How do I determine:
import os cwd = os.getcwd()
to pwd within python - anyone os.getcwd()
" after literally hours of searching. Ugh. - anyone To get the full path to the directory a Python file is contained in, write this in that file:
import os
dir_path = os.path.dirname(os.path.realpath(__file__))
(Note that the incantation above won't work if you've already used os.chdir()
to change your current working directory, since the value of the __file__
constant is relative to the current working directory and is not changed by an os.chdir()
call.)
To get the current working directory use
import os
cwd = os.getcwd()
Documentation references for the modules, constants and functions used above:
os
and os.path
modules.__file__
constantos.path.realpath(path)
(returns "the canonical path of the specified filename, eliminating any symbolic links encountered in the path")os.path.dirname(path)
(returns "the directory name of pathname path
")os.getcwd()
(returns "a string representing the current working directory")os.chdir(path)
("change the current working directory to path
")Answered 2023-09-20 20:22:52
realpath
with dirname
, but it will differ when the file (or its directory) is actually a symbolic link. - anyone NameError: name '__file__' is not defined
. How to solve this? - anyone Current working directory: os.getcwd()
And the __file__
attribute can help you find out where the file you are executing is located. This Stack Overflow post explains everything: How do I get the path of the current executed file in Python?
Answered 2023-09-20 20:22:52
You may find this useful as a reference:
import os
print("Path at terminal when executing this file")
print(os.getcwd() + "\n")
print("This file path, relative to os.getcwd()")
print(__file__ + "\n")
print("This file full path (following symlinks)")
full_path = os.path.realpath(__file__)
print(full_path + "\n")
print("This file directory and name")
path, filename = os.path.split(full_path)
print(path + ' --> ' + filename + "\n")
print("This file directory only")
print(os.path.dirname(full_path))
Answered 2023-09-20 20:22:52
__file__
signifies here? It does not work for me. - anyone __file__
is an attribute of the module object. You need run the code inside a Python file, not on the REPL. - anyone The pathlib
module, introduced in Python 3.4 (PEP 428 — The pathlib module — object-oriented filesystem paths), makes the path-related experience much much better.
pwd
/home/skovorodkin/stack
tree
.
└── scripts
├── 1.py
└── 2.py
In order to get the current working directory, use Path.cwd()
:
from pathlib import Path
print(Path.cwd()) # /home/skovorodkin/stack
To get an absolute path to your script file, use the Path.resolve()
method:
print(Path(__file__).resolve()) # /home/skovorodkin/stack/scripts/1.py
And to get the path of a directory where your script is located, access .parent
(it is recommended to call .resolve()
before .parent
):
print(Path(__file__).resolve().parent) # /home/skovorodkin/stack/scripts
Remember that __file__
is not reliable in some situations: How do I get the path of the current executed file in Python?.
Please note, that Path.cwd()
, Path.resolve()
and other Path
methods return path objects (PosixPath
in my case), not strings. In Python 3.4 and 3.5 that caused some pain, because open
built-in function could only work with string or bytes objects, and did not support Path
objects, so you had to convert Path
objects to strings or use the Path.open()
method, but the latter option required you to change old code:
from pathlib import Path
p = Path(__file__).resolve()
with p.open() as f: pass
with open(str(p)) as f: pass
with open(p) as f: pass
print('OK')
python3.5 scripts/2.py
Traceback (most recent call last):
File "scripts/2.py", line 11, in <module>
with open(p) as f:
TypeError: invalid file: PosixPath('/home/skovorodkin/stack/scripts/2.py')
As you can see, open(p)
does not work with Python 3.5.
PEP 519 — Adding a file system path protocol, implemented in Python 3.6, adds support of PathLike
objects to the open
function, so now you can pass Path
objects to the open
function directly:
python3.6 scripts/2.py
OK
Answered 2023-09-20 20:22:52
app_path = Path(__file__).resolve().parent.parent.parent
as a parallel to ../../../
if you need to. - anyone python3.5
" and "python3.6
"? Ubuntu Ubuntu MATE 20.04 (Focal Fossa) doesn't (at least not by default). It has executables by the name "python3
" and "python2
" (but not "python
" - which causes some things to break) - anyone python3.x
symlinks that time. Maybe I thought it would make snippets a bit clearer to the reader. - anyone To get the current directory full path
>>import os
>>print os.getcwd()
Output: "C :\Users\admin\myfolder"
To get the current directory folder name alone
>>import os
>>str1=os.getcwd()
>>str2=str1.split('\\')
>>n=len(str2)
>>print str2[n-1]
Output: "myfolder"
Answered 2023-09-20 20:22:52
os.getcwd().split('\\')[-1]
- anyone Pathlib can be used this way to get the directory containing the current script:
import pathlib
filepath = pathlib.Path(__file__).resolve().parent
Answered 2023-09-20 20:22:52
app_path = Path(__file__).resolve().parent.parent.parent
as a parallel to ../../../
if you need to. - anyone If you are trying to find the current directory of the file you are currently in:
OS agnostic way:
dirname, filename = os.path.split(os.path.abspath(__file__))
Answered 2023-09-20 20:22:52
If you're using Python 3.4, there is the brand new higher-level pathlib
module which allows you to conveniently call pathlib.Path.cwd()
to get a Path
object representing your current working directory, along with many other new features.
More info on this new API can be found here.
Answered 2023-09-20 20:22:52
pathlib2
: pypi.python.org/pypi/pathlib2 - anyone To get the current directory full path:
os.path.realpath('.')
Answered 2023-09-20 20:22:52
os.getcwd()
and `os.path.realpath('.') return exactly the same string path. - anyone Answer to #1:
If you want the current directory, do this:
import os
os.getcwd()
If you want just any folder name and you have the path to that folder, do this:
def get_folder_name(folder):
'''
Returns the folder name, given a full folder path
'''
return folder.split(os.sep)[-1]
Answer to #2:
import os
print os.path.abspath(__file__)
Answered 2023-09-20 20:22:52
I think the most succinct way to find just the name of your current execution context would be:
current_folder_path, current_folder_name = os.path.split(os.getcwd())
Answered 2023-09-20 20:22:52
If you're searching for the location of the currently executed script, you can use sys.argv[0]
to get the full path.
Answered 2023-09-20 20:22:52
sys.argv[0]
needn't contain the full path to the executing script. - anyone For question 1, use os.getcwd() # Get working directory
and os.chdir(r'D:\Steam\steamapps\common') # Set working directory
I recommend using sys.argv[0]
for question 2 because sys.argv
is immutable and therefore always returns the current file (module object path) and not affected by os.chdir()
. Also you can do like this:
import os
this_py_file = os.path.realpath(__file__)
# vvv Below comes your code vvv #
But that snippet and sys.argv[0]
will not work or will work weird when compiled by PyInstaller, because magic properties are not set in __main__
level and sys.argv[0]
is the way your executable was called (it means that it becomes affected by the working directory).
Answered 2023-09-20 20:22:52