Jeremy wrote:
I have a module that, when loaded, reads and parses a supporting
file. The supporting file contains all the data for the module and
the function that reads/parses the file sets up the data structure for
the module.
How can I locate the file during the import statement. The supporting
file is located in the same directory as the module, but when I import
I get a No such file or directory error. I could hard code the path
to the filename, but that would make it only work on my machine.
A related question: Can I parse the data once and keep it somewhere
instead of reading the supporting file every time? I tried pickling
but that wouldn't work because I have custom classes. (Either that or
I just don't know how to pickle—this is a highly probable event.)
Thanks,
Jeremy
By supporting file, presumably you mean a text or binary file, not a
python module. That's why import won't work.
Now, if the data supports it, you could make that supporting file simply
be an importable python program. I've seen code generators to create
python code from a binary file such as a jpeg, so that you can just
embed such code into your program.
But I'll guess that this isn't reasonable for your data, that the format
is something you want to separately manipulate.
I think the secret you're looking for is the __file__ attribute of a
module. That contains the full path to the module source file (.py or
.pyc, usually). So you use dirname on that string, to get the directory
name, and os.path.join to combine with your own filename. Now you have a
string to pass to open().
DaveA
--
http://mail.python.org/mailman/listinfo/python-list
