Read file from current path in Python

Question

I had to write a custom function to load a yaml file from the current working directory. The function itself works and my intention was to write it in a pure fashion but my senior colleague told me that the way I wrote this function is utterly bad and I have to rewrite it.

Which commandment in Python did I violate? Can anyone tell me what I did wrong here and how a "professional" solution would look like?

from typing import Dict
import yaml
from yaml import SafeLoader
from pathlib import Path
import os
def read_yaml_from_cwd(file: str) -> Dict:
 """[reads a yaml file from current working directory]
 Parameters
 ----------
 file : str
 [.yaml or .yml file]
 Returns
 -------
 Dict
 [Dictionary]
 """
 path = os.path.join(Path.cwd().resolve(), file)
 if os.path.isfile(path):
 with open(path) as f:
 content = yaml.load(f, Loader=SafeLoader)
 return content
 else:
 return None
content = read_yaml_from_cwd("test.yaml")
print(content)

Answer 1

Which commandment did you violate? Using os.path and pathlib in the same breath! pathlib is an object-oriented replacement to os.path.

 path = os.path.join(Path.cwd().resolve(), file)
 if os.path.isfile(path):
 with open(path) as f:

could be written as:

 path = Path.cwd().joinpath(file)
 if path.is_file():
 with path.open() as f:

or since you're starting at the current directory, simply:

 path = Path(file)
 if path.is_file():
 with path.open() as f:

• N3buchadnezzar: I am using ruamel yaml in my own project, so I might be missremembering the details. I thought one could simply do content = yaml.load(path, Loader=SafeLoader) directly on the path without having to explicitly open it?
• AJNeufeld: @N3buchadnezzar That may be true; I haven't used yaml at all. My answer focused on the mashup between os.path and pathlib, stopping just before the yaml.load(...) line for that very reason.