CODEWITHSUNDEEP
filepython-3.4
user3002473
user3002473

Reputation: 5074

Python 3.4 - Opening a file in a module with a different directory

I have a package that looks like the following

package/
    __init__.py
    module.py

In module.py I have something like this

def function(file_name):
    with open(file_name) as f:
        # do stuff

Somewhere else in an arbitrary directory I have a python file that looks something like this

import package

package.function("some_file.txt")

But upon running it, it's giving me FileNotFoundError: [Errno 2] No such file or directory: "some_file.txt".

The problem is that the absolute path of some_file.txt may look something like C:\Users\USER\Documents\some_file.txt, but in package.function the path absolute path is something like C:\Users\USER\Documents\package\some_file.txt. Is there any way I can make it so that calling package.function from some file outside the package directory automatically includes the absolute path of the file I want to open?

Sorry if my terminology is really ambiguous, I'm really unfamiliar with os stuff.

Edit: The exact file setup I have looks like this:

directory/
    foo.py
    package/
        __init__.py
        module.py
    another_directory/
        bar.txt

And foo.py looks exactly like this

import package

package.function("another_directory/bar.txt")

Upvotes: 0

Views: 125

Answers (1)

Marcus Müller
Marcus Müller

Reputation: 36346

I think you're missing the point.

It's not important where the source code lies, relative paths (and pure file names are relative paths) are always interpreted relative to the directory that the program process runs in (ie. the directory you are when you type python C:\Path\to\my\python\code\code.py )

Upvotes: 1

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