How to get the parent dir location

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__)),"..",".."))

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'))

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'

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

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

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'

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

I tried:

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

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.

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]

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