setuptools: package data folder location

Question:

I use setuptools to distribute my python package. Now I need to distribute additional datafiles.

From what I’ve gathered fromt the setuptools documentation, I need to have my data files inside the package directory. However, I would rather have my datafiles inside a subdirectory in the root directory.

What I would like to avoid:

/ #root
|- src/
|  |- mypackage/
|  |  |- data/
|  |  |  |- resource1
|  |  |  |- [...]
|  |  |- __init__.py
|  |  |- [...]
|- setup.py

What I would like to have instead:

/ #root
|- data/
|  |- resource1
|  |- [...]
|- src/
|  |- mypackage/
|  |  |- __init__.py
|  |  |- [...]
|- setup.py

I just don’t feel comfortable with having so many subdirectories, if it’s not essential. I fail to find a reason, why I /have/ to put the files inside the package directory. It is also cumbersome to work with so many nested subdirectories IMHO. Or is there any good reason that would justify this restriction?

Asked By: phant0m

||

Answers:

I think that you can basically give anything as an argument *data_files* to setup().

Answered By: lgautier

Option 1: Install as package data

The main advantage of placing data files inside the root of your Python package
is that it lets you avoid worrying about where the files will live on a user’s
system, which may be Windows, Mac, Linux, some mobile platform, or inside an Egg. You can
always find the directory data relative to your Python package root, no matter where or how it is installed.

For example, if I have a project layout like so:

project/
    foo/
        __init__.py
        data/
            resource1/
                foo.txt

You can add a function to __init__.py to locate an absolute path to a data
file:

import os

_ROOT = os.path.abspath(os.path.dirname(__file__))
def get_data(path):
    return os.path.join(_ROOT, 'data', path)

print get_data('resource1/foo.txt')

Outputs:

/Users/pat/project/foo/data/resource1/foo.txt

After the project is installed as an Egg the path to data will change, but the code doesn’t need to change:

/Users/pat/virtenv/foo/lib/python2.6/site-packages/foo-0.0.0-py2.6.egg/foo/data/resource1/foo.txt

Option 2: Install to fixed location

The alternative would be to place your data outside the Python package and then
either:

  1. Have the location of data passed in via a configuration file,
    command line arguments or
  2. Embed the location into your Python code.

This is far less desirable if you plan to distribute your project. If you really want to do this, you can install your data wherever you like on the target system by specifying the destination for each group of files by passing in a list of tuples:

from setuptools import setup
setup(
    ...
    data_files=[
        ('/var/data1', ['data/foo.txt']),
        ('/var/data2', ['data/bar.txt'])
        ]
    )

Updated: Example of a shell function to recursively grep Python files:

atlas% function grep_py { find . -name '*.py' -exec grep -Hn $* {} ; }
atlas% grep_py ": ["
./setup.py:9:    package_data={'foo': ['data/resource1/foo.txt']}
Answered By: samplebias

I Think I found a good compromise which will allow you to mantain the following structure:

/ #root
|- data/
|  |- resource1
|  |- [...]
|- src/
|  |- mypackage/
|  |  |- __init__.py
|  |  |- [...]
|- setup.py

You should install data as package_data, to avoid the problems described in samplebias answer, but in order to mantain the file structure you should add to your setup.py:

try:
    os.symlink('../../data', 'src/mypackage/data')
    setup(
        ...
        package_data = {'mypackage': ['data/*']}
        ...
    )
finally:
    os.unlink('src/mypackage/data')

This way we create the appropriate structure “just in time”, and mantain our source tree organized.

To access such data files within your code, you ‘simply’ use:

data = resource_filename(Requirement.parse("main_package"), 'mypackage/data')

I still don’t like having to specify ‘mypackage’ in the code, as the data could have nothing to do necessarally with this module, but i guess its a good compromise.

Answered By: polvoazul

I could use importlib_resources or importlib.resources (depending on python version).

https://importlib-resources.readthedocs.io/en/latest/using.html

Answered By: AbbasTari
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