三级目录列表理解

Question

我目前的实施方案是否有一个班轮替代品？

Top Level
-- Sub Level
--- Folder
-- Sub Level
--- Folder
.
.
.

要获得-- Folder每个-- Sub Level -- Folder -- Sub Level我具有以下内容：

dir_ = '/path/to/Top Level'
sub_lvl = [os.path.join(dir_, sub) for sub in os.listdir(dir_)]
# "folder" is always a fixed name
folders = [os.path.join(sub, "folder") for sub in sub_lvl] 

Answer 1

os.path.join接受任意数量的字符串。 并且您应该检查项目是否是文件夹：

sub_lvl = [os.path.join(dir_, sub, "folder") for sub in os.listdir(dir_) if os.path.isdir(os.path.join(dir_,sub))]

Answer 2

你去了：

folders = [os.path.join(os.path.join(dir_, sub), "folder") for sub in os.listdir(dir_)]