如何从列表中的多个项目分割路径和文件名

Question

这是下面的列表

list = ['C:/Users/user/Desktop/1.docx', 'C:/Users/user/Desktop/2.docx', 
'C:/Users/user/Desktop/3.docx']

我想像这样从列表中拆分路径和文件名：

pathlist = ['C:/Users/user/Desktop/'], and filelist = ['1.docx', '2.docx', '3.docx']

任何帮助和建议将不胜感激

Answer 1

您可以使用定界符rsplit "/" {@ 1}}，然后append组成路径和文件列表的各个部分：

pathlist = []
filelist = []

for item in list:
     pathlist.append(item.rsplit("/",1)[0])
     filelist.append(item.rsplit("/",1)[1])

Answer 2

import os
pathlist = list(set(os.path.dirname(name) for name in list))
filelist = list(set(os.path.basename(name) for name in list))