How do I tell a Python script to use a particular version

You can add a shebang line the to the top of the script:

#!/usr/bin/env python2.7

But that will only work when executing as ./my_program.py.

If you execute as python my_program.py, then the whatever Python version that which python returns will be used.

In re: to virtualenv use: virtualenv -p /usr/bin/python3.2 or whatever to set it up to use that Python executable.

Perhaps not exactly what you asked, but I find this to be useful to put at the start of my programs:

import sys

if sys.version_info[0] < 3:
    raise Exception("Python 3 or a more recent version is required.")

I would use the shebang #!/usr/bin/python (first line of code) with the serial number of Python at the end ;)

Then run the Python file as a script, e.g., ./main.py from the command line, rather than python main.py.

It is the same when you want to run Python from a Linux command line.

While the OP may be working on a nix platform this answer could help non-nix platforms. I have not experienced the shebang approach work in Microsoft Windows.

Rephrased: The shebang line answers your question of "within my script" but I believe only for Unix-like platforms. Even though it is the Unix shell, outside the script, that actually interprets the shebang line to determine which version of Python interpreter to call. I am not sure, but I believe that solution does not solve the problem for Microsoft Windows platform users.

In the Microsoft Windows world, the simplify the way to run a specific Python version, without environment variables setup specifically for each specific version of Python installed, is just by prefixing the python.exe with the path you want to run it from, such as C:\Python25\python.exe mymodule.py or D:\Python27\python.exe mymodule.py

However you'd need to consider the PYTHONPATH and other PYTHON... environment variables that would point to the wrong version of Python libraries.

For example, you might run: C:\Python2.5.2\python.exe mymodule

Yet, the environment variables may point to the wrong version as such:

PYTHONPATH = D:\Python27

PYTHONLIB = D:\Python27\lib

Loads of horrible fun!

So a non-virtualenv way, in Windows, would be to use a batch file that sets up the environment and calls a specific Python executable via prefixing the python.exe with the path it resides in. This way has additional details you'll have to manage though; such as using command line arguments for either of the "start" or "cmd.exe" command to "save and replace the "console" environment" if you want the console to stick around after the application exits.

Your question leads me to believe you have several Python modules, each expecting a certain version of Python. This might be solvable "within" the script by having a launching module which uses the subprocess module. Instead of calling mymodule.py you would call a module that calls your module; perhaps launch_mymodule.py

launch_mymodule.py

import sys
import subprocess
if sys.argv[2] == '272':
  env272 = {
    'PYTHONPATH': 'blabla',
    'PYTHONLIB': 'blabla', }
  launch272 = subprocess.Popen('D:\\Python272\\python.exe mymodule.py', env=env272)

if sys.argv[1] == '252'
  env252 = {
    'PYTHONPATH': 'blabla',
    'PYTHONLIB': 'blabla', }
  launch252 = subprocess.Popen('C:\\Python252\\python.exe mymodule.py', env=env252)

I have not tested this.