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Import module with non-standard file name

Howdy all,

Question: I have Python modules named without '.py' as the extension,
and I'd like to be able to import them. How can I do that?

Background:

On Unix, I write programs intended to be run as commands to a file
with no extension. This allows other programs to use the command as an
interface, and I can re-write the program in some other language
without obsoleting the commandline interface.

e.g., I might write 'frobnicate-foo' as a shell program so that other
programs can 'frobnicate-foo --bar baz'. If I later decide to
re-implement 'frobnicate-foo' in Python, I'll save the top level
module to the same file name since it implements the same command-line
interface.

Now that I've got it written as a Python module, I'd like to write
unit tests for that module, which of course will need to import the
program module to test it. The unit test can explicitly add the
directory where the program module lives to 'sys.path' for the purpose
of importing that module.

However, the Python reference tells me that 'import' (specifically,
'__import__()') needs modules to live in files named a particular way:
with a '.py' suffix. But my module is in a file called
'frobnicate-foo', with no suffix, and that's part of the definition of
the program interface.

I don't want symbolic links, or anything else that presents two
filenames for the same module, because there's no need for that except
for Python's apparent insistence on a particular naming
convention. Also, avoiding symbolic links inside the source code tree
makes version control smoother.
What are my options to import a module from a file whose name can't
change?

--
\ "[W]e are still the first generation of users, and for all that |
`\ we may have invented the net, we still don't really get it." |
_o__) -- Douglas Adams |
Ben Finney

Aug 8 '06
10 3772
"Martin v. Löwis" <ma****@v.loewi s.dewrites:
Ben Finney schrieb:
Question: I have Python modules named without '.py' as the extension,
and I'd like to be able to import them. How can I do that?

I recommend to use imp.load_module .
I've tried this; as Patrick Maupin alludes to, it compiles the module
leaving a strangely-named file behind.

Program in a file named 'frob_foo'; no other file names needed nor
desired.

import imp

file_name = "frob_foo"
module_name = 'frob_foo'

module_file = open(file_name, 'r')
module_desc = ("", 'r', imp.PY_SOURCE)
module = imp.load_module (module_name, module_file, file_name, module_desc)

Result: two files, 'frob_foo' and 'frob_fooc'. I can see why this
happens, but it's not what's desired. Currently I'm going with:

file_name = "frob_foo"
module_name = 'frob_foo'

from types import ModuleType
module = ModuleType(modu le_name)

module_file = open(file_name, 'r')
exec module_file in module.__dict__

Still, the purpose is simply to get a module object out, with a named
file as input. If the 'imp' module can do that without leaving
unwanted turds behind, it seems more elegant. Can anyone suggest a way
to get the same result as the above 'exec' method, using the 'imp'
module?

--
\ "...one of the main causes of the fall of the Roman Empire was |
`\ that, lacking zero, they had no way to indicate successful |
_o__) termination of their C programs." -- Robert Firth |
Ben Finney

Aug 18 '06 #11

This thread has been closed and replies have been disabled. Please start a new discussion.

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