Find the current directory and file's directory

In Python, what commands can I use to find:

1. the current directory (where I was in the terminal when I ran the Python script), and
2. where the file I am executing is?

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.)

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To get the current working directory use

import os
cwd = os.getcwd()

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Documentation references for the modules, constants and functions used above:

• The os and os.path modules.
• The __file__ constant
• os.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")

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: https://stackoverflow.com/questions/2632199/how-do-i-get-the-path-of-the-current-executed-file-in-python

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))

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?.

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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:

File scripts/2.py

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')

Output

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

1. To get the current directory full path

   >>import os
   >>print os.getcwd()
   

   Output: "C :\Users\admin\myfolder"

2. 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"

Pathlib can be used this way to get the directory containing the current script:

import pathlib
filepath = pathlib.Path(__file__).resolve().parent

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__))

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.

To get the current directory full path:

os.path.realpath('.')

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__)

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())

If you're searching for the location of the currently executed script, you can use sys.argv[0] to get the full path.

For question 1, use os.getcwd() # Get working directory and os.chdir(r'D:\Steam\steamapps\common') # Set working directory

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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).