<p>类似于tobiasék的答案,但以你想要的格式,排序和所有。<strike>(我想)好吧,它已经过测试,现在似乎可以工作了。在</p>
<p>我们将路径列表转换成具有<code>defaultdict</code>的树,然后递归地将基于<code>defaultdict</code>的树转换为基于排序列表的形式。在</p>
<pre><code>from collections import defaultdict
def tree():
# A dict-based tree that automatically creates nodes when you access them.
# Note that this does not store a name for itself; it's closer to a dropdown
# menu than the little button you click to display the menu, or the contents
# of a directory rather than the directory itself.
return defaultdict(tree)
def paths_to_tree(paths):
# Make a tree representing the menu.
menu = tree()
for path in myList:
add_to = menu
# Traverse the tree to automatically create new tree nodes.
for name in path:
add_to = add_to[name]
return menu
def sort_key(item):
if isinstance(item, list):
return 1, item[0]
else:
# It's a string
return 0, item
# Recursively turn the tree into nested lists.
def tree_to_lists(menu):
lists = [[item, tree_to_lists(menu[item])] if menu[item] else item
for item in menu]
lists.sort(key=sort_key)
return lists
# Note the [0].
output = tree_to_lists(paths_to_tree(myList))[0]
</code></pre>