Rllib docs provide some information about how to create and train a custom environment. There is some information about registering that environment, but I guess it needs to work differently than gym registration.
I'm testing this out working with the SimpleCorridor environment. If I add the registration code to the file like so:
from ray.tune.registry import register_env
class SimpleCorridor(gym.Env):
...
def env_creator(env_config):
return SimpleCorridor(env_config)
register_env("corridor", env_creator)
Then I am able to train an algorithm using the string name no problem:
if __name__ == "__main__":
ray.init()
tune.run(
"PPO",
stop={
"timesteps_total": 10000,
},
config={
"env": "corridor", # <--- This works fine!
"env_config": {
"corridor_length": 5,
},
},
)
However
It is kinda pointless to register the environment in the same file that you define the environment because you can just use the class. OpenAI gym registration is nice because if you install the environment, then you can use it anywhere just by writing
include gym_corridor
It's not clear to me if there is a way to do the same thing for registering environments for rllib. Is there a way to do this?
The registry functions in ray are a massive headache; I don't know why they can't recognize other environments like OpenAI Gym.
Anyway, the way I've solved this is by wrapping my custom environments in another function that imports the environment automatically so I can re-use code. For example:
def env_creator(env_name):
if env_name == 'CustomEnv-v0':
from custom_gym.envs.custom_env import CustomEnv0 as env
elif env_name == 'CustomEnv-v1':
from custom_gym.envs.custom_env import CustomEnv1 as env
else:
raise NotImplementedError
return env
Then, to get it to work with the tune.register_env()
, you can use your custom env with a lambda
function:
env = env_creator('CustomEnv-v0')
tune.register_env('myEnv', lambda: config, env(config))
From there, tune.run()
should work. It's annoying, but that's the best way I've found to work around this registry issue.