How to get the nth subdirectory in Python?

Question

Suppose I have a file known as xyz in a folder as shown below:

Directory
   |
   |+ Subdirectory_n
           |
           |+ Subdirectory_(n-1)
                   |
                   |+ Subdirectory_(n-2)
                           |
                           |+ Subdirectory_(n-3)
                                    ........
                                           |+ Subdirectory_1
                                                   |
                                                   |+ xyz

And if I want the nth subdirectory from the file, how do I obtain it python? For e.g. for the 2nd sub directory I can do something like:

import os
file = open('xyz.dat', 'w+')
print os.path.dirname(os.path.dirname(__file__))

for the third

import os
file = open('xyz.dat', 'w+')
print os.path.dirname(os.path.dirname(os.path.dirname(__file__)))

This method looks a bit awkward. Is there a better way to do this, so that I can generalize it for an nth subdirectory in Python?

@JeremyKemball The nth subdirectory starting 0 from the file. — Tom Kurushingal, Jul 08 '15 at 19:11

Sam · Answer 1 · 2015-07-08T19:24:46.727

3

A recursive approach:

def nth_parent(path, n): return path if n <= 0 else os.path.dirname(nth_parent(path, n-1))

n is the number of levels to go up the directory tree.

edited Jul 08 '15 at 19:24

answered Jul 08 '15 at 19:16

Sam

8,330
2
26
51

score 1 · Answer 2 · answered Jul 08 '15 at 19:00

1

You can get the absolute path of the file using os.path.abspath() That should give you something like 'C:\folder1\folder2\folder3\...'. You can then use string operations to find the nth directory.

answered Jul 08 '15 at 19:00

MrocKK

51
5

Jeremy Kemball · Answer 3 · 2015-07-08T19:16:49.317

1

You know this calls for a clever-clogs recursive solution but perhaps something like

n=5
while(n):
    path=os.path.dirname(path)
    n -=1

would be clear and preserve the os.path localization magic you would lose by just splitting os.path.abspath().

You could always try something like

os.path.dirname(os.path.normpath(__file__+"../"*n))

but that's not guaranteed to work for all symlinks, etc.

edited Jul 08 '15 at 19:16

answered Jul 08 '15 at 19:08

Jeremy Kemball

1,175
1
9
14

score 1 · Answer 4 · answered Jul 09 '15 at 19:02

1

This is a bit awkward, but short enough to use directly:

reduce(lambda x, y: y(x), [os.path.dirname]*n, path)

answered Jul 09 '15 at 19:02

rezca

116
6

score 0 · Answer 5 · answered Jun 24 '20 at 13:01

Python ≥ 3.4 implement PEP 428 and brings you pathlib, giving you more flexibility.

The "modern" way to do it could be resumed with:

pathlib.Path('xyz.dat').resolve().parents[nth_parent - 1]

# equivalent of:
filename = os.path.realpath('xyz.dat')
for _ in range(nth_parent):
    filename = os.path.dirname(filename)

Where nth_parent is the number of ancestors and zero mean the immediate parent (hence the - 1 in the example).

Equivalences (with os.path):

filename = 'xyz.dat'
pathlib.Path(filename).resolve(), os.path.realpath(filename)

# PurePath does not have resolve() nor other features added by the Path subclass;
# but when the paths are absolutes, those two are similar
pathlib.Path(__file__) == pathlib.PurePath(__file__)

pathlib.Path(__file__).parents[0], pathlib.Path(__file__).parent, os.path.dirname(__file__)

pathlib.Path(__file__).parents[1], pathlib.Path(__file__).parent.parent, os.path.dirname(os.path.dirname(__file__))

How to get the nth subdirectory in Python?

5 Answers5