将路径拆分为子文件夹的“三角形”

Question

I want to split a path into its components and handle each directory in order from the bottom. For

path = 'a/b/c/d'

I want to get

components = [
    ('', 'a'),
    ('a', 'b'),
    ('a/b', 'c'),
    ('a/b/c', 'd')
]

Is there something in the standard library to help me with the task?

Answer 1

This works, although I would prefer something better looking and hopefully less error prone.

>>> comp = os.path.normpath('a/b/c/d').split(os.sep)
>>> [(os.path.join(*comp[:i]) if comp[:i] else '', comp[i]) for i in range(len(comp))]
[('', 'a'), ('a', 'b'), ('a/b', 'c'), ('a/b/c', 'd')]