How to get a Hydra config without using @hydra.main()

问题

Let's say we have following setup (copied & shortened from the Hydra docs):

Configuration file: config.yaml

db:
  driver: mysql
  user: omry
  pass: secret

Python file: my_app.py

import hydra
@hydra.main(config_path="config.yaml")
def my_app(cfg):
    print(cfg.pretty())
if __name__ == "__main__":
    my_app()

This works well when we can use a decorator on the function my_app. Now I would like (for small scripts and testing purposes, but that is not important) to get this cfg object outside of any function, just in a plain python script. From what I understand how decorators work, it should be possible to call

import hydra
cfg = hydra.main(config_path="config.yaml")(lambda x: x)()
print(cfg.pretty())

but then cfg is just None and not the desired configuration object. So it seems that the decorator does not pass on the returned values. Is there another way to get to that cfg ?

回答1:

Use the experimental compose API:

from hydra.experimental import compose, initialize
from omegaconf import OmegaConf

initialize(config_path="conf", job_name="test_app")
cfg = compose(config_name="config", overrides=["db=mysql", "db.user=me"])
print(OmegaConf.to_yaml(cfg))

This will only compose the config and will not have side effects like changing the working directory or configuring the Python logging system.

回答2:

I found a rather ugly answer but it works - if anyone finds a more elegant solution please let us know!

We can use a closure or some mutable object. In this example we define a list outside and append the config object:

import hydra
c = []
hydra.main(config_path="config.yaml")(c.append)()
cfg = c[0]
print(cfg.pretty())

回答3:

anther ugly answer, but author said this may be crush in next version

Blockquote

from omegaconf import DictConfig
from hydra.utils import instantiate
from hydra._internal.utils import _strict_mode_strategy, split_config_path, create_automatic_config_search_path
from hydra._internal.hydra import Hydra
from hydra.utils import get_class 

class SomeThing:
...
def load_from_yaml(self, config_path, strict=True):
    config_dir, config_file = split_config_path(config_path)
    strict = _strict_mode_strategy(strict, config_file)
    search_path = create_automatic_config_search_path(
        config_file, None, config_dir
    )
    hydra = Hydra.create_main_hydra2(
        task_name='sdfs', config_search_path=search_path, strict=strict
    )
    config = hydra.compose_config(config_file, [])
    config.pop('hydra')
    self.config = config
    print(self.config.pretty())

来源：https://stackoverflow.com/questions/60674012/how-to-get-a-hydra-config-without-using-hydra-main

标签

python-3.x

hydra-python

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