在Python中提取一部分文件路径（目录）

Question

我需要提取某个路径的父目录的名称。看起来是这样的：

c:\stuff\directory_i_need\subdir\file

我正在使用使用文件directory_i_need名（而不是路径）的东西来修改“文件”的内容。我创建了一个函数，该函数会给我所有文件的列表，然后…

for path in file_list:
   #directory_name = os.path.dirname(path)   # this is not what I need, that's why it is commented
   directories, files = path.split('\\')

   line_replace_add_directory = line_replace + directories  
   # this is what I want to add in the text, with the directory name at the end 
   # of the line.

我怎样才能做到这一点？

Answer 1

import os
## first file in current dir (with full path)
file = os.path.join(os.getcwd(), os.listdir(os.getcwd())[0])
file
os.path.dirname(file) ## directory of file
os.path.dirname(os.path.dirname(file)) ## directory of directory of file
...

您可以根据需要继续执行多次…

编辑：
从os.path，您可以使用os.path.split或os.path.basename：

dir = os.path.dirname(os.path.dirname(file)) ## dir of dir of file
## once you're at the directory level you want, with the desired directory as the final path node:
dirname1 = os.path.basename(dir) 
dirname2 = os.path.split(dir)[1] ## if you look at the documentation, this is exactly what os.path.basename does.