Relative paths in Python
In the file that has the script, you want to do something like this:
import osdirname = 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.pyimport osprint os.getcwd()print __file__#in the interactive interpreterimport foo/Users/jasonfoo.py#and finally, at the shell:~ % python foo.py/Users/jasonfoo.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 2.5collections.__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.
As mentioned in the accepted answer
import osdir = 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 osdir = 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