Is there a simple way to move up one directory in python using a single line of code? Something similar to cd .. in command line
>>> import os
>>> print os.path.abspath(os.curdir)
C:\Python27
>>> os.chdir("..")
>>> print os.path.abspath(os.curdir)
C:\
Using os.chdir should work:
import os
os.chdir('..')
Obviously that os.chdir('..') is the right answer here. But just FYI, if in the future you come across situation when you have to extensively manipulate directories and paths, here is a great package (Unipath) which lets you treat them as Python objects: https://pypi.python.org/pypi/Unipath
so that you could do something like this:
>>> from unipath import Path
>>> p = Path("/usr/lib/python2.5/gopherlib.py")
>>> p.parent
Path("/usr/lib/python2.5")
>>> p.name
Path("gopherlib.py")
>>> p.ext
'.py'
Well.. I'm not sure how portable os.chdir('..') would actually be. Under Unix those are real filenames. I would prefer the following:
import os
os.chdir(os.path.dirname(os.getcwd()))
That gets the current working directory, steps up one directory, and then changes to that directory.
In Python 3.4 pathlib was introduced:
>>> from pathlib import Path
>>> p = Path('/etc/usr/lib')
>>> p
PosixPath('/etc/usr/lib')
>>> p.parent
PosixPath('/etc/usr')
It also comes with many other helpful features e.g. for joining paths using slashes or easily walking the directory tree.
For more information refer to the docs or this blog post, which covers the differences between os.path and pathlib.
Although this is not exactly what OP meant as this is not super simple, however, when running scripts from Notepad++ the os.getcwd() method doesn't work as expected. This is what I would do:
import os
# get real current directory (determined by the file location)
curDir, _ = os.path.split(os.path.abspath(__file__))
print(curDir) # print current directory
Define a function like this:
def dir_up(path,n): # here 'path' is your path, 'n' is number of dirs up you want to go
for _ in range(n):
path = dir_up(path.rpartition("\\")[0], 0) # second argument equal '0' ensures that
# the function iterates proper number of times
return(path)
The use of this function is fairly simple - all you need is your path and number of directories up.
print(dir_up(curDir,3)) # print 3 directories above the current one
The only minus is that it doesn't stop on drive letter, it just will show you empty string.
A convenient way to move up multiple directories is pathlib:
from pathlib import Path
full_path = "C:\Program Files\Python37\lib\pathlib.py"
print(Path(full_path).parents[0])
print(Path(full_path).parents[1])
print(Path(full_path).parents[2])
print(Path(full_path).parents[3])
print([str(Path(full_path).parents[i]) for i in range(4)])
output:
C:\Program Files\Python37\lib
C:\Program Files\Python37
C:\Program Files
C:\
['C:\\Program Files\\Python37\\lib', 'C:\\Program Files\\Python37', 'C:\\Program Files', 'C:\\']
Combine Kim's answer with os:
p=Path(os.getcwd())
os.chdir(p.parent)
Related
This question already has answers here:
How do you properly determine the current script directory?
(16 answers)
Closed 6 months ago.
I'm building a simple helper script for work that will copy a couple of template files in our code base to the current directory. I don't, however, have the absolute path to the directory where the templates are stored. I do have a relative path from the script but when I call the script it treats that as a path relative to the current working directory. Is there a way to specify that this relative url is from the location of the script instead?
In the file that has the script, you want to do something like this:
import os
dirname = os.path.dirname(__file__)
filename = os.path.join(dirname, 'relative/path/to/file/you/want')
This will give you the absolute path to the file you're looking for. Note that if you're using setuptools, you should probably use its package resources API instead.
UPDATE: I'm responding to a comment here so I can paste a code sample. :-)
Am I correct in thinking that __file__ is not always available (e.g. when you run the file directly rather than importing it)?
I'm assuming you mean the __main__ script when you mention running the file directly. If so, that doesn't appear to be the case on my system (python 2.5.1 on OS X 10.5.7):
#foo.py
import os
print os.getcwd()
print __file__
#in the interactive interpreter
>>> import foo
/Users/jason
foo.py
#and finally, at the shell:
~ % python foo.py
/Users/jason
foo.py
However, I do know that there are some quirks with __file__ on C extensions. For example, I can do this on my Mac:
>>> import collections #note that collections is a C extension in Python 2.5
>>> collections.__file__
'/System/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/lib-
dynload/collections.so'
However, this raises an exception on my Windows machine.
It's 2018 now, and Python has already evolved to the __future__ long time ago. So how about using the amazing pathlib coming with Python 3.4 to accomplish the task instead of struggling with os, os.path, glob , shutil, etc.
So we have 3 paths here (possibly duplicated):
mod_path: which is the path of the simple helper script
src_path: which contains a couple of template files waiting to be copied.
cwd: current directory, the destination of those template files.
and the problem is: we don't have the full path of src_path, only know its relative path to the mod_path.
Now let's solve this with the amazing pathlib:
# Hope you don't be imprisoned by legacy Python code :)
from pathlib import Path
# `cwd`: current directory is straightforward
cwd = Path.cwd()
# `mod_path`: According to the accepted answer and combine with future power
# if we are in the `helper_script.py`
mod_path = Path(__file__).parent
# OR if we are `import helper_script`
mod_path = Path(helper_script.__file__).parent
# `src_path`: with the future power, it's just so straightforward
relative_path_1 = 'same/parent/with/helper/script/'
relative_path_2 = '../../or/any/level/up/'
src_path_1 = (mod_path / relative_path_1).resolve()
src_path_2 = (mod_path / relative_path_2).resolve()
In the future, it's just that simple.
Moreover, we can select and check and copy/move those template files with pathlib:
if src_path != cwd:
# When we have different types of files in the `src_path`
for template_path in src_path.glob('*.ini'):
fname = template_path.name
target = cwd / fname
if not target.exists():
# This is the COPY action
with target.open(mode='wb') as fd:
fd.write(template_path.read_bytes())
# If we want MOVE action, we could use:
# template_path.replace(target)
you need os.path.realpath (sample below adds the parent directory to your path)
import sys,os
sys.path.append(os.path.realpath('..'))
As mentioned in the accepted answer
import os
dir = os.path.dirname(__file__)
filename = os.path.join(dir, '/relative/path/to/file/you/want')
I just want to add that
the latter string can't begin with the backslash , infact no string
should include a backslash
It should be something like
import os
dir = os.path.dirname(__file__)
filename = os.path.join(dir, 'relative','path','to','file','you','want')
The accepted answer can be misleading in some cases , please refer to this link for details
Consider my code:
import os
def readFile(filename):
filehandle = open(filename)
print filehandle.read()
filehandle.close()
fileDir = os.path.dirname(os.path.realpath('__file__'))
print fileDir
#For accessing the file in the same folder
filename = "same.txt"
readFile(filename)
#For accessing the file in a folder contained in the current folder
filename = os.path.join(fileDir, 'Folder1.1/same.txt')
readFile(filename)
#For accessing the file in the parent folder of the current folder
filename = os.path.join(fileDir, '../same.txt')
readFile(filename)
#For accessing the file inside a sibling folder.
filename = os.path.join(fileDir, '../Folder2/same.txt')
filename = os.path.abspath(os.path.realpath(filename))
print filename
readFile(filename)
See sys.path
As initialized upon program startup, the first item of this list, path[0], is the directory containing the script that was used to invoke the Python interpreter.
Use this path as the root folder from which you apply your relative path
>>> import sys
>>> import os.path
>>> sys.path[0]
'C:\\Python25\\Lib\\idlelib'
>>> os.path.relpath(sys.path[0], "path_to_libs") # if you have python 2.6
>>> os.path.join(sys.path[0], "path_to_libs")
'C:\\Python25\\Lib\\idlelib\\path_to_libs'
Instead of using
import os
dirname = os.path.dirname(__file__)
filename = os.path.join(dirname, 'relative/path/to/file/you/want')
as in the accepted answer, it would be more robust to use:
import inspect
import os
dirname = os.path.dirname(os.path.abspath(inspect.stack()[0][1]))
filename = os.path.join(dirname, 'relative/path/to/file/you/want')
because using __file__ will return the file from which the module was loaded, if it was loaded from a file, so if the file with the script is called from elsewhere, the directory returned will not be correct.
These answers give more detail: https://stackoverflow.com/a/31867043/5542253 and https://stackoverflow.com/a/50502/5542253
From what suggest others and from pathlib documentation, a simple (but not ideal) solution is the following (suppose the file we need to refer to is Test/data/users.csv):
# Current file location: Tests/src/long/module/subdir/some_script.py
from pathlib import Path
# back to Tests/
PROJECT_ROOT = Path(__file__).parents[4]
# then down to Test/data/users.csv
CSV_USERS_PATH = PROJECT_ROOT / 'data' / 'users.csv'
with CSV_USERS_PATH.open() as users:
print(users.read())
This works but looks a bit odd because if you move some_script.py around, the path to the root of our project may change (and we would therefore need to change the parents[4] part).
I think I found a better solution that, based on the same idea.
We will use a file paths.py to store where the root of the project is, this file will remain at the same location compared to the root directory.
Tests
├── data
│ └── users.csv
└── src
├── long
│ └── module
│ └── subdir
│ └── some_script.py
├── main.py
└── paths.py
Where paths.py's only responsability is to provide PROJECT_ROOT:
from pathlib import Path
PROJECT_ROOT = Path(__file__).parents[1]
All scripts can now use paths.PROJECT_ROOT to express absolute paths from the root of the project. For example in src/long/module/subdir/some_script.py we could have:
from paths import PROJECT_ROOT
CSV_USERS_PATH = PROJECT_ROOT / 'data' / 'users.csv'
def hello():
with CSV_USERS_PATH.open() as f:
print(f.read())
And everything goes as expected:
~/Tests/src/$ python main.py
/Users/cglacet/Tests/data/users.csv
hello, user
~/Tests/$ python src/main.py
/Users/cglacet/Tests/data/users.csv
hello, user
The main.py script simply is:
from long.module.subdir import some_script
some_script.hello()
summary of the most important commands
>>> import os
>>> os.path.join('/home/user/tmp', 'subfolder')
'/home/user/tmp/subfolder'
>>> os.path.normpath('/home/user/tmp/../test/..')
'/home/user'
>>> os.path.relpath('/home/user/tmp', '/home/user')
'tmp'
>>> os.path.isabs('/home/user/tmp')
True
>>> os.path.isabs('/tmp')
True
>>> os.path.isabs('tmp')
False
>>> os.path.isabs('./../tmp')
False
>>> os.path.realpath('/home/user/tmp/../test/..') # follows symbolic links
'/home/user'
A detailed description is found in the docs.
These are linux paths. Windows should work analogous.
Hi first of all you should understand functions os.path.abspath(path) and os.path.relpath(path)
In short os.path.abspath(path) makes a relative path to absolute path. And if the path provided is itself a absolute path then the function returns the same path.
similarly os.path.relpath(path) makes a absolute path to relative path. And if the path provided is itself a relative path then the function returns the same path.
Below example can let you understand the above concept properly:
suppose i have a file input_file_list.txt which contains list of input files to be processed by my python script.
D:\conc\input1.dic
D:\conc\input2.dic
D:\Copyioconc\input_file_list.txt
If you see above folder structure, input_file_list.txt is present in Copyofconc folder and the files to be processed by the python script are present in conc folder
But the content of the file input_file_list.txt is as shown below:
..\conc\input1.dic
..\conc\input2.dic
And my python script is present in D: drive.
And the relative path provided in the input_file_list.txt file are relative to the path of input_file_list.txt file.
So when python script shall executed the current working directory (use os.getcwd() to get the path)
As my relative path is relative to input_file_list.txt, that is "D:\Copyofconc", i have to change the current working directory to "D:\Copyofconc".
So i have to use os.chdir('D:\Copyofconc'), so the current working directory shall be "D:\Copyofconc".
Now to get the files input1.dic and input2.dic, i will read the lines "..\conc\input1.dic" then shall use the command
input1_path= os.path.abspath('..\conc\input1.dic') (to change relative path to absolute path. Here as current working directory is "D:\Copyofconc", the file ".\conc\input1.dic" shall be accessed relative to "D:\Copyofconc")
so input1_path shall be "D:\conc\input1.dic"
This code will return the absolute path to the main script.
import os
def whereAmI():
return os.path.dirname(os.path.realpath(__import__("__main__").__file__))
This will work even in a module.
An alternative which works for me:
this_dir = os.path.dirname(__file__)
filename = os.path.realpath("{0}/relative/file.path".format(this_dir))
Example
Here's an example, tested in Python '3.9.5`:
your current directory: 'c:\project1\code\'
and you want to access the following folder: 'c:\project1\dataset\train\'.
Then you can access the folder using the following address: '../dataset/train/'
References
If you want some more information about path in Python, read this:
PEP - 355
PEP - 519
What worked for me is using sys.path.insert. Then I specified the directory I needed to go. For example I just needed to go up one directory.
import sys
sys.path.insert(0, '../')
I think to work with all systems use "ntpath" instead of "os.path". Today, it works well with Windows, Linux and Mac OSX.
import ntpath
import os
dirname = ntpath.dirname(__file__)
filename = os.path.join(dirname, 'relative/path/to/file/you/want')
A simple solution would be
import os
os.chdir(os.path.dirname(__file__))
From C:\Users\xyz\myFolder to C:\Users\xyz\testdata :
import os
working_dir = os.path.abspath(os.path.dirname(os.path.dirname(__file__)))
# C:\Users\xyz\myFolder
print(working_dir)
updated_working_dir = os.path.join(os.path.realpath(working_dir + '/../'), 'testdata')
# C:\Users\xyz\testdata
print(updated_working_dir)
Output
C:\Users\xyz\myFolder
C:\Users\xyz\testdata
Here is my sumup:
First, define the tool function named relpath, which convert a relative path to current file into a relative path to cwd
import os
relpath = lambda p: os.path.normpath(os.path.join(os.path.dirname(__file__), p))
Then we use it to wrap paths which is relative to current file
path1 = relpath('../src/main.py')
And you can also call sys.path.append() to import file relative to current file position
sys.path.append(relpath('..')) # so that you can import from upper dir
The full example code : https://gist.github.com/luochen1990/9b1ffa30f5c4a721dab5991e040e3eb1
Say the current archive named "Helper" and the upper directory named "Workshop", and the template files are in \Workshop\Templates, then the relative path in Python is "..\Templates".
This a simple way to add a relative path to the system path set . For example, for frequent case when the target directory is one level above (thus, '/../') the working directory:
import os
import sys
workingDir = os.getcwd()
targetDir = os.path.join(os.path.relpath(workingDir + '/../'),'target_directory')
sys.path.insert(0,targetDir)
This solution was tested for:
Python 3.9.6 | packaged by conda-forge | (default, Jul 11 2021,
03:37:25) [MSC v.1916 64 bit (AMD64)]
I'm not sure if this applies to some of the older versions, but I believe Python 3.3 has native relative path support.
For example the following code should create a text file in the same folder as the python script:
open("text_file_name.txt", "w+t")
(note that there shouldn't be a forward or backslash at the beginning if it's a relative path)
I have written a code in python which uses / to make a particular file in a folder, if I want to use the code in windows it will not work, is there a way by which I can use the code in Windows and Linux.
In python I am using this code:
pathfile=os.path.dirname(templateFile)
rootTree.write(''+pathfile+'/output/log.txt')
When I will use my code in suppose windows machine my code will not work.
How do I use "/" (directory separator) in both Linux and Windows?
Use os.path.join().
Example: os.path.join(pathfile,"output","log.txt").
In your code that would be: rootTree.write(os.path.join(pathfile,"output","log.txt"))
Use:
import os
print os.sep
to see how separator looks on a current OS.
In your code you can use:
import os
path = os.path.join('folder_name', 'file_name')
You can use os.sep:
>>> import os
>>> os.sep
'/'
os.path.normpath(pathname) should also be mentioned as it converts / path separators into \ separators on Windows. It also collapses redundant uplevel references... i.e., A/B and A/foo/../B and A/./B all become A/B. And if you are Windows, these all become A\B.
If you are fortunate enough to be running Python 3.4+, you can use pathlib:
from pathlib import Path
path = Path(dir, subdir, filename) # returns a path of the system's path flavour
or, equivalently,
path = Path(dir) / subdir / filename
Some useful links that will help you:
os.sep
os.path
os.pathsep
Do a import os and then use os.sep
You can use "os.sep "
import os
pathfile=os.path.dirname(templateFile)
directory = str(pathfile)+os.sep+'output'+os.sep+'log.txt'
rootTree.write(directory)
Don't build directory and file names your self, use python's included libraries.
In this case the relevant one is os.path. Especially join which creates a new pathname from a directory and a file name or directory and split that gets the filename from a full path.
Your example would be
pathfile=os.path.dirname(templateFile)
p = os.path.join(pathfile, 'output')
p = os.path.join( p, 'log.txt')
rootTree.write(p)
If someone is looking for something like this:
He/she wants to know the parent directory and then go to the sub-folders and maybe than to a specific file. If so, I use the following approach.
I am using python 3.9 as of now. So in that version, we have the os module for handling such tasks. So, for getting the parent directory:
parent_dir = os.path.pardir
It's a good coding practice to not hardcode the file path separators (/ or \). Instead, use the operating system dependant mechanism provided by the above-mentioned os module. It makes your code very much reusable for other purposes/people. It goes like this (just an example) :
path = os.path.pardir + os.sep + 'utils' + os.sep + 'properties.ini'
print(f'The path to my global properties file is :: {path}')
Output:
..\utils\properties.ini
You can surely look at the whole documentation here : https://docs.python.org/3/library/os.html
I use pathlib for most things, so I like: pathlib.os.sep.
Usually pathlib is the better choice if you don't need os!
I have a program in Python that during the processes it creates some files. I want the program to recognize the current directory and then then creates a folder inside the directory, so that the created files will be put in that directory.
I tried this:
current_directory = os.getcwd()
final_directory = os.path.join(current_directory, r'/new_folder')
if not os.path.exists(final_directory):
os.makedirs(final_directory)
But it doesn't give me what I wanted. It seems that the second line is not working as I wanted. Can anybody help me to solve the problem?
think the problem is in r'/new_folder' and the slash (refers to the root directory) used in it.
Try it with:
current_directory = os.getcwd()
final_directory = os.path.join(current_directory, r'new_folder')
if not os.path.exists(final_directory):
os.makedirs(final_directory)
That should work.
One thing to note is that (per the os.path.join documentation) if an absolute path is provided as one of the arguments, the other elements are thrown away. For instance (on Linux):
In [1]: import os.path
In [2]: os.path.join('first_part', 'second_part')
Out[2]: 'first_part/second_part'
In [3]: os.path.join('first_part', r'/second_part')
Out[3]: '/second_part'
And on Windows:
>>> import os.path
>>> os.path.join('first_part', 'second_part')
'first_part\\second_part'
>>> os.path.join('first_part', '/second_part')
'/second_part'
Since you include a leading / in your join argument, it is being interpreted as an absolute path and therefore ignoring the rest. Therefore you should remove the / from the beginning of the second argument in order to have the join perform as expected. The reason you don't have to include the / is because os.path.join implicitly uses os.sep, ensuring that the proper separator is used (note the difference in the output above for os.path.join('first_part', 'second_part').
Can’t be hard, but I’m having a mental block.
import os
os.listdir("path") # returns list
One way:
import os
os.listdir("/home/username/www/")
Another way:
glob.glob("/home/username/www/*")
Examples found here.
The glob.glob method above will not list hidden files.
Since I originally answered this question years ago, pathlib has been added to Python. My preferred way to list a directory now usually involves the iterdir method on Path objects:
from pathlib import Path
print(*Path("/home/username/www/").iterdir(), sep="\n")
os.walk can be used if you need recursion:
import os
start_path = '.' # current directory
for path,dirs,files in os.walk(start_path):
for filename in files:
print os.path.join(path,filename)
glob.glob or os.listdir will do it.
The os module handles all that stuff.
os.listdir(path)
Return a list containing the names of the entries in the directory given by path.
The list is in arbitrary order. It does not include the special entries '.' and
'..' even if they are present in the directory.
Availability: Unix, Windows.
In Python 3.4+, you can use the new pathlib package:
from pathlib import Path
for path in Path('.').iterdir():
print(path)
Path.iterdir() returns an iterator, which can be easily turned into a list:
contents = list(Path('.').iterdir())
Since Python 3.5, you can use os.scandir.
The difference is that it returns file entries not names. On some OSes like windows, it means that you don't have to os.path.isdir/file to know if it's a file or not, and that saves CPU time because stat is already done when scanning dir in Windows:
example to list a directory and print files bigger than max_value bytes:
for dentry in os.scandir("/path/to/dir"):
if dentry.stat().st_size > max_value:
print("{} is biiiig".format(dentry.name))
(read an extensive performance-based answer of mine here)
Below code will list directories and the files within the dir. The other one is os.walk
def print_directory_contents(sPath):
import os
for sChild in os.listdir(sPath):
sChildPath = os.path.join(sPath,sChild)
if os.path.isdir(sChildPath):
print_directory_contents(sChildPath)
else:
print(sChildPath)
Given a path such as "mydir/myfile.txt", how do I find the file's absolute path in Python? E.g. on Windows, I might end up with:
"C:/example/cwd/mydir/myfile.txt"
>>> import os
>>> os.path.abspath("mydir/myfile.txt")
'C:/example/cwd/mydir/myfile.txt'
Also works if it is already an absolute path:
>>> import os
>>> os.path.abspath("C:/example/cwd/mydir/myfile.txt")
'C:/example/cwd/mydir/myfile.txt'
You could use the new Python 3.4 library pathlib. (You can also get it for Python 2.6 or 2.7 using pip install pathlib.) The authors wrote: "The aim of this library is to provide a simple hierarchy of classes to handle filesystem paths and the common operations users do over them."
To get an absolute path in Windows:
>>> from pathlib import Path
>>> p = Path("pythonw.exe").resolve()
>>> p
WindowsPath('C:/Python27/pythonw.exe')
>>> str(p)
'C:\\Python27\\pythonw.exe'
Or on UNIX:
>>> from pathlib import Path
>>> p = Path("python3.4").resolve()
>>> p
PosixPath('/opt/python3/bin/python3.4')
>>> str(p)
'/opt/python3/bin/python3.4'
Docs are here: https://docs.python.org/3/library/pathlib.html
import os
os.path.abspath(os.path.expanduser(os.path.expandvars(PathNameString)))
Note that expanduser is necessary (on Unix) in case the given expression for the file (or directory) name and location may contain a leading ~/(the tilde refers to the user's home directory), and expandvars takes care of any other environment variables (like $HOME).
Install a third-party path module (found on PyPI), it wraps all the os.path functions and other related functions into methods on an object that can be used wherever strings are used:
>>> from path import path
>>> path('mydir/myfile.txt').abspath()
'C:\\example\\cwd\\mydir\\myfile.txt'
Update for Python 3.4+ pathlib that actually answers the question:
from pathlib import Path
relative = Path("mydir/myfile.txt")
absolute = relative.absolute() # absolute is a Path object
If you only need a temporary string, keep in mind that you can use Path objects with all the relevant functions in os.path, including of course abspath:
from os.path import abspath
absolute = abspath(relative) # absolute is a str object
This always gets the right filename of the current script, even when it is called from within another script. It is especially useful when using subprocess.
import sys,os
filename = sys.argv[0]
from there, you can get the script's full path with:
>>> os.path.abspath(filename)
'/foo/bar/script.py'
It also makes easier to navigate folders by just appending /.. as many times as you want to go 'up' in the directories' hierarchy.
To get the cwd:
>>> os.path.abspath(filename+"/..")
'/foo/bar'
For the parent path:
>>> os.path.abspath(filename+"/../..")
'/foo'
By combining "/.." with other filenames, you can access any file in the system.
Today you can also use the unipath package which was based on path.py: http://sluggo.scrapping.cc/python/unipath/
>>> from unipath import Path
>>> absolute_path = Path('mydir/myfile.txt').absolute()
Path('C:\\example\\cwd\\mydir\\myfile.txt')
>>> str(absolute_path)
C:\\example\\cwd\\mydir\\myfile.txt
>>>
I would recommend using this package as it offers a clean interface to common os.path utilities.
You can use this to get absolute path of a specific file.
from pathlib import Path
fpath = Path('myfile.txt').absolute()
print(fpath)
Given a path such as mydir/myfile.txt, how do I find the file's absolute path relative to the current working directory in Python?
I would do it like this,
import os.path
os.path.join( os.getcwd(), 'mydir/myfile.txt' )
That returns '/home/ecarroll/mydir/myfile.txt'
if you are on a mac
import os
upload_folder = os.path.abspath("static/img/users")
this will give you a full path:
print(upload_folder)
will show the following path:
>>>/Users/myUsername/PycharmProjects/OBS/static/img/user
In case someone is using python and linux and looking for full path to file:
>>> path=os.popen("readlink -f file").read()
>>> print path
abs/path/to/file