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 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.