Making Python run a few lines before my script

Question

I need to run a script foo.py, but I need to also insert some debugging lines to run before the code in foo.py. Currently I just put those lines in foo.py and I'm careful not to commit that to Git, but I don't like this solution.

What I want is a separate file bar.py that I don't commit to Git. Then I want to run:

python /somewhere/bar.py /somewhere_else/foo.py

What I want this to do is first run some lines of code in bar.py, and then run foo.py as __main__. It should be in the same process that the bar.py lines ran in, otherwise the debugging lines won't help.

Is there a way to make bar.py do this?

Someone suggested this:

import imp
import sys

# Debugging code here

fp, pathname, description = imp.find_module(sys.argv[1])
imp.load_module('__main__', fp, pathname, description)

The problem with that is that because it uses import machinery, I need to be on the same folder as foo.py to run that. I don't want that. I want to simply put in the full path to foo.py.

Also: The solution needs to work with .pyc files as well.

Answer 1

Python has a mechanism for running code at startup; the site module.

"This module is automatically imported during initialization."

The site module will attempt to import a module named sitecustomize before __main__ is imported. It will also attempt to import a module named usercustomize if your environment instructs it to.

For example, you could put a sitecustomize.py file in your site-packages folder that contains this:

import imp

import os

if 'MY_STARTUP_FILE' in os.environ:
    try:
        file_path = os.environ['MY_STARTUP_FILE']
        folder, file_name = os.path.split(file_path)
        module_name, _ = os.path.splitext(file_name)
        fp, pathname, description = imp.find_module(module_name, [folder])
    except Exception as e:
        # Broad exception handling since sitecustomize exceptions are ignored
        print "There was a problem finding startup file", file_path
        print repr(e)
        exit()

    try:
        imp.load_module(module_name, fp, pathname, description)
    except Exception as e:
        print "There was a problem loading startup file: ", file_path
        print repr(e)
        exit()
    finally:
        # "the caller is responsible for closing the file argument" from imp docs
        if fp:
            fp.close()

Then you could run your script like this:

MY_STARTUP_FILE=/somewhere/bar.py python /somewhere_else/foo.py

• You could run any script before foo.py without needing to add code to reimport __main__.
• Run export MY_STARTUP_FILE=/somewhere/bar.py and not need to reference it every time

Answer 2

You can use execfile() if the file is .py and uncompyle2 if the file is .pyc.

Let's say you have your file structure like:

test|-- foo.py
    |-- bar
         |--bar.py   

foo.py

import sys

a = 1
print ('debugging...')

# run the other file
if sys.argv[1].endswith('.py'): # if .py run right away
    execfile(sys.argv[1], globals(), locals())
elif sys.argv[1].endswith('.pyc'): # if .pyc, first uncompyle, then run
    import uncompyle2
    from StringIO import StringIO
    f = StringIO()
    uncompyle2.uncompyle_file(sys.argv[1], f)
    f.seek(0)
    exec(f.read(), globals(), locals())

bar.py

print a
print 'real job'

And in test/, if you do:

$ python foo.py bar/bar.py
$ python foo.py bar/bar.pyc

Both, outputs the same:

debugging...
1
real job

Please also see this answer.