How to Get the Path of the Python Script I am Running In

Getting the full path the script is currently running in [Python]

Let me explain to you by explaining the outputs of all the commands you mentioned by running them in Windows.

Nothing special here, just importing some modules. Importing numpy to test the commands.

In [1]: import os
   ...: import sys
   ...: import numpy

This gives the current working directory in which your process/program is executing. (detailed tutorial here)

In [2]: os.getcwd()
Out[2]: 'C:\\Users\\<username>'

This tells which file the module is loaded from which file.

In [3]: numpy.__file__
Out[3]: 'C:\\Users\\<username>\\anaconda3\\lib\\site-packages\\numpy\\__init__.py'

The path of the loaded file for the module, it can be a relative path.

In [4]: os.path.dirname(numpy.__file__)
Out[4]: 'C:\\Users\\Punit Singh\\anaconda3\\lib\\site-packages\\numpy'

The actual path of the loaded file for the module. See the difference from previous in lib and Lib. The folder name actually starts from the capital letter.

In [5]: os.path.dirname(os.path.realpath(numpy.__file__))
Out[5]: 'C:\\Users\\Punit Singh\\anaconda3\\Lib\\site-packages\\numpy'

This is the absolute path, you'll see the difference when you'll run this in Linux. It starts with the root folder denoted by \.

In [6]: os.path.dirname(os.path.abspath(numpy.__file__))
Out[6]: 'C:\\Users\\Punit Singh\\anaconda3\\lib\\site-packages\\numpy'

[7] and [8] respectively give the path and absolute path of the file which is currently executing, here sys.argv[0] returns the file which is currently executing, since currently, I am running ipython, it is ipython.exe shown in [9]. (detailed tutorial here)

In [7]: os.path.dirname(sys.argv[0])
Out[7]: 'C:\\Users\\<username>\\anaconda3\\Scripts'

In [8]: os.path.abspath(os.path.dirname(sys.argv[0]))
Out[8]: 'C:\\Users\\<username>\\anaconda3\\Scripts'

In [9]: sys.argv[0]
Out[9]: 'C:\\Users\\<username>\\anaconda3\\Scripts\\ipython'

This gives the absolute path of the executable binary for the Python interpreter (documentation here)

In [10]: sys.executable
Out[10]: 'C:\\Users\\Punit Singh\\anaconda3\\python.exe'

Now coming to your requirement, I understand that you need to access files where this ipython.exe as shown in [9] is located, so you can use commands mentioned in [7] or [8].

How do I get the path of the current executed file in Python?

You can't directly determine the location of the main script being executed. After all, sometimes the script didn't come from a file at all. For example, it could come from the interactive interpreter or dynamically generated code stored only in memory.

However, you can reliably determine the location of a module, since modules are always loaded from a file. If you create a module with the following code and put it in the same directory as your main script, then the main script can import the module and use that to locate itself.

some_path/module_locator.py:

def we_are_frozen():
    # All of the modules are built-in to the interpreter, e.g., by py2exe
    return hasattr(sys, "frozen")

def module_path():
    encoding = sys.getfilesystemencoding()
    if we_are_frozen():
        return os.path.dirname(unicode(sys.executable, encoding))
    return os.path.dirname(unicode(__file__, encoding))

some_path/main.py:

import module_locator
my_path = module_locator.module_path()

If you have several main scripts in different directories, you may need more than one copy of module_locator.

Of course, if your main script is loaded by some other tool that doesn't let you import modules that are co-located with your script, then you're out of luck. In cases like that, the information you're after simply doesn't exist anywhere in your program. Your best bet would be to file a bug with the authors of the tool.

Find path to currently running file

__file__ is NOT what you are looking for. Don't use accidental side-effects

sys.argv[0] is always the path to the script (if in fact a script has been invoked) -- see http://docs.python.org/library/sys.html#sys.argv

__file__ is the path of the currently executing file (script or module). This is accidentally the same as the script if it is accessed from the script! If you want to put useful things like locating resource files relative to the script location into a library, then you must use sys.argv[0].

Example:

C:\junk\so>type \junk\so\scriptpath\script1.py
import sys, os
print "script: sys.argv[0] is", repr(sys.argv[0])
print "script: __file__ is", repr(__file__)
print "script: cwd is", repr(os.getcwd())
import whereutils
whereutils.show_where()

C:\junk\so>type \python26\lib\site-packages\whereutils.py
import sys, os
def show_where():
    print "show_where: sys.argv[0] is", repr(sys.argv[0])
    print "show_where: __file__ is", repr(__file__)
    print "show_where: cwd is", repr(os.getcwd())

C:\junk\so>\python26\python scriptpath\script1.py
script: sys.argv[0] is 'scriptpath\\script1.py'
script: __file__ is 'scriptpath\\script1.py'
script: cwd is 'C:\\junk\\so'
show_where: sys.argv[0] is 'scriptpath\\script1.py'
show_where: __file__ is 'C:\\python26\\lib\\site-packages\\whereutils.pyc'
show_where: cwd is 'C:\\junk\\so'

How can I find script's directory?

You need to call os.path.realpath on __file__, so that when __file__ is a filename without the path you still get the dir path:

import os
print(os.path.dirname(os.path.realpath(__file__)))

How do I get the path and name of the file that is currently executing?

p1.py:

execfile("p2.py")

p2.py:

import inspect, os
print (inspect.getfile(inspect.currentframe())) # script filename (usually with path)
print (os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe())))) # script directory

How do i get the path of the python script i am running in for multiple people

import inspect, os
print inspect.getfile(inspect.currentframe()) # file name and path of script
print os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe()))) # directory script is running in

See this answered question

How do I find directory of the Python running script from inside the script?

To get the directory that contains the module you are running:

import os
path = os.path.dirname(os.path.realpath(__file__))

Or if you want the directory from which the script was invoked:

import os
path = os.getcwd()

From the docs:

__file__ is the pathname of the file from which the module was loaded, if it was loaded from a file.

Depending on how the script is called, this may be a relative path from os.getcwd(), so os.path.realpath(__file__) will convert this to an absolute path (or do nothing is the __file__ is already an absolute path). os.path.dirname() will then return the full directory by stripping off the filename.