Import python module NOT on path

I have a module foo, containing util.py and bar.py.

I want to import it in IDLE or python session. How do I go about this?

I could find no documentation on how to import modules not in the current directory or the default python PATH. After trying import "<full path>/foo/util.py", and from "<full path>" import util

The closest I could get was

import imp
imp.load_source('foo.util','C:/.../dir/dir2/foo')

Which gave me Permission denied on windows 7.

One way is to simply amend your http://docs.python.org/library/sys.html#sys.path">path</a>;:

import sys
sys.path.append('C:/full/path')
from foo import util,bar

Note that this requires foo to be a python package, i.e. contain a __init__.py file. If you don't want to modify sys.path, you can http://docs.python.org/tutorial/modules.html#the-module-search-path">also modify the PYTHONPATH environment variable or http://docs.python.org/install/index.html">install the module on your system. Beware that this means that other directories or .py files in that directory may be loaded inadvertently.

Therefore, you may want to use http://docs.python.org/library/imp.html#imp.load_source">`imp.load_source`</a> instead. It needs the filename, not a directory (to a file which the current user is allowed to read):

import imp
util = imp.load_source('util', 'C:/full/path/foo/util.py')

You could customize the module search path using the PYTHONPATH environment variable, or manually modify the sys.path directory list.

See Module Search Path documentation on python.org.

Give this a try

import sys
sys.path.append('c:/.../dir/dir2')
import foo

Following phihag's tip, I have this solution. Just give the path of a source file to load_src and it will load it. You must also provide a name, so you can import this module using this name. I prefer to do it this way because it's more explicit:

def load_src(name, fpath):
    import os, imp
    return imp.load_source(name, os.path.join(os.path.dirname(__file__), fpath))

load_src("util", "../util.py")
import util

print util.method()

Another (less explicit) way is this:

util = load_src("util", "../util.py")    # "import util" is implied here

print util.method()    # works, util was imported by the previous line

Edit: the method is rewritten to make it clearer.