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I want to pass my program a file and get a function out of it.

For example, I have a file, foo.py, who's location is not known until run time (it will be passed to to code by the command line or something like that), can be anywhere on my system and looks like this:

def bar():
    return "foobar"

how can I get my code to run the function bar?

If the location was known before run time I could do this:

import sys
sys.path.append("path_to_foo")
import foo

foo.bar()

I could create an init.py file in the folder where foo.py is and use importlib or imp but it seems messy. I can't use __import__ as I get ImportError: Import by filename is not supported.

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  • 2
    "I could [...] use importlib [...] but it seems messy." You have to resort to messy solutions when you have messy problems. Why isn't importlib sufficient? Commented May 18, 2016 at 14:49
  • Also, you don't need a __init__.py file at all. In Py3, the more powerful importlib functions are the correct way to import arbitrary files. In earlier Python, you can just add the folder containing foo.py to sys.path, then run import foo without using a special importing function at all. Commented May 18, 2016 at 14:56
  • Is foo.py in the same directory as your file? Commented May 18, 2016 at 15:00
  • I'm using Python 2.7. I can't use import foo as I don't know the file name until run time. Commented May 18, 2016 at 15:04
  • I don't want to create an init.py file and leave it lying around, I could clean it up afterwards. I don't know how I can be sure to get rid of it but that's another topic. It just seems perverse that I have to create an empty file to do something. Commented May 18, 2016 at 15:13

3 Answers 3

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11

You could open the file and execute it using exec. 

f = open('foo.py')
source = f.read()
exec(source)
print bar()

You could even look for the specific function using re

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4 Comments

I ended up using something like this. I actually did sys.path.append("path_to_foo"); file_name = foo.py; exec(import file_name as foo); foo.bar()
@R.Mckemey You said who's location is not known until run time, can be anywhere on my system.... How are you going to find the file?
Sorry I was unclear, the whole file path will be passed to the program. I'll run it something like python program.py path_to_foo/foo.py
To be noted: I you try to put this in a function, you need to have your function return whatever functions or variables you are interested in and assign the result of the function call, otherwise, the functions and variables created by exec(source) will only live in the function scope.
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You can write a dummy function and call it everywhere you need, to bypass checking:

def bar():
    pass

at run-time override the function with the one you actually intend to, and automatically everywhere it will be used.

Note: This is more messy than using the importlib.

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2 Comments

how I would "override the function"?
@R.Mckemey dynamically when you figure out the location of your function code you can use, eval or exec to load/run your function code.
0

If foo.py is in the same directory as your main file, you can use 

from . import foo

(Python 3). This works because . is the directory of your file, and Python will import files in the same directory for you. You can then use 

foo.bar()

If it is not, you need to find it, and then execute it:

import os
from os.path import join

lookfor = "foo.py"
for root, dirs, files in os.walk('C:\\'): # or '/' for Linux / OSX
    if lookfor in files:
        execfile(join(root, lookfor)) # Python 2, or
        with open(join(root, lookfor)) as file:
            exec(''.join(file.readlines()))
        break
foo.bar()

Credits: Martin Stone. I did not just copy the code, I understand the code, and could've made it myself.

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