# A python question **URL:** \ **Category:** Ruby\ **Created:** [April 6, 2007, 7:33pm UTC](https://www.ruby-forum.com/t/a-python-question/91325 "2007-04-06T19:33:21Z")\ **Posts on this page:** 3\ **Page:** 1
**Author:** ![tsuraan](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/t/8797f3/32.png) [@tsuraan](https://www.ruby-forum.com/u/tsuraan)\ **Post date:** [April 6, 2007, 7:33pm UTC](https://www.ruby-forum.com/t/a-python-question/91325/1 "2007-04-06T19:33:21Z")
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 > > > \ > > > > 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? ---
**Author:** ![tsuraan](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/t/8797f3/32.png) [@tsuraan](https://www.ruby-forum.com/u/tsuraan)\ **Post date:** [April 6, 2007, 8:42pm UTC](https://www.ruby-forum.com/t/a-python-question/91325/2 "2007-04-06T20:42:11Z")
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](http://www.python.org/dev/peps/pep-0273/) ---
**Author:** ![tsuraan](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/t/8797f3/32.png) [@tsuraan](https://www.ruby-forum.com/u/tsuraan)\ **Post date:** [April 6, 2007, 8:52pm UTC](https://www.ruby-forum.com/t/a-python-question/91325/3 "2007-04-06T20:52:14Z")
> 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 🙂 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 🙂