查找父文件夹

Question

我有这样的文件夹列表：

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/', 

现在，我要的是唯一的父文件夹列表中。

即在上面的例子中我想，要减少：

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

，因为它们都包含子文件夹。

我在我的数据库中递归添加文件夹，所以如果我有testfolder1那么脚本会自动递归其子文件夹。所以我不需要列表中的子文件夹，如果他们的父母也在列表中。

我该怎么做？

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

怎么样使用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

伴侣我下面beleive是你正在寻找

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 

输出

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

这个程序反复将删除列表中的子文件夹的解决方案，徒留父夹。

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'] 

它不是非常聪明，但它有效的地方有一个共同的根。