python: get directory two levels up
Question:
Ok…I dont know where module x
is, but I know that I need to get the path to the directory two levels up.
So, is there a more elegant way to do:
import os
two_up = os.path.dirname(os.path.dirname(__file__))
Solutions for both Python 2 and 3 are welcome!
Answers:
You can use pathlib
. Unfortunately this is only available in the stdlib for Python 3.4. If you have an older version you’ll have to install a copy from PyPI here. This should be easy to do using pip
.
from pathlib import Path
p = Path(__file__).parents[1]
print(p)
# /absolute/path/to/two/levels/up
This uses the parents
sequence which provides access to the parent directories and chooses the 2nd one up.
Note that p
in this case will be some form of Path
object, with their own methods. If you need the paths as string then you can call str
on them.
Very easy:
Here is what you want:
import os.path as path
two_up = path.abspath(path.join(__file__ ,"../.."))
Personally, I find that using the os module is the easiest method as outlined below. If you are only going up one level, replace (‘../..’) with (‘..’).
import os
os.chdir('../..')
--Check:
os.getcwd()
I don’t yet see a viable answer for 2.7 which doesn’t require installing additional dependencies and also starts from the file’s directory. It’s not nice as a single-line solution, but there’s nothing wrong with using the standard utilities.
import os
grandparent_dir = os.path.abspath( # Convert into absolute path string
os.path.join( # Current file's grandparent directory
os.path.join( # Current file's parent directory
os.path.dirname( # Current file's directory
os.path.abspath(__file__) # Current file path
),
os.pardir
),
os.pardir
)
)
print grandparent_dir
And to prove it works, here I start out in ~/Documents/notes
just so that I show the current directory doesn’t influence outcome. I put the file grandpa.py
with that script in a folder called “scripts”. It crawls up to the Documents dir and then to the user dir on a Mac.
(testing)AlanSE-OSX:notes AlanSE$ echo ~/Documents/scripts/grandpa.py
/Users/alancoding/Documents/scripts/grandpa.py
(testing)AlanSE-OSX:notes AlanSE$ python2.7 ~/Documents/scripts/grandpa.py
/Users/alancoding
This is the obvious extrapolation of the answer for the parent dir. Better to use a general solution than a less-good solution in fewer lines.
I have found that the following works well in 2.7.x
import os
two_up = os.path.normpath(os.path.join(__file__,'../'))
You can use this as a generic solution:
import os
def getParentDir(path, level=1):
return os.path.normpath( os.path.join(path, *([".."] * level)) )
For getting the directory 2 levels up:
import os.path as path
curr_dir=Path(os.path.dirname(os.path.abspath(__file__)))
two_dir_up_=os.fspath(Path(curr_dir.parent.parent).resolve())
I have done the following to go up two and drill down on other dir
default_config_dir=os.fspath(Path(curr_dir.parent.parent,
'data/config').resolve())
I was going to add this just to be silly, but also because it shows newcomers the potential usefulness of aliasing functions and/or imports.
Having written it, I think this code is more readable (i.e. lower time to grasp intention) than the other answers to date, and readability is (usually) king.
from os.path import dirname as up
two_up = up(up(__file__))
Note: you only want to do this kind of thing if your module is very small, or contextually cohesive.
More cross-platform implementation will be:
import pathlib
two_up = (pathlib.Path(__file__) / ".." / "..").resolve()
Using parent
is not supported on Windows. Also need to add .resolve()
, to:
Make the path absolute, resolving all symlinks on the way and also normalizing it (for example turning slashes into backslashes under Windows)
The best solution (for python >= 3.4) when executing from any directory is:
from pathlib import Path
two_up = Path(__file__).resolve().parents[1]
Assuming you want to access folder named xzy two folders up your python file. This works for me and platform independent.
“.././xyz”
(pathlib.Path(‘../../’) ).resolve()
100% working answer:
os.path.abspath(os.path.join(os.getcwd() ,"../.."))
There is already an accepted answer, but for two levels up I think a chaining approach is arguably more readable:
pathlib.Path(__file__).parent.parent.resolve()
Surprisingly it seems no one has yet explored this nice one-liner option:
import os
two_up = os.path.normpath(__file__).rsplit(os.sep, maxsplit=2)[0]
rsplit
is interesting since the maxsplit
parameter directly represents how many parent folders to move up and it always returns a result in just one pass through the path.
With Pathlib (recommended after Python 3.5, the/a general solution that works not only in file.py files, but also in Jupyter (or other kind of) notebook and Python shell is:
p = Path.cwd().resolve().parents[1]
You only need to substitute (__file__)
for cwd()
(current working directory).
Indeed it would even work just with:
p = Path().resolve().parents[1]
(and of course with .parent.parent
instead of parents[1]
)
Ok…I dont know where module x
is, but I know that I need to get the path to the directory two levels up.
So, is there a more elegant way to do:
import os
two_up = os.path.dirname(os.path.dirname(__file__))
Solutions for both Python 2 and 3 are welcome!
You can use pathlib
. Unfortunately this is only available in the stdlib for Python 3.4. If you have an older version you’ll have to install a copy from PyPI here. This should be easy to do using pip
.
from pathlib import Path
p = Path(__file__).parents[1]
print(p)
# /absolute/path/to/two/levels/up
This uses the parents
sequence which provides access to the parent directories and chooses the 2nd one up.
Note that p
in this case will be some form of Path
object, with their own methods. If you need the paths as string then you can call str
on them.
Very easy:
Here is what you want:
import os.path as path
two_up = path.abspath(path.join(__file__ ,"../.."))
Personally, I find that using the os module is the easiest method as outlined below. If you are only going up one level, replace (‘../..’) with (‘..’).
import os
os.chdir('../..')
--Check:
os.getcwd()
I don’t yet see a viable answer for 2.7 which doesn’t require installing additional dependencies and also starts from the file’s directory. It’s not nice as a single-line solution, but there’s nothing wrong with using the standard utilities.
import os
grandparent_dir = os.path.abspath( # Convert into absolute path string
os.path.join( # Current file's grandparent directory
os.path.join( # Current file's parent directory
os.path.dirname( # Current file's directory
os.path.abspath(__file__) # Current file path
),
os.pardir
),
os.pardir
)
)
print grandparent_dir
And to prove it works, here I start out in ~/Documents/notes
just so that I show the current directory doesn’t influence outcome. I put the file grandpa.py
with that script in a folder called “scripts”. It crawls up to the Documents dir and then to the user dir on a Mac.
(testing)AlanSE-OSX:notes AlanSE$ echo ~/Documents/scripts/grandpa.py
/Users/alancoding/Documents/scripts/grandpa.py
(testing)AlanSE-OSX:notes AlanSE$ python2.7 ~/Documents/scripts/grandpa.py
/Users/alancoding
This is the obvious extrapolation of the answer for the parent dir. Better to use a general solution than a less-good solution in fewer lines.
I have found that the following works well in 2.7.x
import os
two_up = os.path.normpath(os.path.join(__file__,'../'))
You can use this as a generic solution:
import os
def getParentDir(path, level=1):
return os.path.normpath( os.path.join(path, *([".."] * level)) )
For getting the directory 2 levels up:
import os.path as path
curr_dir=Path(os.path.dirname(os.path.abspath(__file__)))
two_dir_up_=os.fspath(Path(curr_dir.parent.parent).resolve())
I have done the following to go up two and drill down on other dir
default_config_dir=os.fspath(Path(curr_dir.parent.parent,
'data/config').resolve())
I was going to add this just to be silly, but also because it shows newcomers the potential usefulness of aliasing functions and/or imports.
Having written it, I think this code is more readable (i.e. lower time to grasp intention) than the other answers to date, and readability is (usually) king.
from os.path import dirname as up
two_up = up(up(__file__))
Note: you only want to do this kind of thing if your module is very small, or contextually cohesive.
More cross-platform implementation will be:
import pathlib
two_up = (pathlib.Path(__file__) / ".." / "..").resolve()
Using parent
is not supported on Windows. Also need to add .resolve()
, to:
Make the path absolute, resolving all symlinks on the way and also normalizing it (for example turning slashes into backslashes under Windows)
The best solution (for python >= 3.4) when executing from any directory is:
from pathlib import Path
two_up = Path(__file__).resolve().parents[1]
Assuming you want to access folder named xzy two folders up your python file. This works for me and platform independent.
“.././xyz”
(pathlib.Path(‘../../’) ).resolve()
100% working answer:
os.path.abspath(os.path.join(os.getcwd() ,"../.."))
There is already an accepted answer, but for two levels up I think a chaining approach is arguably more readable:
pathlib.Path(__file__).parent.parent.resolve()
Surprisingly it seems no one has yet explored this nice one-liner option:
import os
two_up = os.path.normpath(__file__).rsplit(os.sep, maxsplit=2)[0]
rsplit
is interesting since the maxsplit
parameter directly represents how many parent folders to move up and it always returns a result in just one pass through the path.
With Pathlib (recommended after Python 3.5, the/a general solution that works not only in file.py files, but also in Jupyter (or other kind of) notebook and Python shell is:
p = Path.cwd().resolve().parents[1]
You only need to substitute (__file__)
for cwd()
(current working directory).
Indeed it would even work just with:
p = Path().resolve().parents[1]
(and of course with .parent.parent
instead of parents[1]
)