How to get the parent dir location

Question

this code is get the templates/blog1/page.html in b.py:

path = os.path.join(os.path.dirname(__file__), os.path.join('templates', 'blog1/page.html'))

but i want to get the parent dir location:

aParent
   |--a
   |  |---b.py
   |      |---templates
   |              |--------blog1
   |                         |-------page.html
   |--templates
          |--------blog1
                     |-------page.html

and how to get the aParent location

thanks

updated:

this is right:

dirname=os.path.dirname
path = os.path.join(dirname(dirname(__file__)), os.path.join('templates', 'blog1/page.html'))

or

path = os.path.abspath(os.path.join(os.path.dirname(__file__),".."))

So you want to get `blog1` or `a` ? And where is your current file located? — Felix Kling, May 12 '10 at 09:03
`os.path.join('templates', 'blog1/page.html')` looks strange to me. You are mixing things up. Either `os.path.join('templates', 'blog1', 'page.html')` or `'templates/blog1/page.html'`. And much easier would be `os.path.abspath(os.path.join('templates', 'blog1', 'page.html'))` then — Felix Kling, May 12 '10 at 09:15
@zjm: no, you don't *get* that page. It's not some blackbox that you could just use to get the template file. It performs a series of trivial small steps, and if you could understand them, you wouldn't have this question. — SilentGhost, May 12 '10 at 09:20

Marcelo Cantos · Accepted Answer · 2014-02-01T01:11:00.513

222

You can apply dirname repeatedly to climb higher: dirname(dirname(file)). This can only go as far as the root package, however. If this is a problem, use os.path.abspath: dirname(dirname(abspath(file))).

edited Feb 01 '14 at 01:11

answered May 12 '10 at 09:03

Marcelo Cantos

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I know the OP knows about `dirname`. It isn't obvious to everyone that applying dirname to a directory yields the parent directory. – Marcelo Cantos May 12 '10 at 09:09
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`dirname` does **NOT** always return the parent directory; http://twitter.com/#!/ActiveState/status/671049326788608 – Sridhar Ratnakumar Nov 05 '10 at 22:12
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@Sridhar: That depends on your perspective; I consider a path ending in `/` as not representing the leaf directory entry itself, but its contents, which is why, e.g., `mv xxx yyy/` fails if `yyy` isn't a preexisting directory. In any case, even if we take your point as a given, it is irrelevant in the context of my answer. Neither `file` nor the result of `dirname` will ever end in a `/`. – Marcelo Cantos Nov 06 '10 at 00:56
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Minor correction: `dirname` may return `'/'`, which clearly ends in a `/`. That is the only exception, AFAIK. – Marcelo Cantos Feb 06 '13 at 23:19

Gavriel Cohen · Answer 2 · 2019-08-26T06:44:43.180

74

Use relative path with the pathlib module in Python 3.4+:

from pathlib import Path

Path(__file__).parent

You can use multiple calls to parent to go further in the path:

Path(__file__).parent.parent

As an alternative to specifying parent twice, you can use:

Path(__file__).parents[1]

edited Aug 26 '19 at 06:44

answered Sep 21 '16 at 13:18

Gavriel Cohen

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If you need the path as a string because you want to modify it, you can just use `str(Path(__file__).parent)`. – Philipp May 10 '20 at 11:16
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It seems that in python 3.8.5 you need `Path(__file__).resolve().parent` – Markus Kaukonen Dec 22 '20 at 08:41
2

Thanks, I knew about `.parent` but not about `.parents`. IMO, that's much nicer than chaining `.parent` or `os.path.dirname`s. – kennyvh Mar 15 '21 at 17:44

score 68 · Answer 3 · answered Jan 04 '13 at 03:38

os.path.abspath doesn't validate anything, so if we're already appending strings to __file__ there's no need to bother with dirname or joining or any of that. Just treat __file__ as a directory and start climbing:

# climb to __file__'s parent's parent:
os.path.abspath(__file__ + "/../../")

That's far less convoluted than os.path.abspath(os.path.join(os.path.dirname(__file__),"..")) and about as manageable as dirname(dirname(__file__)). Climbing more than two levels starts to get ridiculous.

But, since we know how many levels to climb, we could clean this up with a simple little function:

uppath = lambda _path, n: os.sep.join(_path.split(os.sep)[:-n])

# __file__ = "/aParent/templates/blog1/page.html"
>>> uppath(__file__, 1)
'/aParent/templates/blog1'
>>> uppath(__file__, 2)
'/aParent/templates'
>>> uppath(__file__, 3)
'/aParent'

This is nice, but it would also be cool if the standard library added a convenience function that accomplished this...don't want to come to SO every time I need this func — Julian Irwin, Jan 27 '16 at 17:26
Would `os.path.abspath(os.path.join(__file__, "..", "..")` be more portable? — slowD, Aug 31 '18 at 00:28

score 13 · Answer 4 · answered May 12 '10 at 09:36

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os.path.dirname(os.path.abspath(__file__))

Should give you the path to a.

But if b.py is the file that is currently executed, then you can achieve the same by just doing

os.path.abspath(os.path.join('templates', 'blog1', 'page.html'))

answered May 12 '10 at 09:36

Felix Kling

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o ,i know ,you are right ,and i want to get the a's parent folder. how to get it – zjm1126 May 12 '10 at 09:43
@zjm1126: See Marcelo Cantos' answer. Apply `dirname()` twice. Everything you need now should be on this site. – Felix Kling May 12 '10 at 09:51

score 9 · Answer 5 · answered Nov 25 '13 at 14:32

9

os.pardir is a better way for ../ and more readable.

import os
print os.path.abspath(os.path.join(given_path, os.pardir))

This will return the parent path of the given_path

answered Nov 25 '13 at 14:32

Sun Liwen

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score 6 · Answer 6 · answered Aug 07 '13 at 21:59

6

A simple way can be:

import os
current_dir =  os.path.abspath(os.path.dirname(__file__))
parent_dir = os.path.abspath(current_dir + "/../")
print parent_dir

answered Aug 07 '13 at 21:59

Muneeb Ali

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score 3 · Answer 7 · answered May 12 '10 at 09:26

3

May be join two .. folder, to get parent of the parent folder?

path = os.path.abspath(os.path.join(os.path.dirname(os.path.abspath(__file__)),"..",".."))

answered May 12 '10 at 09:26

YOU

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score 2 · Answer 8 · answered Mar 19 '14 at 16:03

Here is another relatively simple solution that:

does not use dirname() (which does not work as expected on one level arguments like "file.txt" or relative parents like "..")
does not use abspath() (avoiding any assumptions about the current working directory) but instead preserves the relative character of paths

it just uses normpath and join:

def parent(p):
    return os.path.normpath(os.path.join(p, os.path.pardir))

# Example:
for p in ['foo', 'foo/bar/baz', 'with/trailing/slash/', 
        'dir/file.txt', '../up/', '/abs/path']:
    print parent(p)

Result:

.
foo/bar
with/trailing
dir
..
/abs

score 2 · Answer 9 · edited Apr 04 '15 at 00:12

2

Use the following to jump to previous folder:

os.chdir(os.pardir)

If you need multiple jumps a good and easy solution will be to use a simple decorator in this case.

edited Apr 04 '15 at 00:12

Zulu

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answered Apr 03 '15 at 21:22

Marco smdm

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score 1 · Answer 10 · answered Apr 03 '22 at 18:53

1

from os.path import basename, dirname
basename(dirname('foo/bar/foo_bar'))

answered Apr 03 '22 at 18:53

Ernest

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Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community Apr 03 '22 at 21:54

score 0 · Answer 11 · answered Feb 15 '14 at 15:39

I think use this is better:

os.path.realpath(__file__).rsplit('/', X)[0]


In [1]: __file__ = "/aParent/templates/blog1/page.html"

In [2]: os.path.realpath(__file__).rsplit('/', 3)[0]
Out[3]: '/aParent'

In [4]: __file__ = "/aParent/templates/blog1/page.html"

In [5]: os.path.realpath(__file__).rsplit('/', 1)[0]
Out[6]: '/aParent/templates/blog1'

In [7]: os.path.realpath(__file__).rsplit('/', 2)[0]
Out[8]: '/aParent/templates'

In [9]: os.path.realpath(__file__).rsplit('/', 3)[0]
Out[10]: '/aParent'

Not quite, it's OS dependent (will not work on Windows). Also it does not allow to use relative paths. — kgadek, Oct 22 '14 at 17:49

score 0 · Answer 12 · answered May 07 '14 at 06:03

0

I tried:

import os
os.path.abspath(os.path.join(os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe()))), os.pardir))

answered May 07 '14 at 06:03

gogasca

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How to get the parent dir location

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