I've written a python program. And if I have a shebang like this one:
#!/usr/bin/python
and I make the file executable with:
$ chmod 755 program.py
I can run the program like so:
$ ./program.py
Here is the issue. I use the conda virtual environments. When I run the program like above, the system creates a subshell that does not recognize the active environment:
(my_env) $ ./program.py
ImportError: No module named pymongo
If I do it this way, however...
(my_env) $ python program.py
# blah blah... runs great
How do I specify the right environment for use in the subshell? Is it possible? I'd like to save my fingers the effort of typing the six character string that is python
.
Another post, Shebangs in conda managed environments, briefly touches on this but does not provide the right answer. Instead of activating the environment in the subshell, it just says, go ahead and ignore the shebang... just use the $ python program.py
syntax.
In your script, change...
#!/usr/bin/python
...to:
#!/usr/bin/env python
The python used by an activated conda environment is ${CONDA_PREFIX}/bin/python
and not /usr/bin/python
Notice the difference?
(root) ~/condaexpts$ which python
/home/ubuntu/condaexpts/m3/bin/python
(root) ~/condaexpts$ /usr/bin/env python
Python 3.5.2 |Continuum Analytics, Inc.| (default, Jul 2 2016, 17:53:06)
[GCC 4.4.7 20120313 (Red Hat 4.4.7-1)] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>>
(root) ~/condaexpts$ source deactivate
~/condaexpts$ which python
/usr/bin/python
~/condaexpts$ /usr/bin/env python
Python 2.7.6 (default, Oct 26 2016, 20:30:19)
[GCC 4.8.4] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>>