Import python module NOT on path

One way is to simply amend your path:

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 also modify the PYTHONPATH environment variable or 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 imp.load_source 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.

• python
• import
• external
• python-module