when set to cook out of process the python script allows you to set a venv path. So I tried to set this up with an anaconda venv.

however even with a simple script with no module imports I get this error:


Traceback (most recent call last):
File "C:\PROGRA~1/SIDEEF~1/HOUDIN~1.547/houdini/pdg/types\schedulers\local.py", line 69, in spawn
proc = start_async(item_command, output_file, job_env, extra)
File "C:\PROGRA~1/SIDEEF~1/HOUDIN~1.547/houdini/pdg/types\schedulers\local.py", line 1228, in start_async
proc = subprocess.Popen(argv,
File "C:\PROGRA~1\SIDEEF~1\HOUDIN~1.547\python310\lib\subprocess.py", line 971, in __init__
self._execute_child(args, executable, preexec_fn, close_fds,
File "C:\PROGRA~1\SIDEEF~1\HOUDIN~1.547\python310\lib\subprocess.py", line 1440, in _execute_child
hp, ht, pid, tid = _winapi.CreateProcess(executable, args,
FileNotFoundError: The system cannot find the file specified

has anyone been successful in getting an anaconda environment to work here?

The Venv Path parameter is intended to be used with environments created using the Python Virtual Environment TOP, which uses Python's built-in venv module to initialize the environment. It currrently does not work with virtual environments that require initializing a custom shell environment, e.g. Conda.

Depending on the purpose of the python script, you may be able to use a generic generator to call a command on the system shell.
On macOS, I've used an .sh script to run an external python script in a conda environment. On Windows you can use a bat or ps1 script.
You may need to clear the PYTHINHOME andd PYTHONPATH environment variables inside the .sh script before activating the conda environment.