如何獲取文件的父目錄和子目錄

Question

我想列出特定文件所屬目錄的名稱。下面是我的文件樹的例子：

root_folder 
├── topic_one 
│   ├── one-a.txt 
│   └── one-b.txt 
└── topic_two 
    ├── subfolder_one 
    │   └── sub-two-a.txt 
    ├── two-a.txt 
    └── two-b.txt 

理想的情況下，想什麼，我已經打印出來的是：

"File: file_name belongs in parent directory" 
"File: file_name belongs in sub directory, parent directory" 

我寫這個劇本：

for root, dirs, files in os.walk(root_folder): 

# removes hidden files and dirs 
    files = [f for f in files if not f[0] == '.'] 
    dirs = [d for d in dirs if not d[0] == '.'] 

    if files: 
     tag = os.path.relpath(root, os.path.dirname(root)) 
     for file in files: 
      print file, "belongs in", tag 

這給我這個輸出：

one-a.txt belongs in topic_one 
one-b.txt belongs in topic_one 
two-a.txt belongs in topic_two 
two-b.txt belongs in topic_two 
sub-two-a.txt belongs in subfolder_one 

我不能se他們想知道如何獲取包含在子目錄中的文件的父目錄。任何幫助或替代方法將不勝感激。

Answer 1

由於Jean-François Fabre和Jjpx對於此解決方案：

for root, dirs, files in os.walk(root_folder): 

    # removes hidden files and dirs 
    files = [f for f in files if not f[0] == '.'] 
    dirs = [d for d in dirs if not d[0] == '.'] 

    if files: 
     tag = os.path.relpath(root, root_folder) 

     for file in files: 
      tag_parent = os.path.dirname(tag) 

      sub_folder = os.path.basename(tag) 

      print "File:",file,"belongs in",tag_parent, sub_folder if sub_folder else "" 

打印出：

File: one-a.txt belongs in topic_one 
File: one-b.txt belongs in topic_one 
File: two-a.txt belongs in topic_two 
File: two-b.txt belongs in topic_two 
File: sub-two-a.txt belongs in topic_two subfolder_one