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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.

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.

In the file that has the script, you want to do something like this:

import os
dir = os.path.dirname(__file__)
filename = os.path.join(dir, '/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.

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.

In the file that has the script, you want to do something like this:

import os
dir = os.path.dirname(__file__)
filename = os.path.join(dir, '/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.

In the file that has the script, you want to do something like this:

import os
dir = os.path.dirname(__file__)
filename = os.path.join(dir, '/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.