How do I find the current module name?
A module can find out its own module name by looking at the predefined global variable __name__. If this has the value '__main__', the program is running as a script. Many modules that are usually used by importing them also provide a command-line interface or a self-test, and only execute this code after checking __name__:

def main():
print 'Running test...'
...

if __name__ == '__main__':
main()
__import__('x.y.z') returns

Try:
__import__('x.y.z').y.z

For more realistic situations, you may have to do something like
m = __import__(s)
for i in s.split(".")[1:]:
m = getattr(m, i)

When I edit an imported module and reimport it, the changes don't show up. Why does this happen?
For reasons of efficiency as well as consistency, Python only reads the module file on the first time a module is imported. If it didn't, in a program consisting of many modules where each one imports the same basic module, the basic module would be parsed and re-parsed many times. To force rereading of a changed module, do this:

import modname
reload(modname)

Warning: this technique is not 100% fool-proof. In particular, modules containing statements like

from modname import some_objects

will continue to work with the old version of the imported objects. If the module contains class definitions, existing class instances will not be updated to use the new class definition. This can result in the following paradoxical behavior:

>>> import cls
>>> c = cls.C() # Create an instance of C
>>> reload(cls)
<module 'cls' from 'cls.pyc'>
>>> isinstance(c, cls.C) # isinstance is false?!?
False

The nature of the problem is made clear if you print out the class objects:

>>> c.__class__
<class cls.C at 0x7352a0>
>>> cls.C
<class cls.C at 0x4198d0>

Where is the math.py (socket.py, regex.py, etc.) source file?
There are (at least) three kinds of modules in Python:

1. modules written in Python (.py);
2. modules written in C and dynamically loaded (.dll, .pyd, .so, .sl, etc);
3. modules written in C and linked with the interpreter; to get a list of these, type:
import sys
print sys.builtin_module_names

How do I make a Python script executable on Unix?
You need to do two things: the script file's mode must be executable and the first line must begin with #! followed by the path of the Python interpreter.

The first is done by executing chmod +x scriptfile or perhaps chmod 755 scriptfile.
The second can be done in a number of ways. The most straightforward way is to write

#!/usr/local/bin/python

as the very first line of your file, using the pathname for where the Python interpreter is installed on your platform.

If you would like the script to be independent of where the Python interpreter lives, you can use the "env" program. Almost all Unix variants support the following, assuming the python interpreter is in a directory on the user's $PATH:

#! /usr/bin/env python

Don't do this for CGI scripts. The $PATH variable for CGI scripts is often very minimal, so you need to use the actual absolute pathname of the interpreter.

Occasionally, a user's environment is so full that the /usr/bin/env program fails; or there's no env program at all. In that case, you can try the following hack (due to Alex Rezinsky):

#! /bin/sh
""":"
exec python $0 ${1+"$@"}
"""

The minor disadvantage is that this defines the script's __doc__ string. However, you can fix that by adding

__doc__ = """...Whatever..."""