How can I find the full path to the currently running Python script? That is to say, what do I have to do to achieve this:
$ pwd
/tmp
$ python baz.py
running from /tmp
file is baz.py
How can I find the full path to the currently running Python script? That is to say, what do I have to do to achieve this:
$ pwd
/tmp
$ python baz.py
running from /tmp
file is baz.py
__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'
__file__
is the one to use. (Such as, an __init__.py
where you don't know where it's going to live and need to add it's dirname()
to sys.path
when that module is imported...)
__file__
is defined to be). Your 2nd sentence is nothing to with the question, and is extremely puzzling: if you don't know where a module is, you can't import it, and if you can't import it you can't get its __file__
...
Commented
Mar 31, 2012 at 7:09
__main__
? Which is why I clarify that if you intend currently running line, __file__
is correct.
sys.path
without knowing it's path. HOWEVER, some modules (such as p4python
) require it's own path to be in sys.path
. I felt it silly to go around and modify everyone's systems to have that in their environment, so instead I made an __init.py__
that will do it automatically no matter where their project root is. We do not require our developers to have identical locations for the project root, as different contractors have different hardware / harddrive capacities.
This will print the directory in which the script lives (as opposed to the working directory):
import os
dirname, filename = os.path.split(os.path.abspath(__file__))
print "running from", dirname
print "file is", filename
Here's how it behaves, when I put it in c:\src
:
> cd c:\src
> python so-where.py
running from C:\src
file is so-where.py
> cd c:\
> python src\so-where.py
running from C:\src
file is so-where.py
__file__
altogether and following this: py2exe.org/index.cgi/WhereAmI ??
Commented
Aug 19, 2009 at 2:08
__file__
is a KLUDGE. Importing a dummy module is a KLUDGE. "you'll have to split it a bit differently" == "without any conditional checks"?? Hard to find?? here's another link: http://www.voidspace.org.uk/python/pathutils.html#get-main-dir
Commented
Aug 23, 2009 at 10:36
import sys, os
file = sys.argv[0]
pathname = os.path.dirname(file)
print 'running from %s' % os.path.abspath(pathname)
print 'file is %s' % file
Check os.getcwd() (docs)
The running file is always __file__
.
Here's a demo script, named identify.py
print __file__
Here's the results
MacBook-5:Projects slott$ python StackOverflow/identify.py
StackOverflow/identify.py
MacBook-5:Projects slott$ cd StackOverflow/
MacBook-5:StackOverflow slott$ python identify.py
identify.py
I would suggest
import os, sys
print os.path.split(os.path.abspath(os.path.realpath(sys.argv[0])))[0]
This way you can safely create symbolic links to the script executable and it will still find the correct directory.
The script name will (always?) be the first index of sys.argv:
import sys
print sys.argv[0]
An even easier way to find the path of your running script:
os.path.dirname(sys.argv[0])
r"C:\myfiles\fubar.py"
, the basename is "fubar.py"
-- is that what you call "the path of your running script"?
Commented
Aug 20, 2009 at 13:41
The directory of the script which python is executing is added to sys.path
This is actually an array (list) which contains other paths.
The first element contains the full path where the script is located (for windows).
Therefore, for windows, one can use:
import sys
path = sys.path[0]
print(path)
Others have suggested using sys.argv[0]
which works in a very similar way and is complete.
import sys
path = os.path.dirname(sys.argv[0])
print(path)
Note that sys.argv[0]
contains the full working directory (path) + filename whereas sys.path[0]
is the current working directory without the filename.
I have tested sys.path[0]
on windows and it works. I have not tested on other operating systems outside of windows, so somebody may wish to comment on this.
Aside from the aforementioned sys.argv[0]
, it is also possible to use the __main__
:
import __main__
print(__main__.__file__)
Beware, however, this is only useful in very rare circumstances;
and it always creates an import loop, meaning that the __main__
will not be fully executed at that moment.