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

Question

I have scripts calling other script files but I need to get the filepath of the file that is currently running within the process    For example  let s say I have three files  Using execfile    script 1 py calls script 2 py   In turn  script 2 py calls script 3 py     How can I get the file name and path of script 3 py  from code within script 3 py  without having to pass that information as arguments from script 2 py    Executing os getcwd   returns the original starting script s filepath not the current file s

User · Answer

file     as others have said  You may also want to use os path realpath to eliminate symlinks   import os  os path realpath   file

User · Answer

To get directory of executing script     print os path dirname  inspect getfile inspect currentframe

User · Answer

Here is what I use so I can throw my code anywhere without issue    name   is always defined  but   file   is only defined when the code is run as a file  e g  not in IDLE iPython   if    file    in globals        self name   globals      file     elif    file    in locals        self name   locals      file     else      self name     name    Alternatively  this can be written as  self name   globals   get    file     locals   get    file       name

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Since Python 3 is fairly mainstream  I wanted to include a pathlib answer  as I believe that it is probably now a better tool for accessing file and path information   from pathlib import Path  current file  Path   Path   file    resolve     If you are seeking the directory of the current file  it is as easy as adding  parent to the Path   statement   current path  Path   Path   file    parent resolve

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I have always just used the os feature of Current Working Directory  or CWD   This is part of the standard library  and is very easy to implement  Here is an example        import os     base directory   os getcwd

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Simplest way is   in script 1 py   import subprocess subprocess call   python3   lt path to script 2 py gt      in script 2 py   sys argv 0    P S   I ve tried execfile  but since it reads script 2 py as a string  sys argv 0  returned  lt string gt

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You can use inspect stack    import inspect os inspect stack   0     gt    lt frame object at 0x00AC2AC0 gt    g   Python  Test   GetCurrentProgram py   15    lt module gt      print inspect stack   0  n    0  os path abspath  inspect stack   0  1     gt   g   Python  Test   GetCurrentProgram py

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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

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import os os path dirname   file      relative directory path os path abspath   file      absolute file path os path basename   file      the file name only

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I used the approach with   file   os path abspath   file    but there is a little trick  it returns the  py file  when the code is run the first time   next runs give the name of   pyc file so I stayed with  inspect getfile inspect currentframe    or sys  getframe   f code co filename

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print   file    print   import    pathlib   Path   file    parent

User · Answer

To keep the migration consistency across platforms  macOS Windows Linux   try   path   r  s    os getcwd   replace

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I have a script that must work under windows environment  This code snipped is what I ve finished with   import os sys PROJECT PATH   os path abspath os path split sys argv 0   0     it s quite a hacky decision  But it requires no external libraries and it s the most important thing in my case

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import sys print sys argv 0

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I wrote a function which take into account eclipse debugger and unittest  It return the folder of the first script you launch  You can optionally specify the   file   var  but the main thing is that you don t have to share this variable across all your calling hierarchy   Maybe you can handle others stack particular cases I didn t see  but for me it s ok   import inspect  os def getRootDirectory  file  None               Get the directory of the root execution file     Can help  http   stackoverflow com questions 50499 how-do-i-get-the-path-and-name-of-the-file-that-is-currently-executing     For eclipse user with unittest or debugger  the function search for the correct folder in the stack     You can pass   file    with 4 underscores  if you want the caller directory               If we don t have the   file         if  file  is None            We get the last           rootFile   inspect stack   -1  1          folder   os path abspath rootFile            If we use unittest           if    pysrc  in folder   amp    org python pydev  in folder               previous   None               We search from left to right the case py               for el in inspect stack                    currentFile   os path abspath el 1                   if   unittest case py  in currentFile      org python pydev  in currentFile                       break                 previous   currentFile             folder   previous           We return the folder           return os path dirname folder      else            We return the folder according to specified   file             return os path dirname os path realpath  file

User · Answer

I think this is cleaner   import inspect print inspect stack   0  1    and gets the same information as   print inspect getfile inspect currentframe      Where  0  is the current frame in the stack  top of stack  and  1  is for the file name  increase to go backwards in the stack i e   print inspect stack   1  1    would be the file name of the script that called the current frame  Also  using  -1  will get you to the bottom of the stack  the original calling script

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Most of these answers were written in Python version 2 x or earlier  In Python 3 x the syntax for the print function has changed to require parentheses  i e  print     So  this earlier high score answer from user13993 in Python 2 x   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   Becomes in Python 3 x   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

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Update 2018-11-28   Here is a summary of experiments with Python 2 and 3  With   main py - runs foo py foo py - runs lib bar py lib bar py - prints filepath expressions      Python   Run statement         Filepath expression                       -------- --------------------- ----------------------------------------         2   execfile              os path abspath inspect stack   0  1            2   from lib import bar     file                                          3   exec                   wasn t able to obtain it                       3   import lib bar          file                                     For Python 2  it might be clearer to switch to packages so can use from lib import bar - just add empty   init   py files to the two folders   For Python 3  execfile doesn t exist - the nearest alternative is exec open  lt filename gt   read     though this affects the stack frames  It s simplest to just use import foo and import lib bar - no   init   py files needed   See also Difference between import and execfile    Original Answer   Here is an experiment based on the answers in this thread - with Python 2 7 10 on Windows   The stack-based ones are the only ones that seem to give reliable results  The last two have the shortest syntax  i e  -  print os path abspath inspect stack   0  1                       C  filepaths lib bar py print os path dirname os path abspath inspect stack   0  1       C  filepaths lib   Here s to these being added to sys as functions  Credit to  Usagi and  pablog  Based on the following three files  and running main py from its folder with python main py  also tried execfiles with absolute paths and calling from a separate folder     C  filepaths main py  execfile  foo py   C  filepaths foo py  execfile  lib bar py   C  filepaths lib bar py   import sys import os import inspect  print  Python     sys version print  print   file                                            main py print sys argv 0                                        main py print inspect stack   0  1                              lib bar py print sys path 0                                        C  filepaths print  print os path realpath   file                           C  filepaths main py print os path abspath   file                            C  filepaths main py print os path basename   file                           main py print os path basename os path realpath sys argv 0      main py print  print sys path 0                                        C  filepaths print os path abspath os path split sys argv 0   0      C  filepaths print os path dirname os path abspath   file            C  filepaths print os path dirname os path realpath sys argv 0       C  filepaths print os path dirname   file                             empty string  print  print inspect getfile inspect currentframe              lib bar py  print os path abspath inspect getfile inspect currentframe       C  filepaths lib bar py print os path dirname os path abspath inspect getfile inspect currentframe        C  filepaths lib print  print os path abspath inspect stack   0  1              C  filepaths lib bar py print os path dirname os path abspath inspect stack   0  1       C  filepaths lib print

User · Answer

I think it s just   file     Sounds like you may also want to checkout the inspect module

User · Answer

import sys  print sys path 0    this would print the path of the currently executing script

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The suggestions marked as best are all true if your script consists of only one file    If you want to find out the name of the executable  i e  the root file passed to the python interpreter for the current program  from a file that may be imported as a module  you need to do this  let s assume this is in a file named foo py    import inspect  print inspect stack   -1  1   Because the last thing   -1   on the stack is the first thing that went into it  stacks are LIFO FILO data structures    Then in file bar py if you import foo it ll print bar py  rather than foo py  which would be the value of all of these      file   inspect getfile inspect currentframe    inspect stack   0  1

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The   file   attribute works for both the file containing the main execution code as well as imported modules   See https   web archive org web 20090918095828 http   pyref infogami com   file

User · Answer

import os os path dirname os path abspath   file       No need for inspect or any other library   This worked for me when I had to import a script  from a different directory then the executed script   that used a configuration file residing in the same folder as the imported script

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This should work   import os sys filename os path basename os path realpath sys argv 0    dirname os path dirname os path realpath sys argv 0

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import os print os path basename   file      this will give us the filename only  i e  if abspath of file is c  abcd abc py then 2nd line will print abc py

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It s not entirely clear what you mean by  the filepath of the file that is currently running within the process   sys argv 0  usually contains the location of the script that was invoked by the Python interpreter  Check the sys documentation for more details   As  Tim and  Pat Notz have pointed out  the   file   attribute provides access to     the file from which the module was   loaded  if it was loaded from a file

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Try this   import os os path dirname os path realpath   file

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import os  import wx     return the full path of this file print os getcwd     icon   wx Icon os getcwd       img image png   wx BITMAP TYPE PNG  16  16     put the icon on the frame self SetIcon icon

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if you want just the filename without    or  py you can try this  filename   testscript py file name     file   2 -3    file name will print testscript you can generate whatever you want by changing the index inside

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

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