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How to refer to relative paths of resources when working with a code repository in Python

Posted by: admin November 1, 2017 Leave a comment

Questions:

We are working with a code repository which is deployed both to Windows and Linux – sometimes on different directories. How should one of the modules inside the project refer to one of the non-Python resources in the project (CSV files, etc.)?

If we do something like:

thefile=open('test.csv')

or:

thefile=open('../somedirectory/test.csv')

It will work only when the script is run from one specific directory, or a subset of the directories.

What I would like to do is something like:

path=getBasePathOfProject()+'/somedirectory/test.csv'
thefile=open(path)

Is this the right way? Is it possible?

Answers:

Try to use a filename relative to the current files path. Example for ‘./my_file’:

fn = os.path.join(os.path.dirname(__file__), 'my_file')

Questions:
Answers:

If you are using setup tools or distribute (a setup.py install) then the “right” way to access these packaged resources seem to be using package_resources.

In your case the example would be

import pkg_resources
my_data = pkg_resources.resource_string(__name__, "foo.dat")

Which of course reads the resource and the read binary data would be the value of my_data

If you just need the filename you could also use

resource_filename(package_or_requirement, resource_name)

Example:

resource_filename("MyPackage","foo.dat")

The advantage is that its guaranteed to work even if it is an archive distribution like an egg.

See http://packages.python.org/distribute/pkg_resources.html#resourcemanager-api

Questions:
Answers:

I often use something similar to this:

import os
DATA_DIR = os.path.abspath(os.path.join(os.path.dirname(__file__), 'datadir'))

# if you have more paths to set, you might want to shorten this as
here = lambda x: os.path.abspath(os.path.join(os.path.dirname(__file__), x))
DATA_DIR = here('datadir') 

pathjoin = os.path.join
# ...
# later in script
for fn in os.listdir(DATA_DIR):
    f = open(pathjoin(DATA_DIR, fn))
    # ...

The variable

__file__

holds the file name of the script you write that code in, so you can make paths relative to script, but still written with absolute paths. It works quite well for several reasons:

  • path is absolute, but still relative
  • the project can still be deployed in a relative container

But you need to watch for platform compatibility – Windows’ os.pathsep is different than UNIX.

Questions:
Answers:
import os
cwd = os.getcwd()
path = os.path.join(cwd, "my_file")
f = open(path)

You also try to normalize your cwd using os.path.abspath(os.getcwd()). More info here.

Questions:
Answers:

In Python, paths are relative to the current working directory, which in most cases is the directory from which you run your program. The current working directory is very likely not as same as the directory of your module file, so using a path relative to your current module file is always a bad choice.

Using absolute path should be the best solution:

import os
package_dir = os.path.dirname(os.path.abspath(__file__))
thefile = os.path.join(package_dir,'test.cvs')

Questions:
Answers:

You can use the build in __file__ variable. It contains the path of the current file. I would implement getBaseOfProject in a module in the root of your project. There I would get the path part of __file__ and would return that. This method can then be used everywhere in your project.

Questions:
Answers:

I spent a long time figuring out the answer to this, but I finally got it (and it’s actually really simple):

import sys
import os
sys.path.append(os.getcwd() + '/your/subfolder/of/choice')

# now import whatever other modules you want, both the standard ones,
# as the ones supplied in your subfolders

This will append the relative path of your subfolder to the directories for python to look in
It’s pretty quick and dirty, but it works like a charm 🙂