def convert(dictionary我的类中有一个辅助函数 ( )__init__来协助配置设置。定义如下：class Configuration: def __init__(self, config_file=None, config=None): if config_file is not None: with open(config_file) as in_file: self._config = yaml.load(in_file, Loader=yaml.FullLoader) elif config is not None: self._config = config else: raise ValueError("Could not create configuration. Must pass either location of config file or valid " "config.") def convert(dictionary): return namedtuple('Config', dictionary.keys())(**dictionary)这使我可以按如下方式拨打电话__init__： self.input = convert(self._config["input"]) self.output = convert(self._config["output"]) self.build = convert(self._config["build_catalog"])由于我要设置多个配置，因此我想从他的类中继承如下：class BuildConfiguration(Configuration): def __init__(self, config_file=None, config=None): super().__init__(config_file, config) self.input = convert(self._config["input"]) self.output = convert(self._config["output"]) self.build = convert(self._config["build_catalog"])convert但是，我无法从父类访问。我也试过这个：self.input = super().__init__.convert(self._config["input"])这似乎也行不通。所以问题是如何访问super().__init__子类中定义的函数？
1 回答
牧羊人nacy
TA贡献1862条经验 获得超7个赞
你不能。每次调用都会创建一个新函数，__init__然后将其丢弃，它不存在于函数之外。请注意，这也适用于由创建的类namedtuple('Config', dictionary.keys())(**dictionary)。继续创建所有这些不必要的类确实不好，这完全违背了namedtuple创建内存高效记录类型的目的。在这里，每个实例都有自己的类！
以下是您应该如何定义它：
Config = namedtuple('Config', "foo bar baz")
def convert(dictionary): # is this really necessary?
return Config(**dictionary)
class Configuration:
def __init__(self, config_file=None, config=None):
if config_file is not None:
with open(config_file) as in_file:
self._config = yaml.load(in_file, Loader=yaml.FullLoader)
elif config is not None:
self._config = config
else:
raise ValueError("Could not create configuration. Must pass either location of config file or valid "
"config.")
self.input = convert(self._config["input"])
self.output = convert(self._config["output"])
self.build = convert(self._config["build_catalog"])
虽然在这一点上，使用它似乎更干净
Config(**self._config["input"])
etc 而不是 helper convert。
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