[python] How to refer to relative paths of resources when working with a code repository

We are working with a code repository which is deployed to both Windows and Linux - sometimes in 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 it possible?

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 :)

I got stumped here a bit. Wanted to package some resource files into a wheel file and access them. Did the packaging using manifest file, but pip install was not installing it unless it was a sub directory. Hoping these sceen shots will help

+-- cnn_client
¦   +-- image_preprocessor.py
¦   +-- __init__.py
¦   +-- resources
¦   ¦   +-- mscoco_complete_label_map.pbtxt
¦   ¦   +-- retinanet_complete_label_map.pbtxt
¦   ¦   +-- retinanet_label_map.py
¦   +-- tf_client.py

MANIFEST.in

recursive-include cnn_client/resources *

Created a weel using standard setup.py . pip installed the wheel file. After installation checked if resources are installed. They are

ls /usr/local/lib/python2.7/dist-packages/cnn_client/resources

mscoco_complete_label_map.pbtxt
retinanet_complete_label_map.pbtxt 
 retinanet_label_map.py  

In tfclient.py to access these files. from

templates_dir = os.path.join(os.path.dirname(__file__), 'resources')
 file_path = os.path.join(templates_dir, \
            'mscoco_complete_label_map.pbtxt')
        s = open(file_path, 'r').read()

And it works.

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.

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.

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.

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')

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