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Being fairly new to python I only recently discovered the ability to directly execute a .zip file by placing a __main__.py file at the top of the file. This works great for python code, but can I bundle other types of files and access them with my scripts? If so, how?

My ultimate goal would be to bundle some image files along with the python code in a single .zip file, then be able to use those images within the app without having to extract them to disk. I also would like to bundle a copyright notice, release notes, etc so that the entire app and its data files is in a single zip that can be executed without having to extract it somewhere.

Flux
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Bryan Oakley
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4 Answers4

21

You could use pkg_resources functions to access files:

# __main__.py
import pkg_resources
from PIL import Image

print pkg_resources.resource_string(__name__, 'README.txt')

im = Image.open(pkg_resources.resource_stream('app', 'im.png'))
im.rotate(45).show()

Where zipfile contains:

.
|-- app
|   |-- im.png
|   `-- __init__.py
|-- README.txt
`-- __main__.py

To make zipfile executable, run:

$ echo '#!/usr/bin/env python' | cat - zipfile > program-name
$ chmod +x program-name

To test it:

$ cp program-name /another-dir/
$ cd /another-dir && ./program-name
jfs
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    this works only with pkg_resources module distributed with setuptools installed – Xavier Combelle Mar 22 '11 at 19:53
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    there is [`zipapp` module in Python 3.5](https://docs.python.org/3/library/zipapp.html) that automates the archive creation. – jfs Jan 18 '16 at 05:57
6

At least on my Linux box there is no open filehandle or mapped memory by the process to its own zipfile, so presumably there is no way to "magically" access it.

However, creating your own access is not that hard. Create a __main__.py like so:

import os, zipfile

me = zipfile.ZipFile(os.path.dirname(__file__), 'r')
f = me.open('other.txt')
print f.read()
f.close()
me.close()

Edit: Somewhat terse, that. For completeness:

$ echo "Hello ZIP" > other.txt
$ zip testo.zip __main__.py other.txt
$ python testo.zip
Hello ZIP
flying sheep
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Bittrance
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  • It never occurred to me to re-open the currently executing zipfile. Cool. Unfortunately, when I run my script on OSX `__file__` is `None`. I think this is a step in the right direction though. – Bryan Oakley Mar 18 '11 at 18:21
  • @Bryan Oakley: `sys.argv[0]` contains the zipfile path if the first line in the zipfile is `#!/usr/bin/env python` or similar. – jfs Mar 18 '11 at 20:20
6

You can simply use pkgutil.get_data() as suggested in this answer.

Chris
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Xavier Combelle
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  • this only works if I put my files in a package, which means it won't work for a README.txt or COPYRIGHT.txt file at the root of the zip file. That's not a critical issue, but I would like to be able to put some files alongside the __main__.py at the top of the zip file hierarchy. – Bryan Oakley Mar 18 '11 at 19:15
  • Normally your main module is accessibles via import `__main__` you can refer to it – Xavier Combelle Mar 18 '11 at 19:44
  • Combell: I wish that were true. At least with my test .zip file on OSX and python 2.6, __main__ is undefined. – Bryan Oakley Mar 18 '11 at 19:49
  • @Bryan Oakley: the zipfile is added as the first item in `sys.path` therefore you should always be able to import `__main__`. I've test my answer and README.txt stored along-side `__main__.py` in the zip-file can be read just fine http://stackoverflow.com/questions/5355694/python-can-executable-zip-files-include-data-files/5356563#5356563 – jfs Mar 18 '11 at 20:02
  • `s/I've test/I've tested/` ^^ – jfs Mar 18 '11 at 20:21
1

pkgutil.get_data(package, resource) takes the name of a package and a resource. That means you have to put your data files inside a package within the zip file.

So for example a zip file containing:

__main__.py
zippeddata/__init__.py
zippeddata/data.txt

The __init__.py file can be empty or just a comment, but you need one to make zippeddata importable.

Then in __main__.py you just call:

data = pkgutil.get_data('zippeddata', 'data.txt')
Duncan
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