Open file in a relative location in Python

Question

With this type of thing you need to be careful what your actual working directory is. For example, you may not run the script from the directory the file is in. In this case, you can't just use a relative path by itself.

If you are sure the file you want is in a subdirectory beneath where the script is actually located, you can use __file__ to help you out here. __file__ is the full path to where the script you are running is located.

So you can fiddle with something like this:

import os
script_dir = os.path.dirname(__file__) #<-- absolute dir the script is in
rel_path = "2091/data.txt"
abs_file_path = os.path.join(script_dir, rel_path)

This code works fine:

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)

I created an account just so I could clarify a discrepancy I think I found in Russ's original response.

For reference, his original answer was:

import os
script_dir = os.path.dirname(__file__)
rel_path = "2091/data.txt"
abs_file_path = os.path.join(script_dir, rel_path)

This is a great answer because it is trying to dynamically creates an absolute system path to the desired file.

Cory Mawhorter noticed that __file__ is a relative path (it is as well on my system) and suggested using os.path.abspath(__file__). os.path.abspath, however, returns the absolute path of your current script (i.e. /path/to/dir/foobar.py)

To use this method (and how I eventually got it working) you have to remove the script name from the end of the path:

import os
script_path = os.path.abspath(__file__) # i.e. /path/to/dir/foobar.py
script_dir = os.path.split(script_path)[0] #i.e. /path/to/dir/
rel_path = "2091/data.txt"
abs_file_path = os.path.join(script_dir, rel_path)

The resulting abs_file_path (in this example) becomes: /path/to/dir/2091/data.txt

It depends on what operating system you're using. If you want a solution that is compatible with both Windows and *nix something like:

from os import path

file_path = path.relpath("2091/data.txt")
with open(file_path) as f:
    <do stuff>

should work fine.

The path module is able to format a path for whatever operating system it's running on. Also, python handles relative paths just fine, so long as you have correct permissions.

Edit:

As mentioned by kindall in the comments, python can convert between unix-style and windows-style paths anyway, so even simpler code will work:

with open("2091/data/txt") as f:
    <do stuff>

That being said, the path module still has some useful functions.