Does ruamel.yaml have a function to do the process with all files in one directory?

Question

Does ruamel.yaml have a function to do the process with all files in one directory? Something like this: data = yaml.load(Path("*.*"))

Answer 1

No, it does not, but you can do it one line (assuming you have the imports and the YAML() instance):

from pathlib import Path
import ruamel.yaml

yaml = ruamel.yaml.YAML()

data = [yaml.load(p) for p in Path('.').glob('*.yaml')]