什么是更好的 yaml 配置加载实践？

Question

您好，

我正在做一个使用 yaml 配置文件的小项目。最近我想知道在 (1) 将先前初始化的配置字典直接加载到模块中，(2) 将配置值作为参数传递或 (3) 其他方式中，哪种做法更好（使代码更具可读性、可扩展性和可维护性）。

示例代码：

config.yml:

my_foo: foo
my_bar: bar

案例（一）：

my_config.py：

import yaml

with open(config.yaml, "r") as stream:
    try:
        my_config = yaml.safe_load(stream)
    except yaml.YAMLError as exc:
        print(exc)

my_foo.py：

from my_config import my_config

class MyFoo:
    def __init__(self):
        self.val = my_config[my_foo]

my_bar.py：

from my_config import my_config

class MyBar:
    def __init__(self):
        self.val = my_config[my_bar]

my_main.py：

from my_foo import MyFoo
from my_bar import MyBar

my_foo_instance = MyFoo()
my_bar_instance = MyBar()

案例（2）：

my_foo.py：

class MyFoo:
    def __init__(self, val):
        self.val = val

my_bar.py：

class MyBar:
    def __init__(self, val):
        self.val = val

my_main.py：

from my_foo import MyFoo
from my_bar import MyBar

with open(config.yaml, "r") as stream:
    try:
        my_config = yaml.safe_load(stream)
    except yaml.YAMLError as exc:
        print(exc)

my_foo_instance = MyFoo(val=my_config[my_foo])
my_bar_instance = MyBar(val=my_config[my_bar])

Answer 1

我会建议对您的第一个案例进行改进，我认为这比您的第二个案例要好。

config.py：

import yaml

def load_config(path):
    with open(path, "r") as stream:
        # I'm leaving the try-except out because normally 
        # you want your program to fail if the config is incorrect!
        my_config = yaml.safe_load(stream)
    return my_config

main.py:

from my_foo import MyFoo
from my_bar import MyBar
from config import load_config

# Giving the path here is more flexible, you will 
# be able to pass a path as an argument to your program for example
my_config = load_config("my/path.yaml")

my_foo_instance = MyFoo(val=my_config[my_foo])
my_bar_instance = MyBar(val=my_config[my_bar])