Path to a file without basename

Question

How can I get the path of a file without the file basename ?

Something like /a/path/to/my/file.txt --> /a/path/to/my/

Tried with .split

Answer 1

You can import os

>>> filepath
'/a/path/to/my/file.txt'
>>> os.path.dirname(filepath)
'/a/path/to/my'
>>> 

Answer 2

Use os.path.dirname(filename).

Answer 3

(dirname, filename) = os.path.split(path)

Answer 4

Check subs of os.path

os.path.dirname('/test/one')