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?
anaconda Virtual environments in Python Script TOP
888 1 0- ltendava
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- tpetrick
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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.
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