查找父文件夾

Question

我有這樣的文件夾列表：

u'Magazines/testfolder1', 
u'Magazines/testfolder1/folder1/folder2/folder3', 
u'Magazines/testfolder1/folder1/', 
u'Magazines/testfolder1/folder1/folder2/', 
u'Magazines/testfolder2', 
u'Magazines/testfolder2/folder1/folder2/folder3', 
u'Magazines/testfolder2/folder1/', 
u'Magazines/testfolder2/folder1/folder2/', 
u'Magazines/testfolder3', 
u'Magazines/testfolder3/folder1/folder2/folder3', 
u'Magazines/testfolder3/folder1/', 
u'Magazines/testfolder3/folder1/folder2/', 

現在，我要的是唯一的父文件夾列表中。

即在上面的例子中我想，要減少：

u'Magazines/testfolder1', 
u'Magazines/testfolder2', 
u'Magazines/testfolder3', 

，因爲它們都包含子文件夾。

我在我的數據庫中遞歸添加文件夾，所以如果我有testfolder1那麼腳本會自動遞歸其子文件夾。所以我不需要列表中的子文件夾，如果他們的父母也在列表中。

我該怎麼做？

Answer 1

使用set：

>>> list_of_folders = [ 
...  u'Magazines/testfolder1', 
...  u'Magazines/testfolder1/folder1/folder2/folder3', 
...  u'Magazines/testfolder1/folder1/', 
...  u'Magazines/testfolder1/folder1/folder2/', 
...  u'Magazines/testfolder2', 
...  u'Magazines/testfolder2/folder1/folder2/folder3', 
...  u'Magazines/testfolder2/folder1/', 
...  u'Magazines/testfolder2/folder1/folder2/', 
...  u'Magazines/testfolder3', 
...  u'Magazines/testfolder3/folder1/folder2/folder3', 
...  u'Magazines/testfolder3/folder1/', 
...  u'Magazines/testfolder3/folder1/folder2/', 
... ] 
>>> result = set() 
>>> for folder in list_of_folders: 
...  for parent in result: 
...   if folder.startswith(parent): 
...    break 
...  else: 
...   result.add(folder) 
... 
>>> result 
{'Magazines/testfolder3', 'Magazines/testfolder2', 'Magazines/testfolder1'} 

---

UPDATE

list_of_folders = [ 
    ... 
] 
result = set() 
for folder in list_of_folders: 
    if all(not folder.startswith(parent) for parent in result): 
     result.add(folder) 
print result 

Answer 2

怎麼樣使用regular expression。

import re 

l = [ 
    u'Magazines/testfolder1', 
    u'Magazines/testfolder1/folder1/folder2/folder3', 
    u'Magazines/testfolder1/folder1/', 
    u'Magazines/testfolder1/folder1/folder2/', 
    u'Magazines/testfolder2', 
    u'Magazines/testfolder2/folder1/folder2/folder3', 
    u'Magazines/testfolder2/folder1/', 
    u'Magazines/testfolder2/folder1/folder2/', 
    u'Magazines/testfolder3', 
    u'Magazines/testfolder3/folder1/folder2/folder3', 
    u'Magazines/testfolder3/folder1/', 
    u'Magazines/testfolder3/folder1/folder2/', 
] 

expect = [ 
    u'Magazines/testfolder1', 
    u'Magazines/testfolder2', 
    u'Magazines/testfolder3', 
] 

result = filter(lambda x: re.match('^[^\/]+\/[^\/]+$', x), l) 

assert expect == result 

Answer 3

伴侶我下面beleive是你正在尋找

lst = [ 
u'Magazines/testfolder1', 
u'Magazines/testfolder1/folder1/folder2/folder3', 
u'Magazines/testfolder1/folder1/', 
u'Magazines/testfolder1/folder1/folder2/', 
u'Magazines/testfolder2', 
u'Magazines/testfolder2/folder1/folder2/folder3', 
u'Magazines/testfolder2/folder1/', 
u'Magazines/testfolder2/folder1/folder2/', 
u'Magazines/testfolder3', 
u'Magazines/testfolder3/folder1/folder2/folder3', 
u'Magazines/testfolder3/folder1/', 
u'Magazines/testfolder3/folder1/folder2/' 
] 

    for x in lst: 
     for y in lst[:]: 
      if x in y and len(x)<len(y): 
       lst.remove(y) 
    print lst 

輸出

[u'Magazines/testfolder1', u'Magazines/testfolder2', u'Magazines/testfolder3'] 

這個程序反覆將刪除列表中的子文件夾的解決方案，徒留父夾。

Answer 4

l =[u'Magazines/testfolder1', 
    u'Magazines/testfolder1/folder1/folder2/folder3', 
    u'Magazines/testfolder1/folder1/', 
    u'Magazines/testfolder1/folder1/folder2/', 
    u'Magazines/testfolder2', 
    u'Magazines/testfolder2/folder1/folder2/folder3', 
    u'Magazines/testfolder2/folder1/', 
    u'Magazines/testfolder2/folder1/folder2/', 
    u'Magazines/testfolder3', 
    u'Magazines/testfolder3/folder1/folder2/folder3', 
    u'Magazines/testfolder3/folder1/', 
    u'Magazines/testfolder3/folder1/folder2/', ] 

mincount = min(s.count('/') for s in l) 
[d for d in sorted(l) if d.count('/') <= mincount] 
#=> [u'Magazines/testfolder1', u'Magazines/testfolder2', u'Magazines/testfolder3'] 

它不是非常聰明，但它有效的地方有一個共同的根。