@author jackzhenguo @desc @date 2019/10/3
104 获取路径中的文件名
In [11]: import os ...: file_ext = os.path.split('./data/py/test.py') ...: ipath,ifile = file_ext ...: In [12]: ipath Out[12]: './data/py' In [13]: ifile Out[13]: 'test.py'
python-small-examples/md/104.md at master · data-python/python-small-examples