A python question

Ruby
1

So, I’m just going on the hope that the Ruby community, with its great
number of other-language enthusiasts, might have a member or two who’s a
Python guru. I asked this question about a week ago on the python list
and
got all of zero responses, so I’ll try here.

In python, each directory can be a library if it has a file in it called
init.py. So, you can have the following directory structure:

a/init.py
/b/init.py
/c.py

And c.py could have some method in it, like say:

def d():
return ‘d’

And, from the python command line, you can do

from a.b.c import d
d()
‘d’

Now, you can also put this structure in a zipfile with the same
structure,
and load data from it similarly:

$ unzip -Z a.zip
Archive: a.zip 1302 bytes 8 files
drwxr-xr-x 2.3 unx 0 bx stor 1-Apr-07 10:41 a/
drwxr-xr-x 2.3 unx 0 bx stor 1-Apr-07 10:41 a/b/
-rw-r–r-- 2.3 unx 22 tx stor 1-Apr-07 10:41 a/b/c.py
-rw-r–r-- 2.3 unx 0 bx stor 1-Apr-07 10:41 a/b/init.py
-rw-r–r-- 2.3 unx 95 bx defN 1-Apr-07 10:41 a/b/init.pyc
-rw-r–r-- 2.3 unx 186 bx defN 1-Apr-07 10:41 a/b/c.pyc
-rw-r–r-- 2.3 unx 0 bx stor 1-Apr-07 10:41 a/init.py
-rw-r–r-- 2.3 unx 93 bx defN 1-Apr-07 10:41 a/init.pyc
8 files, 396 bytes uncompressed, 238 bytes compressed: 39.9%
$ python
Python 2.4.3 (#1, Oct 11 2006, 23:25:18)
[GCC 4.1.1 (Gentoo 4.1.1)] on linux2
Type “help”, “copyright”, “credits” or “license” for more information.

import sys; sys.path.insert(0, ‘a.zip’)
import a; a
<module ‘a’ from ‘a.zip/a/init.pyc’>

from a.b.c import d
d()
‘d’

So, basically importing from a zip file looks just like importing from a
standard directory. Now, what I can’t figure out how to do is import
from a
zipfile that’s nested in a directory. My layout is like this:

$ find a
a
a/init.py
a/init.pyc
a/b.zip
$ unzip -Z a/b.zip
Archive: a/b.zip 416 bytes 3 files
drwxr-xr-x 2.3 unx 0 bx stor 6-Apr-07 12:29 b/
-rw-r–r-- 2.3 unx 0 bx stor 31-Mar-07 19:58 b/init.py
-rw-r–r-- 2.3 unx 22 tx stor 6-Apr-07 12:29 b/c.py
3 files, 22 bytes uncompressed, 22 bytes compressed: 0.0%
$ python
Python 2.4.3 (#1, Oct 11 2006, 23:25:18)
[GCC 4.1.1 (Gentoo 4.1.1)] on linux2
Type “help”, “copyright”, “credits” or “license” for more information.

import a.b
Traceback (most recent call last):
File “”, line 1, in ?
ImportError: No module named b

import a
dir(a)
[‘builtins’, ‘doc’, ‘file’, ‘name’, ‘path’, ‘b’,
‘here’,
‘mod’, ‘os’, ‘zip’, ‘zipimport’]

dir(a.b)
[‘builtins’, ‘doc’, ‘file’, ‘loader’, ‘name’,
‘path’]

Does anyone here know how to import a zipfile library’s contents into a
normal library? Any Python gurus around?

2

Hello, I happen to enjoy Python and Ruby.

Importing modules from zip files was proposed in PEP-273 [1]

Here is how the spec of PEP-273 begins:

‘’’
Currently, sys.path is a list of directory names as strings. If this
PEP is implemented, an item of sys.path can be a string naming a zip
file archive.
‘’’

My interpretation of the above is that, to be importable, a zip file
must be explicitly named in sys.path, so the mere fact that a zip file
lies somewhere in a directory which is part of the sys.path does not
make it importable.

Cheers,

Luciano

[1] PEP 273 – Import Modules from Zip Archives | peps.python.org

3

My interpretation of the above is that, to be importable, a zip file
must be explicitly named in sys.path, so the mere fact that a zip file
lies somewhere in a directory which is part of the sys.path does not
make it importable.

I just realized that I left out the tricky part of what I’m doing.
Within
a/init.py, I have the following:

import zipimport
import os
here = os.path.join(os.getcwd(), path[0])
zip = os.path.join(here, ‘b.zip’)
mod = os.path.split(zip)[-1][:-4]
globals()[mod] = zipimport.zipimporter(zip).load_module(mod)

So, this explicitly uses zipimport to manually load the b.zip module
into
a’s namespace. I sort of forgot that was important :slight_smile: Anyhow, b does
exist
within a as a module, but it has no contents, and there doesn’t seem to
be
any way to get them, unless anyone knows a cool trick. Thanks for the
help
though :slight_smile: