CODEWITHSUNDEEP
pythonfb-hydraomegaconf
miccio
miccio

Reputation: 153

Hydra: access name of config file from code

I have a config tree such as:

config.yaml
model/
  model_a.yaml
  model_b.yaml
  model_c.yaml

Where config.yaml contains:

# @package _global_
defaults:
  - _self_
  - model: model_a.yaml

some_var: 42

I would like to access the name of the model config file used (either the default or overridden) from my python code or from the file itself. Something like:

@hydra.main(...)
def main(config):
  model_name = config.model.__filename__

or (from e.g. model_a.yaml)

dropout: true
dense_layers: 128
model_name: ${__filename__}

Thanks in advance!

Upvotes: 6

Views: 4501

Answers (1)

Jasha
Jasha

Reputation: 7639

The a look at the hydra.runtime.choices variable mentioned in the Configuring Hydra - Introduction page of the Hydra docs. This variable stores a mapping that describes each of the choices that Hydra has made in composing the output config.

Using your example from above with model: model_a.yaml in the defaults list:

# my_app.py
import hydra
from pprint import pprint
from hydra.core.hydra_config import HydraConfig
from omegaconf import OmegaConf

@hydra.main(config_path=".", config_name="config")
def main(config):
    hydra_cfg = HydraConfig.get()
    print("choice of model:")
    pprint(OmegaConf.to_container(hydra_cfg.runtime.choices))

main()

At the command line:

$ python3 app.py
choices used:
{'hydra/callbacks': None,
 'hydra/env': 'default',
 'hydra/help': 'default',
 'hydra/hydra_help': 'default',
 'hydra/hydra_logging': 'default',
 'hydra/job_logging': 'default',
 'hydra/launcher': 'basic',
 'hydra/output': 'default',
 'hydra/sweeper': 'basic',
 'model': 'model_a.yaml'}

As you can see, in this example the config option model_a.yaml is stored in the Hydra config at hydra_cfg.runtime.choices.model.

Upvotes: 9

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