How do you properly determine the current script directory

Question

I would like to see what is the best way to determine the current script directory in Python  I discovered that  due to the many ways of calling Python code  it is hard to find a good solution  Here are some problems     file   is not defined if the script is executed with exec  execfile   module   is defined only in modules  Use cases     myfile py python myfile py   somedir myfile py python somedir myfile py execfile  myfile py    from another script  that can be located in another directory and that can have another current directory   I know that there is no perfect solution  but I m looking for the best approach that solves most of the cases  The most used approach is os path dirname os path abspath   file     but this really doesn t work if you execute the script from another one with exec    Warning Any solution that uses current directory will fail  this can be different based on the way the script is called or it can be changed inside the running script

User · Answer

Just use os path dirname os path abspath   file     and examine very carefully whether there is a real need for the case where exec is used   It could be a sign of troubled design if you are not able to use your script as a module     Keep in mind Zen of Python  8  and if you believe there is a good argument for a use-case where it must work for exec  then please let us know some more details about the background of the problem

User · Answer

os path dirname os path abspath   file       is indeed the best you re going to get   It s unusual to be executing a script with exec execfile  normally you should be using the module infrastructure to load scripts  If you must use these methods  I suggest setting   file   in the globals you pass to the script so it can read that filename   There s no other way to get the filename in execed code  as you note  the CWD may be in a completely different place

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If   file   is available    -- script1 py -- import os file path   os path abspath   file    print os path dirname file path    For those we want to be able to run the command from the interpreter or get the path of the place you re running the script from    -- script2 py -- import os print os path abspath       This works from the interpreter  But when run in a script  or imported  it gives the path of the place where you ran the script from  not the path of directory containing the script with the print  Example  If your directory structure is test dir  in the home dir   -- main py  -- test subdir      -- script1 py      -- script2 py  with   -- main py -- import script1 py import script2 py  The output is    test dir test subdir   test dir

User · Answer

First   a couple missing use-cases here if we re talking about ways to inject anonymous code    code compile command   code interact   imp load compiled   imp load dynamic   imp load module     builtin   compile   loading C compiled shared objects  example   socket     But  the real question is  what is your goal - are you trying to enforce some sort of security  Or are you just interested in whats being loaded   If you re interested in security  the filename that is being imported via exec execfile is inconsequential - you should use rexec  which offers the following      This module contains the RExec class    which supports r eval    r execfile      r exec    and r import   methods  which    are restricted versions of the standard   Python functions eval    execfile   and    the exec and import statements  Code    executed in this restricted environment   will only have access to modules and    functions that are deemed safe  you can    subclass RExec add or remove capabilities as   desired    However  if this is more of an academic pursuit   here are a couple goofy approaches that you might be able to dig a little deeper into     Example scripts     deep py  print    gt  gt  level 1  execfile  deeper py   print    lt  lt  level 1      deeper py  print   t  gt  gt  level 2  exec  import sys  sys path append   tmp    import deepest   print   t  lt  lt  level 2     tmp deepest py  print   t t  gt  gt  level 3  print   t t t I can see the earths core   print   t t  lt  lt  level 3      codespy py  import sys  os  def overseer frame  event  arg       print  loaded  s     os path abspath frame f code co filename   sys settrace overseer  execfile  deep py   sys exit 0    Output  loaded  Users synthesizerpatel deep py   gt  gt  level 1 loaded  Users synthesizerpatel deeper py       gt  gt  level 2 loaded  Users synthesizerpatel  lt string gt   loaded  tmp deepest py           gt  gt  level 3             I can see the earths core           lt  lt  level 3      lt  lt  level 2  lt  lt  level 1   Of course  this is a resource-intensive way to do it  you d be tracing all your code   Not very efficient  But  I think it s a novel approach since it continues to work even as you get deeper into the nest  You can t override  eval   Although you can override execfile     Note  this approach only coveres exec execfile  not  import   For higher level  module  load hooking you might be able to use use sys path hooks  Write-up courtesy of PyMOTW    Thats all I have off the top of my head

User · Answer

Note  this answer is now a package   pip install locate   gt  gt  gt  from locate import this dir  gt  gt  gt  print this dir    C  Users simon  For  py scripts as well as interactive usage  I frequently use the directory of my scripts  for accessing files stored along side them   but I also frequently run these scripts in an interactive shell for debugging purposes  I define   dirpath   as   When running or importing a  py file  the file s base directory  This is always the correct path  When running an  ipyn notebook  the current working directory  This is always the correct path  since Jupyter sets the working directory as the  ipynb base directory  When running in a REPL  the current working directory  Hmm  what is the actual  quot correct path quot  when the code is detached from a file   Rather  make it your responsibility to change into the  quot correct path quot  before invoking the REPL   Python 3 4  and above   from pathlib import Path   dirpath     Path globals   get  quot   file   quot    quot     quot    absolute   parent  Python 2  and above   import os   dirpath     os path dirname os path abspath globals   get  quot   file   quot    quot     quot      Explanation   globals   returns all the global variables as a dictionary   get  quot   file   quot    quot     quot   returns the value from the key  quot   file   quot  if it exists in globals    otherwise it returns the provided default value  quot     quot   The rest of the code just expands   file    or  quot     quot   into an absolute filepath  and then returns the filepath s base directory

User · Answer

If you really want to cover the case that a script is called via execfile       you can use the inspect module to deduce the filename  including the path    As far as I am aware  this will work for all cases you listed   filename   inspect getframeinfo inspect currentframe    filename path   os path dirname os path abspath filename

User · Answer

This should work in most cases   import os sys dirname os path dirname os path realpath sys argv 0

User · Answer

The os path    approach was the  done thing  in Python 2   In Python 3  you can find directory of script as follows   from pathlib import Path cwd   Path   file    parents 0

User · Answer

Here is a partial solution  still better than all published ones so far   import sys  os  os path  inspect   os chdir        if    file    not in locals          file     inspect getframeinfo inspect currentframe    0   print os path dirname os path abspath   file       Now this works will all calls but if someone use chdir   to change the current directory  this will also fail   Notes     sys argv 0  is not going to work  will return -c if you execute the script with python -c  execfile  path-tester py    I published a complete test at https   gist github com 1385555 and you are welcome to improve it

User · Answer

Would  import os cwd   os getcwd     do what you want   I m not sure what exactly you mean by the  current script directory    What would the expected output be for the use cases you gave

User · Answer

In Python 3 4  you can use the simpler pathlib module  from inspect import currentframe  getframeinfo from pathlib import Path  filename   getframeinfo currentframe    filename parent   Path filename  resolve   parent  You can also use   file   to avoid the inspect module altogether  from pathlib import Path parent   Path   file    resolve   parent

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usr bin env python import inspect import os import sys  def get script dir follow symlinks True       if getattr sys   frozen   False     py2exe  PyInstaller  cx Freeze         path   os path abspath sys executable      else          path   inspect getabsfile get script dir      if follow symlinks          path   os path realpath path      return os path dirname path   print get script dir      It works on CPython  Jython  Pypy  It works if the script is executed using execfile    sys argv 0  and   file   -based solutions would fail here   It works if the script is inside an executable zip file   an egg   It works if the script is  imported   PYTHONPATH  path to library zip python -mscript to run  from a zip file  it returns the archive path in this case  It works if the script is compiled into a standalone executable  sys frozen   It works for symlinks  realpath eliminates symbolic links   It works in an interactive interpreter  it returns the current working directory in this case

User · Answer

Hopefully this helps -  If you run a script module from anywhere you ll be able to access the   file   variable which is a module variable representing the location of the script    On the other hand  if you re using the interpreter you don t have access to that variable  where you ll get a name NameError and os getcwd   will give you the incorrect directory if you re running the file from somewhere else    This solution should give you what you re looking for in all cases   from inspect import getsourcefile from os path import abspath abspath getsourcefile lambda 0     I haven t thoroughly tested it but it solved my problem

[python] How do you properly determine the current script directory?

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