INFO:Order is [ asc ].
2020-08-11 02:44:33,392 commontest.py:18:INFO:sort versions is [ ['1.0.0', '1.0.2', '1.0.12', '1.1.2', '1.3.3'] ].

参考：排序一个以点分隔的数字列表，比如软件版本。 - 问答 - Python中文网

3. 方法三（纯数字），利用python元组的比较

if __name__ == '__main__':
    v1 = ['40.1', '40.3', '40.9', '40.11', '40.10', '40.2']
    v1.sort(key=lambda x: tuple(int(v) for v in x.split(".")), reverse=True)
    print(v1)
 
    version = ['2.0.1', '1.0.2', '1.0.21', '2.2.9', '1.2.11']
    version.sort(key=lambda x: tuple(int(v) for v in x.split(".")), reverse=True)
    print(version)
    print('end!')