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pythonreferencepackagerelative-path
Alex
Alex

Reputation: 44719

Relative file paths in Python packages

How do I reference a file relatively to a package's directory?

My directory structure is:

    /foo
     package1/
      resources/
      __init__.py
     package2/
      resources/
      __init__.py
     script.py

script.py imports packages package1 and package2. Although the packages can be imported by any other script on the system. How should I reference resources inside, say, package1 to ensure it would work in case os.path.curdir is arbitrary?

Upvotes: 21

Views: 20259

Answers (5)

jslatane
jslatane

Reputation: 330

Use importlib_resources.

Adding to the the example in the question, you might have something that looks like this:

    /foo
     package1/
      resources/
      __init__.py
      resource1.txt
     package2/
      resources/
      __init__.py
     script.py

In Python you can use:

from importlib_resources import files
resource_path = files('package1.resources').joinpath('resource1.txt')
resource_contents = resource_path.read_text()

Upvotes: 0

jjs
jjs

Reputation: 594

A simple/safe way to do this is using the resource_filename method from pkg_resources (which is distributed with setuptools) like so:

from pkg_resources import resource_filename
filepath = resource_filename('package1', 'resources/thefile')

Or, if you're implementing this inside package1/___init___.py:

from pkg_resources import resource_filename
filepath = resource_filename(__name__, 'resources/thefile')

This gives you a clean solution that is also (if I'm not mistaken) zip safe.

Upvotes: 14

Glyph
Glyph

Reputation: 31860

You can be zip-safe and at the same time use a nice convenient API if you use twisted.python.modules.

For example, if I have a data.txt with some text in it and and this sample.py in one directory:

from twisted.python.modules import getModule
moduleDirectory = getModule(__name__).filePath.parent()
print repr(moduleDirectory.child("data.txt").open().read())

then importing sample will do this:

>>> import sample
'Hello, data!\n'
>>>

If your module is in a regular directory, getModule(__name__).filePath will be a FilePath; if it's in a zip file, it will be a ZipPath, which supports most, but not all, of the same APIs.

Upvotes: 4

Alexander Artemenko
Alexander Artemenko

Reputation: 22776

This is a bad idea, because, if your package was installed as zipped egg, then resources can be unavailable.

If you use setuptool, don't forget to add zip_safe=False to the setup.py config.

Upvotes: -1

Alex Morega
Alex Morega

Reputation: 4228

If you want to reference files from the foo/package1/resources folder you would want to use the __file__ variable of the module. Inside foo/package1/__init__.py:

from os import path
resources_dir = path.join(path.dirname(__file__), 'resources')

Upvotes: 27

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